The remaining terms obtained by the Leibniz rule in the sum occurring in the definition of $\psi''$ are

$$
\begin{align*}
& - \sum_{p_1, p_2, q, r \ge 0} \tau \underbrace{\langle h, \dots, h \mid d(h), h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p_1 p_2 q r} \\
& - \sum_{p, q, r \ge 0} \tau \underbrace{\langle h, \dots, h \mid d(\psi), h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p q r} \\
& + \sum_{p, q_1, q_2, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', d(h'), h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p q_1 q_2 r} \\
& + \sum_{p, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', d(\psi'), h'', \dots, h'' \rangle}_{p q r} \\
& - \sum_{p, q, r_1, r_2 \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'', d(h''), h'', \dots, h'' \rangle}_{p r r_1 r_2}.
\end{align*}
$$

By using equations (1) and (2), the definition of $\psi''$ and that $h, h', h'' \in \text{Aut}_{\text{Deligne}(A)}(\tau)$, we obtain

$$
\begin{align*}
d(\psi'') ={}& h - h'' - \psi'\langle\tau\rangle - \psi'\langle\tau\rangle - \sum_{p,q\ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h' \rangle}_{p q} \\
& - \sum_{q,r\ge 0} \tau \underbrace{\langle h', \dots, h', \psi', h'', \dots, h'' \rangle}_{q r} \\
& - \sum_{p_1,p_2,q,r\ge 0} \tau \underbrace{\langle h, \dots, h, \tau + h\langle\tau\rangle, h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p_1 q r} \\
& - \sum_{p,q,r\ge 0} \tau \underbrace{\langle h, \dots, h, h - h' - \psi\langle\tau\rangle, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p q r} \\
& + \sum_{p,q_1,q_2,r\ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \tau + h'\langle\tau\rangle, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p q_1 q_2 r} \\
& + \sum_{p,q,r\ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', h' - h'' - \psi'\langle\tau\rangle, h'', \dots, h'' \rangle}_{p q r} \\
& - \sum_{p,q,r_1,r_2\ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'', \tau + h''\langle\tau\rangle, h'', \dots, h'' \rangle}_{p r r_1 r_2} \\
& - \sum_{p,r\ge 0} \tau \underbrace{\langle h, \dots, h, \psi'' - \psi - \psi', h'', \dots, h'' \rangle}_{p r}.
\end{align*}
$$

Now, by some variable substitutions, note that we have the identities

$$
\begin{align*}
& \sum_{p,q,r \ge 0} \tau \underbrace{\langle h, \dots, h, h - h', h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p q r} \\
& = \sum_{p,r \ge 0} \tau \langle h, \dots, h, \psi', h'', \dots, h'' \rangle - \sum_{q,r \ge 0} \tau \langle h', \dots, h', \psi', h'', \dots, h'' \rangle;
\end{align*}
$$
$$
\sum_{p,q,r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', h' - h'', h'', \dots, h'' \rangle}_{p} = \sum_{p,q \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h' \rangle}_{p} - \sum_{p,r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h'', \dots, h'' \rangle}_{r}
$$

This finally gives

$$
\begin{align*}
d(\psi'') = h - h'' - \psi'\langle\tau\rangle - \psi'\langle\tau\rangle & \\
& - \sum_{p_1, p_2, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \tau + h\langle\tau\rangle, h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p_1} \\
& + \sum_{p, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi\langle\tau\rangle, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p} \\
& + \sum_{p, q_1, q_2, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \tau + h'\langle\tau\rangle, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{q_1} \\
& - \sum_{p, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \psi'\langle\tau\rangle, h'', \dots, h'' \rangle}_{p} \\
& - \sum_{p, q, r_1, r_2 \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'' \rangle, \tau + h''\langle\tau\rangle, h'', \dots, h'' \rangle}_{r_1} \\
& - \sum_{p, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi'', h'', \dots, h'' \rangle}_{p} - \sum_{p, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h'', \dots, h'' \rangle}_{r}
\end{align*}
$$

We also have

$$
\begin{align*}
\psi''\langle\tau\rangle = \psi\langle\tau\rangle + \psi'\langle\tau\rangle &+ \sum_{p_1, p_2, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \tau + h\langle\tau\rangle, h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p_1} \\
& - \sum_{p, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi\langle\tau\rangle, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{p} \\
& - \sum_{p, q_1, q_2, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \tau + h'\langle\tau\rangle, h', \dots, h', \psi', h'', \dots, h'' \rangle}_{q_1} \\
& + \sum_{p, q, r \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \psi'\langle\tau\rangle, h'', \dots, h'' \rangle}_{p} \\
& + \sum_{p, q, r_1, r_2 \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h', \psi', h'', \dots, h'', \tau + h''\langle\tau\rangle, h'', \dots, h'' \rangle}_{q_2}
\end{align*}
$$
At the end, we obtain

$$d(\psi'') + \psi''\langle\tau\rangle = h - h'' - \sum_{p,q\ge0} \tau\langle h, \dots, h, \psi'', \underbrace{h'', \dots, h''}_{q} \rangle$$

which proves that $h \sim_{\tau} h''$.

We now prove that if $h \sim_{\tau} h'$, then $h' \sim_{\tau} h$. We use the previous construction. More precisely, let $\psi \in A_1$ be such that

$$h - h' = d(\psi) + \psi\langle\tau\rangle + \sum_{p,q\ge0} \tau\langle h, \dots, h, \psi, \underbrace{h', \dots, h'}_{q} \rangle.$$

We search some element $\psi'$ such that the associated $\psi''$ previously constructed for the transitivity is 0. We set $\psi'_0 = -\psi$ and, for all $n \ge 0$,

$$\psi'_{n+1} = -\psi - \sum_{p,q,r\ge0} \tau\langle h, \dots, h, \psi, \underbrace{\mu', \dots, \mu'}_{q}, \psi'_{n}, \underbrace{h, \dots, h}_{r} \rangle.$$

We obtain a Cauchy sequence $(\psi'_n)_n$. Let $\psi'$ be its limit, which satisfies

$$\psi' = -\psi - \sum_{p,q,r\ge0} \tau\langle h, \dots, h, \psi, \underbrace{h', \dots, h'}_{q}, \psi', \underbrace{h, \dots, h}_{r} \rangle.$$

By the same computations as for the proof of the transitivity, we can check that $\psi'$ satisfies the equation

$$h' - h = d(\psi') + \psi'\langle\tau\rangle + \sum_{p,q\ge0} \tau\langle h', \dots, h', \psi', \underbrace{h, \dots, h}_{q} \rangle,$$

which proves that $h' \sim_{\tau} h$.

We thus have proved that $\sim_{\tau}$ is an equivalence relation. We now prove that the circular product $\odot$ is compatible with $\sim_{\tau}$. Let $h, h_1, h_2 \in A_0$ be such that $h_1 \sim_{\tau} h_2$. Let $\psi \in A_1$ be such that

$$(3) \qquad h_1 - h_2 = d(\psi) + \psi\langle\tau\rangle + \sum_{p,q\ge0} \tau\langle h_1, \dots, h_1, \psi, \underbrace{h_2, \dots, h_2}_{q} \rangle.$$

We prove first that $h_1 \overline{\odot} h \sim_{\tau} h_2 \overline{\odot} h$. Let $\psi' := \psi \odot (1+h)$. We need to show that

$$h_1 \overline{\odot} h - h_2 \overline{\odot} h = d(\psi') + \psi'\langle\tau\rangle + \sum_{p,q\ge0} \tau\langle h_1 \overline{\odot} h, \dots, h_1 \overline{\odot} h, \psi', \underbrace{h_2 \overline{\odot} h, \dots, h_2 \overline{\odot} h}_{q} \rangle.$$

We first compute $d(\psi')$. From [Ver23, Lemma 2.28], we have

$$d(\psi') = d(\psi) \odot (1+h) - \psi \odot (1+h; d(h)),$$
where we have set, following [Ver23, Definition 2.27] in the case of a complete brace algebra,

$$a \odot (1+b; c) := \sum_{n \ge 0} \sum_{k=0}^{n} a \langle b, \dots, b, c, b, \dots, b \rangle.$$

for every $a \in A$ and $b, c \in A_0$. We have

$$d(\psi) \odot (1+h) = h_1 \overline{\odot} h - h_2 \overline{\odot} h - \psi \langle \tau \rangle \odot (1+h) - \sum_{p,q \ge 0} \tau \langle h_1, \dots, h_1, \psi, h_2, \dots, h_2 \rangle \odot (1+h).$$

By the second formula of [Ver23, Lemma 2.28], we have

$$\psi \langle \tau \rangle \odot (1+h) = \psi \odot (1+h; \tau \odot (1+h)).$$

Finally, we have

$$\sum_{p,q \ge 0} \tau \langle h_1, \dots, h_1, \psi, h_2, \dots, h_2 \rangle \odot (1+h) = \sum_{p,q \ge 0} \tau \langle h_1 \overline{\odot} h, \dots, h_1 \overline{\odot} h, \psi \odot (1+h), h_2 \overline{\odot} h, \dots, h_2 \overline{\odot} h \rangle.$$

We thus have

$$d(\psi') = h_1 \overline{\odot} h - h_2 \overline{\odot} h - \psi \odot (1+h; \tau \odot (1+h)) - \sum_{p,q \ge 0} \tau \langle h_1 \overline{\odot} h, \dots, h_1 \overline{\odot} h, \psi', h_2 \overline{\odot} h, \dots, h_2 \overline{\odot} h \rangle - \psi \odot (1+h; d(h)).$$

Since we have, by [Ver23, Lemma 2.28], that

$$\psi' \langle \tau \rangle = \psi \odot (1+h; \tau + h \langle \tau \rangle).$$

we obtain at the end

$$d(\psi') + \psi' \langle \tau \rangle + \sum_{p,q \ge 0} \tau \langle h_1 \overline{\odot} h, \dots, h_1 \overline{\odot} h, \psi', h_2 \overline{\odot} h, \dots, h_2 \overline{\odot} h \rangle = h_1 \overline{\odot} h - h_2 \overline{\odot} h$$

which proves that $h_1 \overline{\odot} h \sim_{\tau} h_2 \overline{\odot} h$.

We now prove that $h \overline{\odot} h_1 \sim_{\tau} h \overline{\odot} h_2$. Let $\psi' = \psi + \sum_{p,q \ge 0} h \langle h_1, \dots, h_1, \psi, h_2, \dots, h_2 \rangle$. We show that

$$h \overline{\odot} h_1 - h \overline{\odot} h_2 = d(\psi') + \psi' \langle \tau \rangle + \sum_{p,q \ge 0} \tau \langle h \overline{\odot} h_1, \dots, h \overline{\odot} h_1, \psi', h \overline{\odot} h_2, \dots, h \overline{\odot} h_2 \rangle.$$
We first compute the sum $\sum_{p,q \ge 0} d(h) \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle$. We use that $d(h) = \tau + h\langle \tau \rangle - \tau \odot (1+h)$ to get

$$ \sum_{p,q \ge 0} d(h) \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle = \sum_{p,q \ge 0} \tau \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle \\ + \sum_{p_1, p_2, q \ge 0} h \langle h_1, \dots, h_1 | \tau \odot (1+h_1), h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle \\ + \sum_{p,q,s,t \ge 0} h \langle h_1, \dots, h_1 | \tau \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle, h_2, \dots, h_2 \rangle \\ - \sum_{p,q_1, q_2 \ge 0} h \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 | \tau \odot (1+h_2), h_2, \dots, h_2 \rangle \\ - \sum_{p,q,s,t \ge 0} \tau \langle h \odot h_1, \dots, h \odot h_1 | \psi + h \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle, h \odot h_2, \dots, h \odot h_2 \rangle. $$

Using the Leibniz rule, we obtain

$$ \begin{aligned} d(\psi') = d(\psi) &+ \sum_{p,q \ge 0} \tau \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle \\ &+ \sum_{p_1, p_2, q \ge 0} h \langle h_1, \dots, h_1 | \tau \odot (1+h_1), h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle \\ &+ \sum_{p,q,s,t \ge 0} h \langle h_1, \dots, h_1 | \tau \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle, h_2, \dots, h_2 \rangle \\ &- \sum_{p,q_1, q_2 \ge 0} h \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 | \tau \odot (1+h_2), h_2, \dots, h_2 \rangle \\ &- \sum_{p,q,s,t \ge 0} \tau \langle h \odot h_1, \dots, h \odot h_1 | \psi + h \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle, h \odot h_2, \dots, h \odot h_2 \rangle \\ &+ \sum_{p_1, p_2, q \ge 0} h \langle h_1, \dots, h_1 | d(h_1), h_1, \dots, h_1 | \psi, h_2, \dots, h_2 \rangle \\ &+ \sum_{p,q \ge 0} h \langle h_1, \dots, h_1 | d(\psi), h_2, \dots, h_2 \rangle \\ &- \sum_{p,q_1, q_2 \ge 0} h \langle h_1, \dots, h_1 | \psi, h_2, \dots, h_2 | d(h_2), h_2, \dots, h_2 \rangle. \end{aligned} $$

Using equation (3) and that $h_1, h_2 \in \mathrm{Aut}_{\mathrm{Deligne}(A)}(\tau)$, we deduce
$$
\begin{aligned}
d(\psi') = h_1 - h_2 - \psi\langle\tau\rangle &+ \sum_{p_1, p_2, q \ge 0} h\langle h_1, \dots, h_1, \tau + h_1\langle\tau\rangle, h_1, \dots, h_1, \psi, h_2, \dots, h_2 \rangle \\
& - \sum_{p, q_1, q_2 \ge 0} h\langle h_1, \dots, h_1, \psi, h_2, \dots, h_2, \tau + h_2\langle\tau\rangle, h_2, \dots, h_2 \rangle \\
& + \sum_{p, q \ge 0} h\langle h_1, \dots, h_1, h_1 - h_2 - \psi\langle\tau\rangle, h_2, \dots, h_2 \rangle \\
& - \sum_{p, q, \ge 0} \tau\langle h\bar{\odot}h_1, \dots, h\bar{\odot}h_1, \psi', h\bar{\odot}h_2, \dots, h\bar{\odot}h_2 \rangle.
\end{aligned}
$$

Since we have

$$
\sum_{p,q \ge 0} h\langle h_1, \dots, h_1, h_1 - h_2, \underbrace{h_2, \dots, h_2}_{q} \rangle = \sum_{p \ge 0} h\langle h_1, \dots, h_1 \underbrace{,}_{p} \rangle - \sum_{r \ge 0} h\langle \underbrace{h_2, \dots, h_2}_{r} \rangle,
$$

we finally obtain

$$
\begin{aligned}
d(\psi') = h\bar{\odot}h_1 - h\bar{\odot}h_2 - \psi\langle\tau\rangle &+ \sum_{p_1, p_2, q \ge 0} h\langle h_1, \dots, h_1, \tau + h_1\langle\tau\rangle, h_1, \dots, h_1, \psi, h_2, \dots, h_2 \rangle \\
& - \sum_{p, q_1, q_2 \ge 0} h\langle h_1, \dots, h_1, \psi, h_2, \dots, h_2, \tau + h_2\langle\tau\rangle, h_2, \dots, h_2 \rangle \\
& - \sum_{p, q \ge 0} h\langle h_1, \dots, h_1, \psi\langle\tau\rangle, h_2, \dots, h_2 \rangle \\
& - \sum_{p, q, \ge 0} \tau\langle h\bar{\odot}h_1, \dots, h\bar{\odot}h_1, \psi', h\bar{\odot}h_2, \dots, h\bar{\odot}h_2 \rangle.
\end{aligned}
$$

We also have

$$
\begin{aligned}
\psi'\langle\tau\rangle = \psi\langle\tau\rangle &- \sum_{p_1, p_2, q \ge 0} h\langle h_1, \dots, h_1, \tau + h_1\langle\tau\rangle, h_1, \dots, h_1, \psi, h_2, \dots, h_2 \rangle \\
&+ \sum_{p, q \ge 0} h\langle h_1, \dots, h_1, \psi\langle\tau\rangle, h_2, \dots, h_2 \rangle \\
&+ \sum_{p, q_1, q_2 \ge 0} h\langle h_1, \dots, h_1, \psi, h_2, \dots, h_2, \tau + h_2\langle\tau\rangle, h_2, \dots, h_2 \rangle.
\end{aligned}
$$

At the end, we obtain

$$
d(\psi') + \psi'\langle\tau\rangle = h\bar{\odot}h_1 - h\bar{\odot}h_2 - \sum_{p,q,\ge 0} \tau \underbrace{\langle h\bar{\odot}h_1, \dots, h\bar{\odot}h_1, \psi', h\bar{\odot}h_2, \dots, h\bar{\odot}h_2 \rangle}_{p} \underbrace{\langle h\bar{\odot}h_2, \dots, h\bar{\odot}h_2, \tau + h_2\langle\tau\rangle, h_2, \dots, h_2 \rangle}_{q}
$$

which proves that $h\bar{\odot}h_1 \sim_{\tau} h\bar{\odot}h_2$. The lemma is proved. $\square$
**Theorem 4.10.** Let $A$ be a complete brace algebra and $\tau \in \mathcal{MC}(A)$. Then

$$ \pi_1(\mathcal{MC}_\bullet(\mathcal{A}), \tau) \simeq \mathrm{Aut}_{\mathrm{Deligne}(\mathcal{A})}(\tau) / \sim_\tau. $$

*Proof.* Recall that $\mathrm{Aut}_{\mathrm{Deligne}(\mathcal{A})}(\tau) = \{h \in A_0 \mid d(h) = \tau + h\langle\tau\rangle - \tau \odot (1+h)\}$. Let $h \in A_0$. By Lemma 4.7, we have that $h \in \mathrm{Aut}_{\mathrm{Deligne}(\mathcal{A})}(\tau)$ if and only if

$$ -\tau \otimes (\underline{0}^{\vee} + \underline{1}^{\vee}) - h \otimes \underline{01}^{\vee} \in \mathcal{MC}(A \otimes \Sigma N^*(\Delta^1)). $$

We thus have a bijection

$$ f: \mathrm{Aut}_{\mathrm{Deligne}(\mathcal{A})}(\tau) \to \mathcal{MC}_1(A)_\tau $$

where we denote by $\mathcal{MC}_1(A)_\tau$ the subset of $\mathcal{MC}_1(A)$ given by elements whose 0 and 1 vertices are given by $\tau$. Consider now $h, h' \in A_0$ such that

$$ d(h) = \tau + h\langle\tau\rangle - \tau \odot (1+h); $$

$$ d(h') = \tau + h'\langle\tau\rangle - \tau \odot (1+h'). $$

Let $\xi \in \mathcal{MC}_2(A)$ be such that $d_1\xi = f(h)$ and $d_2\xi = f(h')$. We write $\xi$ as

$$ \xi = -\tau \otimes (\underline{0}^{\vee} + \underline{1}^{\vee} + \underline{2}^{\vee}) - h' \otimes \underline{01}^{\vee} - h \otimes \underline{02}^{\vee} + \psi \otimes \underline{012}^{\vee} $$

for some $\psi \in A_1$. We make precise the Maurer-Cartan condition on $\xi$. We first have

$$ d(\xi) = -d(\tau) \otimes (\underline{0}^{\vee} + \underline{1}^{\vee} + \underline{2}^{\vee}) - d(h') \otimes \underline{01}^{\vee} - d(h) \otimes \underline{02}^{\vee} + (d(\psi) - h + h') \otimes \underline{012}^{\vee}. $$

By Lemma 3.8, we have

$$ \xi[\xi]_1 = -\tau\langle\tau\rangle \otimes (\underline{0}^{\vee} + \underline{1}^{\vee} + \underline{2}^{\vee}) - (\tau\langle h' \rangle - h'\langle\tau\rangle) \otimes \underline{01}^{\vee} \\ - (\tau\langle h \rangle + h\langle\tau\rangle) \otimes \underline{02}^{\vee} + (\tau\langle\psi\rangle + \psi\langle\tau\rangle) \otimes \underline{012}^{\vee}. $$

Let $r \ge 2$. From the computations of $\partial^1$ and $\partial^2$ in Corollary 3.9, we deduce

$$ \xi[\xi]_r = -\tau \underbrace{\langle h, \dots, h \rangle}_{r} \otimes \underline{01}^{\vee} - \tau \underbrace{\langle h', \dots, h' \rangle}_{r} \otimes \underline{02}^{\vee} + \sum_{p+q=r-1} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h' \rangle}_{p} \otimes \underline{012}^{\vee}. $$

We thus have proved that $\xi$ is a Maurer-Cartan element if and only if

$$ h - h' = d(\psi) + \psi\langle\tau\rangle + \sum_{p,q \ge 0} \tau \underbrace{\langle h, \dots, h, \psi, h', \dots, h' \rangle}_{p} \otimes \underline{012}^{\vee}. $$

Equivalently, we have that $[f(h)] = [f(h')]$ if and only if $h \sim_\tau h'$. We thus have a well defined bijection

$$ \bar{f}: \mathrm{Aut}_{\mathrm{Deligne}(\mathcal{A})}(\tau) / \sim_\tau \to \pi_1(\mathcal{MC}_\bullet(\mathcal{A}), \tau), $$

We now check that $\bar{f}$ is compatible with the group structures. Let $h, h' \in A_0$ be such that $\alpha = -\tau \otimes (\underline{0}^{\vee} + \underline{1}^{\vee}) - h \otimes \underline{01}^{\vee}$ and $\alpha' = -\tau \otimes (\underline{0}^{\vee} + \underline{1}^{\vee}) - h' \otimes \underline{01}^{\vee}$ are Maurer-Cartan elements in $\mathcal{MC}(A \otimes \Sigma N^*(\Delta^1))$. As we have seen before, by Lemma 4.7, it is equivalent to ask

$$ d(\mu) = \tau + h\langle\tau\rangle - \tau \odot (1+h); $$
$$d(h') = \tau + h'\langle\tau\rangle - \tau \odot (1 + h').$$

By Corollary 3.9, we see that

$$-\tau \otimes (\underline{0}^{\vee} + \underline{1}^{\vee} + \underline{2}^{\vee}) - h' \otimes \underline{12}^{\vee} - (h\bar{\odot}h') \otimes \underline{02}^{\vee} - h \otimes \underline{01}^{\vee} \in \mathcal{MC}(A \otimes \Sigma N^*(\Delta^2)).$$

We then have

$$[\alpha] \cdot [\alpha'] = [-\tau \otimes (\underline{0}^{\vee} + \underline{1}^{\vee}) - (h\bar{\odot}h') \otimes \underline{01}^{\vee}]$$

in $\pi_1(\mathcal{MC}_\bullet(A), \tau)$, which gives

$$\bar{f}([\alpha] \cdot [\alpha']) = h\bar{\odot}h' = \bar{f}([\alpha]) \odot \bar{f}([\alpha'])$$

showing that $\bar{f}$ is an isomorphism of groups. $\square$

### 4.2.3 The group $\pi_2(\mathcal{MC}_\bullet(A), \tau)$

We now compute the group $\pi_2(\mathcal{MC}_\bullet(A), \tau)$. We begin by general lemmas that will also be useful for the computations of $\pi_n(\mathcal{MC}_\bullet(A), \tau)$ for $n \ge 3$.

**Lemma 4.11.** Let $T$ be a canonical tree with $|T| \ge 3$, and $n \ge 1$. If the first branch of $T$ has only one vertex, then there is no element of the form $\underline{0}, \dots, \underline{n} \in \Sigma^{-1}N_*(\Delta^n)$ among the non-root vertices in the tensor products produced by $T \otimes \Lambda\mu_T(\underline{0} \cdots \underline{n}) \in \mathcal{Brace}^\vee(\Sigma^{-1}N_*(\Delta^n))$.

*Proof.* For every finite set $E$ and $k \in E$, we denote by $\pi_{\{k\}} : \chi(E) \to \chi(E \setminus \{k\})$ the morphism which forgets the element $k$. If a surjection has multiple occurrences of the element $k$, then its image by $\pi_{\{k\}}$ is 0 by convention. Note that if $A$ and $B$ are disjoint finite sets, then for every $u \in \chi(A), v \in \chi(B)$, we have $\pi_{\{k\}}(u \cdot v) = \pi_{\{k\}}(u) \cdot \pi_{\{k\}}(v)$ in $\chi(A \sqcup B)$.

Let $T$ be a canonical tree with $|T| \ge 3$. By Lemma 3.5, there exists $u_T \in \chi(V_T \setminus \{1\})$ such that

$$TR(\mu_T) = 12 \cdot u_T.$$

We write uniquely $u_T$ as

$$u_T = \sum_{i=1}^{m_T} \lambda_i^T u_T^i,$$

where $\lambda_1^T, \dots, \lambda_{m_T}^T \in \mathbb{K}$ and $u_T^1, \dots, u_T^{m_T}$ are non degenerate surjections. We prove that, for every $1 \le i \le m_T$ and $2 \le k \le |T|,$

$$\pi_{\{k\}}(2 \cdot u_T^i) = 0.$$

It is true for $k = 2$, since $u_T^i \in \chi(2 < \dots < |T|)$ so that there are at least two occurrences of 2 in the surjection $2 \cdot u_T^i$. Suppose now that $k \ge 3$. We prove the statement by induction on $|T|$. If $|T| = 3$, the first tree of Example 3.2 gives $TR(\mu_T) = \pm 1232$ and $\pi_{\{3\}}(232) = 22 = 0$. We now suppose that $|T| \ge 4$. By the proof of Lemma 3.5, we have

$$Tr(\mu_T) = - \sum_{S \subset T} \pm 12 \cdot \pi_2(TR(\mu_{T/S}) \circ_S TR(\mu_S)).$$

Let $S \subset T$ be such that $b_S = b_{T/S} = 1$. Suppose that $|S|, |T/S| \ne 2$. By Lemma 3.5, there exist $u_S \in \chi(V_S \setminus \{r(S)\})$ and $u_{T/S} \in \chi(V_{T/S} \setminus \{r(T/S)\})$ such that
$$
\begin{align*}
\mathrm{TR}(\mu_S) &= r(S)p \cdot u_S; \\
\mathrm{TR}(u_{T/S}) &= r(T/S)q \cdot u_{T/S}
\end{align*}
$$

where $p \in V_S$ and $q \in V_{T/S}$ are the second element of their respective totally ordered set. If $r(S) \neq 1$, then $r(T/S) = 1$ and $q = 2$, since $b_T = 1$, so that

$$
\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S) = 12 \cdot (u_{T/S} \circ_S (r(S)p \cdot u_S))
$$

whose associated term in the sum is 0. Suppose now that $r(S) = 1$. Then $r(T/S) = S$, so that

$$
\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S) = 1p \cdot u_S \cdot q \cdot u_{T/S}.
$$

We write $u_S$ and $u_{T/S}$ in the basis given by non degenerated surjections:

$$
\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S) = \sum_{i=1}^{m_S} \sum_{j=1}^{m_{T/S}} \lambda_i^S \lambda_j^{T/S} 1p \cdot u_S^i \cdot q \cdot u_{T/S}^j.
$$

Since $k \neq 1,2$, we have, for every $1 \le i \le m_S$ and $1 \le j \le m_{T/S}$,

$$
\pi_{\{k\}}(12 \cdot \pi_2(1p \cdot u_S^i \cdot q \cdot u_{T/S}^j)) = 12 \cdot \pi_{\{k\}}(p \cdot u_S^i) \cdot \pi_{\{k\}}(q \cdot u_{T/S}^j).
$$

By induction hypothesis (on $S$ if $k \in V_S$, on $T/S$ else), we obtain 0. Suppose now $|S| = 2$ and $|T/S| \neq 2$. By the same argument as before, we can restrict to the case $r(S) = 1$ (which implies that $r(T/S) = S$), so that

$$
\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S) = 1pq \cdot u_{T/S}.
$$

where $p \in V_S$ and $q \in V_{T/S}$ are the second element of their respective totally ordered set. We have

$$
12 \cdot \pi_2(\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S)) = 12pq \cdot u_{T/S} = \sum_{i=1}^{m_{T/S}} \lambda_i^{T/S} 12pq \cdot u_{T/S}^i.
$$

Let $1 \le i \le m_{T/S}$. If $p = 2$, then the corresponding term in the sum is 0. If $p \ne 2$, then $q = 2$ so
that we need to compute

$$
1 \cdot \pi_{\{k\}}(2p2 \cdot u_{T/S}^{i}).
$$

If $k = p$, then $\pi_{\{p\}}(2p2 \cdot u_{T/S}^i) = 22 \cdot u_{T/S}^i = 0$. If $k \ne p$, then $\pi_{\{k\}}(2p2 \cdot u_{T/S}^i) = 2p2 \cdot \pi_{\{k\}}(u_{T/S}^i)$,
which is 0 by induction hypothesis on $T/S$. Suppose now that $|T/S| = 2$ and $|S| \ne 2$. As before,
we can suppose that $r(S) = 1$ and $2 \notin V_S$. We then have

$$
\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S) = 13 \cdot u_S \cdot 2,
$$

which gives

$$
12 \cdot \pi_2(\mathrm{TR}(\mu_{T/S}) \circ_S \mathrm{TR}(\mu_S)) = 123 \cdot u_S \cdot 2 = \sum_{i=1}^{m_S} \lambda_i^S 123 \cdot u_S^i \cdot 2.
$$

Let $1 \le i \le m_S$. Then

$$
\pi_{\{k\}}(23 \cdot u_S^i \cdot 2) = 2 \cdot \pi_{\{k\}}(3 \cdot u_S^i) \cdot 2 = 0,
$$
by induction hypothesis on S. The case $|S| = |T/S| = 2$ gives $|T| = 3$ which has already be proved in the beginning of the proof.

We thus have proved that $\pi_{\{k\}}(2 \cdot u_T^i) = 0$ for every canonical tree $T$ such that $|T| \ge 3$ and $2 \le k \le |T|$, $1 \le i \le m_T$. We now prove the lemma. Let $2 \le k \le |T|$. By definition of the interval cut operations (see [BF04, §2.2.1]), the tensors with a factor of the form $\underline{0}, \dots, \underline{n}$ at position $k$ occurring in the expansion of $(T \otimes \Lambda \mu_T)(\underline{0} \cdots \underline{n})$ are precisely produced by the surjections $12 \cdot u_T^1, \dots, 12 \cdot u_T^{m_T}$ which contain only one occurrence of $k$. Let $1 \le i \le m_T$ be such that $u_T^i$ contains only one occurrence of $k$. The tensors produced by $2 \cdot u_T^i$ with a degree $-1$ element at position $k$ are given by the insertion of the appropriate degree 0 vertex at position $k$ of the tensors produced by the surjection $\pi_{\{k\}}(2 \cdot u_T^i)$. Since this surjection is 0, the lemma is proved. $\square$

**Lemma 4.12.** Let $n \ge 2$. Let $a, b_1, \dots, b_m \in A$, let $\underline{x}, \underline{y_1}, \dots, \underline{y_m} \in N_*(\Delta^n)$ be basis elements and $r_1, \dots, r_m \ge 0$. Suppose that

$$|\underline{x}| + r_1|\underline{y_1}| + \dots + r_m|\underline{y_m}| > n - 2.$$

$$\text{Then } a \otimes \underline{x}^{\vee} \{ b_1 \otimes \underline{y_1}^{\vee}, \dots, b_m \otimes \underline{y_m}^{\vee} \}_{r_1, \dots, r_m} = 0.$$

*Proof.* Let $r = r_1 + \dots + r_m$. We more generally show that for every $\mu \in \Sigma^{-r}\mathcal{E}(r+1)_{r-1}$ and $\underline{z_1}, \dots, \underline{z_r} \in N_*(\Delta^n)$ such that $|\underline{x}| + |\underline{z_1}| + \dots + |\underline{z_r}| > n-2$, the evaluation of $\mu$ on the tensor $\underline{x}^{\vee} \otimes \underline{z_1}^{\vee} \otimes \dots \otimes \underline{z_r}^{\vee}$ when using the $\mathcal{E}$-algebra structure of $N_*(\Delta^n)$ is 0. One one hand, the evaluation of $\mu$ on the tensor $\underline{x}^{\vee} \otimes \underline{z_1}^{\vee} \otimes \dots \otimes \underline{z_r}^{\vee}$ is an element with degree $-1-|\underline{x}|-|\underline{z_1}|-\dots-|\underline{z_r}| < 1-n$. On the other hand, since the result is an element of $\Sigma N_*(\Delta^n)$, its degree is equal or greater than $|\underline{0} \cdots \underline{n}^{\vee}| = 1-n$. The evaluation of $\mu$ on the tensor $\underline{x}^{\vee} \otimes \underline{z_1}^{\vee} \otimes \dots \otimes \underline{z_r}^{\vee}$ must then be 0.

To obtain the lemma, we apply this result to $\mu = \Lambda \mu_T$ where $T \in \mathcal{PT}(r+1)$ is a canonical tree, and $\underline{z_1}, \dots, \underline{z_r} = \underline{y_1}, \dots, \underline{y_1}, \dots, \underline{y_m}, \dots, \underline{y_m}$ up to a shuffle permutation in $\mathrm{Sh}(r_1, \dots, r_m)$. $\square$

Before stating the next lemma, recall that if $A$ is a brace algebra and $\tau \in \mathcal{MC}(A)$, then we have a differential defined by

$$d_{\tau}(x) = d(x) + \tau\langle x \rangle - (-1)^{|x|}x\langle \tau \rangle.$$

We denote by $A^\tau$ the underlying dg $\mathbb{K}$-module.

**Lemma 4.13.** Let $\tau \in \mathcal{MC}(A)$ and $n \ge 1$. We denote by $\mathcal{MC}_{n+1}(A)_\tau$ the set given by elements $\xi \in \mathcal{MC}_{n+1}(A)$ with faces given by $\tau$. Then we have a bijection

$$f: Z_n(A^\tau) \to \mathcal{MC}_{n+1}(A)_\tau$$

given by

$$f(h) = -\tau \otimes \left( \sum_{k=0}^{n+1} \underline{k}^{\vee} \right) - h \otimes \underline{0} \cdots \underline{(n+1)^{\vee}}.$$

*Proof.* Let $\xi \in \mathcal{MC}_{n+1}(A)_\tau$. Then there exists $h \in A_n$ such that

$$\xi = -\tau \otimes \left( \sum_{k=0}^{n+1} \underline{k}^{\vee} \right) - h \otimes \underline{0} \cdots \underline{(n+1)^{\vee}}.$$
We make precise the Maurer-Cartan condition on $\xi$. Let $p \ge 2$. By Lemma 4.11, we have

$$ \xi[\xi]_p = \sum_{k=0}^{n+1} (-1)^p \tau \otimes \underline{k}^{\vee} [h \otimes \underline{0 \cdots (n+1)^{\vee}}]_p + (-1)^{p+1} h \otimes \underline{0 \cdots (n+1)^{\vee}} [h \otimes \underline{0 \cdots (n+1)^{\vee}}]_p. $$

By Lemma 4.12, and since we have $n + np > -1 + np > n - 1$ because $p \ge 2$, we deduce that $\xi[\xi]_p = 0$. If $p = 1$, then, by Corollary 3.8,

$$ \xi[\xi]_1 = -\tau\langle\tau\rangle \otimes \left(\sum_{k=0}^{n+1} \underline{k}^{\vee}\right) - (\tau\langle h \rangle - (-1)^n h\langle\tau\rangle) \otimes \underline{0 \cdots (n+1)^{\vee}}. $$

We also have

$$ d(\xi) = -d(\tau) \otimes \left( \sum_{k=0}^{n+1} \underline{k}^{\vee} \right) - d(h) \otimes \underline{0 \cdots (n+1)^{\vee}}. $$

The Maurer-Cartan condition on $\xi$ is then equivalent to

$$ d(h) + \tau\langle h \rangle - (-1)^n h\langle\tau\rangle = 0. $$

which gives our desired bijection

$$ f: Z_n(A^\tau) \to \mathcal{MC}(A \otimes \Sigma N^*(\Delta^{n+1}))_\tau. $$

□

We now consider $n = 2$. The computation of $\pi_2$ will emphasize a group structure on $H_1(A^\tau)$ given by the following lemma.

**Lemma 4.14.** Let $A$ be a complete brace algebra and $\tau \in \mathcal{MC}(A)$. Then $(H_1(A^\tau), *_{\tau}, 0)$ is an abelian group with the product $*_{\tau}$ defined by

$$ [\mu] *_{\tau} [\mu'] = [\mu + \mu' + \tau\langle\mu, \mu'\rangle]. $$

*Proof.* We first prove that if $\mu, \mu' \in Z_1(A^\tau)$ then $\mu'' := \mu + \mu' + \tau\langle\mu, \mu'\rangle \in Z_1(A^\tau)$. We have

$$ d(\mu'') = d(\mu) + d(\mu') - \tau\langle\tau, \mu, \mu'\rangle + \tau\langle\mu, \tau, \mu'\rangle - \tau\langle\mu, \mu', \tau\rangle \\ - \tau\langle\tau\langle\mu\rangle, \mu'\rangle + \tau\langle\mu, \tau\langle\mu'\rangle\rangle - \tau\langle\tau\langle\mu, \mu'\rangle\rangle - \tau\langle d(\mu), \mu'\rangle + \tau\langle\mu, d(\mu')\rangle. $$

We also have

$$ \mu''\langle\tau\rangle = \mu\langle\tau\rangle + \mu'\langle\tau\rangle + \tau\langle\tau, \mu, \mu'\rangle - \tau\langle\mu, \tau, \mu'\rangle + \tau\langle\mu, \mu'\langle\tau\rangle\rangle - \tau\langle\mu\langle\tau\rangle, \mu'\rangle + \tau\langle\mu, \mu'\langle\tau\rangle\rangle $$

and

$$ \tau\langle\mu''\rangle = \tau\langle\mu\rangle + \tau\langle\mu'\rangle + \tau\langle\tau\langle\mu, \mu'\rangle\rangle. $$
At the end, we obtain that $d_{\tau}(\mu'') = 0$, which proves that $\mu'' \in Z_1(A^{\tau})$.

We now show that the product $*_\tau$ is well defined on $H_1(A^\tau)$. Let $\mu, \mu_1, \mu_2 \in Z_1(A^\tau)$ and $\psi \in A_2$ be such that

$$\mu_1 - \mu_2 = d(\psi) + \tau\langle\psi\rangle - \psi\langle\tau\rangle.$$

Let $\psi' := \psi + \tau\langle\mu, \psi\rangle$. We show that

$$\mu_1 - \mu_2 + \tau\langle\mu, \mu_1 - \mu_2\rangle = d(\psi') + \tau\langle\psi'\rangle - \psi'\langle\tau\rangle.$$

We first compute $d(\psi')$. We have

$$d(\psi') = d(\psi) - \tau\langle\tau, \mu, \psi\rangle + \tau\langle\mu, \tau, \psi\rangle + \tau\langle\mu, \psi, \tau\rangle - \tau\langle\tau\langle\mu\rangle, \psi\rangle + \tau\langle\mu, \tau\langle\psi\rangle\rangle \\ - \tau\langle\tau\langle\mu, \psi\rangle\rangle - \tau\langle d(\mu), \psi\rangle + \tau\langle\mu, d(\psi)\rangle.$$

We also have

$$\psi'\langle\tau\rangle = \psi\langle\tau\rangle - \tau\langle\tau, \mu, \psi\rangle + \tau\langle\mu, \tau, \psi\rangle + \tau\langle\mu, \psi, \tau\rangle + \tau\langle\mu\langle\tau\rangle, \psi\rangle + \tau\langle\mu, \psi\langle\tau\rangle\rangle.$$

At the end, we obtain

$$d(\psi') + \tau\langle\psi'\rangle - \psi'\langle\tau\rangle = \mu_1 - \mu_2 + \tau\langle\mu, \mu_1 - \mu_2\rangle$$

so that

$$[\mu + \mu_1 + \tau\langle\mu, \mu_1\rangle] = [\mu + \mu_2 + \tau\langle\mu, \mu_2\rangle].$$

By the same computations with $\psi' := \psi - \tau\langle\psi, \mu\rangle$, we can show that

$$[\mu + \mu_1 + \tau\langle\mu_1, \mu\rangle] = [\mu + \mu_2 + \tau\langle\mu_2, \mu\rangle].$$

The product $*_\tau$ is thus well defined on $H_1(A^\tau)$. We now prove that it endows $H_1(A^\tau)$ with an abelian group structure. We prove the associativity of the operation $*_\tau$. We have that

$$([\mu] *_{\tau} [\mu']) *_{\tau} [\mu''] = [\mu + \mu' + \mu'' + \tau\langle\mu, \mu'\rangle + \tau\langle\mu + \mu' + \tau\langle\mu, \mu'\rangle, \mu'')]$$

$$[\mu] *_{\tau} ([\mu'] *_{\tau} [\mu'']) = [\mu + \mu' + \mu'' + \tau\langle\mu', \mu''\rangle + \tau\langle\mu, \mu' + \mu'' + \tau\langle\mu', \mu''\rangle\rangle].$$

The difference between the two representatives is

$$\tau\langle\mu, \tau\langle\mu', \mu''\rangle\rangle - \tau\langle\tau\langle\mu, \mu'\rangle, \mu''\rangle.$$

We show that this element is the image of $\psi := \tau\langle\mu, \mu', \mu''\rangle \in A_2$ under $d_\tau$. First, using that $d(\tau) = -\tau\langle\tau\rangle$ and the brace algebra structure on A, we have

$$d(\tau)\langle\mu, \mu', \mu''\rangle = -\tau\langle\tau, \mu, \mu', \mu''\rangle + \tau\langle\mu, \tau, \mu', \mu''\rangle - \tau\langle\mu, \mu', \tau, \mu''\rangle + \tau\langle\mu, \mu', \mu'', \tau\rangle \\ - \tau\langle\tau\langle\mu\rangle, \mu', \mu''\rangle + \tau\langle\mu, \tau\langle\mu', \mu''\rangle - \tau\langle\mu, \mu', \tau\langle\mu''\rangle\rangle \\ - \tau\langle\tau\langle\mu, \mu', \mu''\rangle\rangle + \tau\langle\mu, \tau\langle\mu', \mu''\rangle\rangle - \tau\langle\tau\langle\mu, \mu', \mu''\rangle\rangle.$$

This gives

$$d(\psi) = -\tau\langle\tau, \mu, \mu', \mu''\rangle + \tau\langle\mu, \tau, \mu', \mu''\rangle - \tau\langle\mu, \mu', \tau, \mu''\rangle + \tau\langle\mu, \mu', \mu'', \tau\rangle \\ - \tau\langle\tau\langle\mu\rangle, \mu', \mu''\rangle + \tau\langle\mu, \tau\langle\mu', \mu''\rangle - \tau\langle\mu, \mu', \tau\langle\mu''\rangle\rangle \\ - \tau\langle\tau\langle\mu, \mu', \mu''\rangle\rangle + \tau\langle\mu, \tau\langle\mu', \mu''\rangle\rangle - \tau\langle\tau\langle\mu, \mu', \mu''\rangle\rangle \\ - \tau\langle d(\mu), \mu', \mu''\rangle + \tau\langle\mu, d(\mu'), \mu''\rangle - \tau\langle\mu, \mu', d(\mu'')\rangle$$
Using that $\mu, \mu', \mu'' \in Z_1(A^\tau)$, we obtain

$$d(\psi) = -\tau\langle\tau, \mu, \mu', \mu''\rangle + \tau\langle\mu, \tau, \mu', \mu''\rangle - \tau\langle\mu, \mu', \tau, \mu''\rangle + \tau\langle\mu, \mu', \mu'', \tau\rangle \\ + \tau\langle\mu\langle\tau\rangle, \mu', \mu''\rangle - \tau\langle\mu, \mu'\langle\tau\rangle, \mu''\rangle + \tau\langle\mu, \mu', \mu''\langle\tau\rangle\rangle \\ - \tau\langle\tau\langle\mu, \mu'\rangle, \mu''\rangle + \tau\langle\mu, \tau\langle\mu', \mu''\rangle\rangle - \tau\langle\tau\langle\mu, \mu', \mu''\rangle\rangle.$$

We also have

$$\psi\langle\tau\rangle = -\tau\langle\tau, \mu, \mu', \mu''\rangle + \tau\langle\mu, \tau, \mu', \mu''\rangle - \tau\langle\mu, \mu', \tau, \mu''\rangle + \tau\langle\mu, \mu', \mu'', \tau\rangle \\ + \tau\langle\mu\langle\tau\rangle, \mu', \mu''\rangle - \tau\langle\mu, \mu'\langle\tau\rangle, \mu''\rangle + \tau\langle\mu, \mu', \mu''\langle\tau\rangle\rangle.$$

which finally gives

$$d(\psi) + \tau\langle\psi\rangle - \psi\langle\tau\rangle = \tau\langle\mu, \tau\langle\mu', \mu''\rangle\rangle - \tau\langle\tau\langle\mu, \mu', \mu''\rangle\rangle$$

so that we have the associativity.

We now prove that every element $[\mu]$ has an inverse under $*_\tau$. We set $\mu'_0 = -\mu$ and, for every $n \ge 0$,

$$\mu'_{n+1} = -\mu - \tau\langle\mu, \mu'_{n}\rangle.$$

We obtain a Cauchy sequence in $A$. Because $A$ is complete, this sequence has a limit denoted by $\mu'$ which satisfies

$$\mu + \mu' + \tau\langle\mu, \mu'\rangle = 0$$

so that $[\mu']$ is the inverse of $[\mu]$ under $*_\tau$. We thus have proved that $*_\tau$ endows $H_1(A^\tau)$ with a group structure.

We now prove that $*_\tau$ is abelian. Let $\mu, \mu' \in A_1$. We set $\psi := \mu\langle\mu'\rangle$, and prove that

$$d(\psi) + \tau\langle\psi\rangle - \psi\langle\tau\rangle = \tau\langle\mu', \mu\rangle - \tau\langle\mu, \mu'\rangle.$$

We have

$$\begin{align*}
d(\psi) &= d(\mu)\langle\mu'\rangle - \mu\langle d(\mu')\rangle \\
&= -\tau\langle\mu\rangle\langle\mu'\rangle - \mu\langle\tau\rangle\langle\mu'\rangle + \mu\langle\tau\langle\mu'\rangle\rangle + \mu\langle\mu'\langle\tau\rangle\rangle \\
&= -\tau\langle\mu(\mu')\rangle - \tau\langle\mu, \mu'\rangle + \tau\langle\mu', \mu\rangle - \mu\langle\tau\langle\mu'\rangle\rangle \\
&\quad -\mu\langle\tau, \mu'\rangle + \mu\langle\mu', \tau\rangle + \mu\langle\tau\langle\mu'\rangle\rangle + \mu\langle\mu'\langle\tau\rangle\rangle \\
&= -\tau\langle\mu(\mu')\rangle - \tau\langle\mu, \mu'\rangle + \tau\langle\mu', \mu\rangle \\
&\quad -\mu\langle\tau, \mu'\rangle + \mu\langle\mu', \tau\rangle + \mu\langle\mu'\langle\tau\rangle\rangle
\end{align*}$$

and

$$\psi\langle\tau\rangle = \mu\langle\mu', \tau\rangle - \mu\langle\tau, \mu'\rangle + \mu\langle\mu'\langle\tau\rangle\rangle.$$

We then have

$$d(\psi) + \tau\langle\psi\rangle - \psi\langle\tau\rangle = \tau\langle\mu', \mu\rangle - \tau\langle\mu, \mu'\rangle$$

which proves that

$$[\mu + \mu' + \tau\langle\mu, \mu'\rangle] = [\mu' + \mu + \tau\langle\mu', \mu\rangle].$$

The operation $*_\tau$ is then commutative.

The lemma is proved. $\square$
**Theorem 4.15.** Let A be a complete brace algebra and $\tau \in \mathcal{MC}(A)$. Then

$$ \pi_2(\mathcal{MC}_\bullet(\mathcal{A}), \tau) \simeq (H_1(\mathcal{A}^\tau), *_{\tau}, 0). $$

*Proof.* By Lemma 4.13, we have a bijection

$$ f: Z_1(A^\tau) \to \mathcal{MC}_2(A)_\tau. $$

We consider its composite $\tilde{f}: Z_1(A^\tau) \to \pi_2(\mathcal{MC}_\bullet(A), \tau)$ with the projection of $\mathcal{MC}_2(A)_\tau$ onto $\pi_2(\mathcal{MC}_\bullet(A), \tau)$. We show that $\tilde{f}$ is compatible with the equivalence relation on $H_1(A^\tau)$ given by Lemma 4.14. Let $\mu, \mu' \in Z_1(A^\tau)$ be such that there exists $\psi \in A_2$ with $\mu - \mu' = d_\tau(\psi)$. Namely,

$$ d(\psi) + \tau\langle\psi\rangle - \psi\langle\tau\rangle = \mu - \mu'. $$

By Corollary 3.8, Corollary 3.9 and Lemma 4.12, we have

$$ -\tau \otimes (\mathbf{0}^{\vee} + \mathbf{1}^{\vee} + \mathbf{2}^{\vee}) - \mu \otimes \mathbf{123}^{\vee} - \mu' \otimes \mathbf{023}^{\vee} + \psi \otimes \mathbf{0123}^{\vee} \in \mathcal{MC}(A \otimes \Sigma N^*(\Delta^3)), $$

which shows that $\tilde{f}(\mu) = \tilde{f}(\mu')$. We thus have a well defined map

$$ \bar{f}: H_1(A^\tau) \to \pi_2(\mathcal{MC}_\bullet(A), \tau). $$

We prove that $\bar{f}$ preserves the group structures. Let $\mu, \mu' \in Z_1(A^\tau)$. Recall that

$$ f(\mu) = -\tau \otimes (\mathbf{0}^{\vee} + \mathbf{1}^{\vee} + \mathbf{2}^{\vee}) - \mu \otimes \mathbf{012}^{\vee}; $$

$$ f(\mu') = -\tau \otimes (\mathbf{0}^{\vee} + \mathbf{1}^{\vee} + \mathbf{2}^{\vee}) - \mu' \otimes \mathbf{012}^{\vee}. $$

We search for $\mu'' \in A_1$ and $\psi \in A_2$ such that

$$ \omega := -\tau \otimes (\mathbf{0}^{\vee} + \mathbf{1}^{\vee} + \mathbf{2}^{\vee} + \mathbf{3}^{\vee}) - \mu' \otimes \mathbf{123}^{\vee} \\ - \mu'' \otimes \mathbf{023}^{\vee} - \mu \otimes \mathbf{013}^{\vee} + \psi \otimes \mathbf{0123}^{\vee} \in \mathcal{MC}(A \otimes \Sigma N^*(\Delta^3)). $$

We have

$$ d(\omega) = -d(\tau) \otimes (\mathbf{0}^{\vee} + \mathbf{1}^{\vee} + \mathbf{2}^{\vee} + \mathbf{3}^{\vee}) - d(\mu') \otimes \mathbf{123}^{\vee} - d(\mu'') \otimes \mathbf{023}^{\vee} - d(\mu) \otimes \mathbf{013}^{\vee} \\ +(d(\psi) - \mu'' + \mu + \mu') \otimes \mathbf{0123}^{\vee}. $$

By Corollary 3.8, we also have

$$ \omega[\omega]_1 = -\tau\langle\tau\rangle \otimes (\mathbf{0}^{\vee} + \mathbf{1}^{\vee} + \mathbf{2}^{\vee} + \mathbf{3}^{\vee}) - (\tau\langle\mu'\rangle + \mu'\langle\tau\rangle) \otimes \mathbf{123}^{\vee} - (\tau\langle\mu''\rangle + \mu''\langle\tau\rangle) \otimes \mathbf{023}^{\vee} \\ -(\tau\langle\mu\rangle + \mu\langle\tau\rangle) \otimes \mathbf{013}^{\vee} + (\tau\langle\psi\rangle - \psi\langle\tau\rangle) \otimes \mathbf{0123}^{\vee}. $$

By Corollary 3.9, we have

$$ \omega[\omega]_2 = \tau\langle\mu, \mu'\rangle \otimes \mathbf{0123}^{\vee}. $$

Finally, for every $p > 2$ and by Lemma 4.11,
We make precise the Maurer-Cartan condition on $\xi$. We first compute $d(\xi)$. Note that we have, for every $0 \le k \le n+2$,

$$d(\underline{k}^{\vee}) = \sum_{k<j \le n+2} \underline{k}j^{\vee} - \sum_{0 \le i < k} \underline{ik}^{\vee},$$

which implies

$$\sum_{k=0}^{n+2} d(\underline{k}^{\vee}) = 0$$

by a variable substitution. We then have

$$d(\xi) = -d(\tau) \otimes \left( \sum_{k=0}^{n+2} \underline{k}^{\vee} \right) - d(\mu) \otimes \underline{1} \cdots \underline{(n+2)^{\vee}} - d(\mu'') \otimes \underline{02} \cdots \underline{(n+2)^{\vee}} \\ - d(\mu') \otimes \underline{013} \cdots \underline{(n+2)^{\vee}} + (d(\psi) + \mu'' - \mu - \mu') \otimes \underline{0} \cdots \underline{(n+2)^{\vee}}.$$

We now compute $\xi[\xi]_1$. By Corollary 3.8, we have

$$\xi[\xi]_1 = -\tau\langle\tau\rangle \otimes \left(\sum_{k=0}^{n+2} \underline{k}^{\vee}\right) - (\tau\langle\mu\rangle - (-1)^n\mu\langle\tau\rangle) \otimes \underline{1}\cdots\underline{(n+2)^{\vee}} \\ - (\tau\langle\mu''\rangle - (-1)^n\mu''\langle\tau\rangle) \otimes \underline{02}\cdots\underline{(n+2)^{\vee}} \\ - (\tau\langle\mu'\rangle - (-1)^n\mu'\langle\tau\rangle) \otimes \underline{013}\cdots\underline{(n+2)^{\vee}} \\ + (\tau\langle\psi\rangle - (-1)^{n+1}\psi\langle\tau\rangle) \otimes \underline{0}\cdots\underline{(n+2)^{\vee}}.$$

We now show that $\xi[\xi]_p = 0$ for every $p \ge 2$. By Lemma 4.11, we have

$$\begin{aligned} \xi[\xi]_p = & - \sum_{s+t=p} \tau \otimes \left( \sum_{k=0}^{n+2} \underline{k}^{\vee} \right) \left\{ \begin{aligned}[t] & \underline{\mu} \otimes \underline{1} \cdots \underline{(n+2)^{\vee}} - \mu'' \otimes \underline{02} \cdots \underline{(n+2)^{\vee}} \\ & - \mu' \otimes \underline{013} \cdots \underline{(n+2)^{\vee}}, \psi \otimes \underline{0} \cdots \underline{(n+2)^{\vee}} \end{aligned} \right\}_{s,t} \\ & + \sum_{s+t=p} \left( \begin{aligned}[t] & \underline{\mu} \otimes \underline{1} \cdots \underline{(n+2)^{\vee}} - \mu'' \otimes \underline{02} \cdots \underline{(n+2)^{\vee}} - \mu' \otimes \underline{013} \cdots \underline{(n+2)^{\vee}} \end{aligned} \right) \left\{ \begin{aligned}[t] & \underline{\mu} \otimes \underline{1} \cdots \underline{(n+2)^{\vee}} \\ & - \mu'' \otimes \underline{02} \cdots \underline{(n+2)^{\vee}} - \mu' \otimes \underline{013} \cdots \underline{(n+2)^{\vee}}, \psi \otimes \underline{0} \cdots \underline{(n+2)^{\vee}} \end{aligned} \right\}_{s,t} \\ & + \sum_{s+t=p} \psi \otimes \underline{0} \cdots \underline{(n+2)^{\vee}} \left\{ \begin{aligned}[t] & \underline{\mu} \otimes \underline{1} \cdots \underline{(n+2)^{\vee}} - \mu'' \otimes \underline{02} \cdots \underline{(n+2)^{\vee}} \\ & - \mu' \otimes \underline{013} \cdots \underline{(n+2)^{\vee}}, \psi \otimes \underline{0} \cdots \underline{(n+2)^{\vee}} \end{aligned} \right\}_{s,t} \end{aligned}$$

Since $n \ge 3$ and $p \ge 2$, we can apply Lemma 4.12 to obtain $\xi[\xi]_p = 0$. At the end, since $\mu, \mu', h \in Z_n(A^\tau)$, we have

$$d(\xi) + \sum_{p \ge 1} \xi[\xi]_p = (d(\psi) + \tau\langle\psi\rangle - (-1)^{n+1}\psi\langle\tau\rangle - \mu - \mu' + \mu'') \otimes \underline{0} \cdots \underline{(n+2)^{\vee}}.$$
If we set $\mu'' := \mu + \mu' \in Z_n(A^\tau)$ and $\psi := 0$, we then obtain that $\xi \in \mathcal{MC}(A \otimes \Sigma N^*(\Delta^{n+2}))$. We thus have proved that

$$[\omega] + [\omega'] = \left[ -\tau \otimes \left( \sum_{k=1}^{n+1} \underline{k}^{\vee} \right) - (\mu + \mu') \otimes \underline{0 \cdots (n+1)}^{\vee} \right],$$

which gives $\tilde{f}(\mu + \mu') = \tilde{f}(\mu) + \tilde{f}(\mu')$. Now, because $f$ is a bijection, we only need to prove that the kernel of $\tilde{f}$ is exactly given by $d_\tau(A_{n+1})$. Let $\mu \in Z_n(A^\tau)$ and $\psi \in A_{n+1}$. By the previous computations, we see that the equation

$$d(\psi) + \tau\langle\psi\rangle - (-1)^{n+1}\psi\langle\tau\rangle = \mu$$

is equivalent to the assumption

$$-\tau \otimes \left( \sum_{k=0}^{n+2} \underline{k}^{\vee} \right) - \mu \otimes \underline{1 \cdots (n+2)}^{\vee} + \psi \otimes \underline{0 \cdots (n+2)}^{\vee} \in \mathcal{MC}(A \otimes \Sigma N^*(\Delta^{n+2})),$$

which shows that $\tilde{f}(\mu) = 0$ if and only if $\mu = d_\tau(\psi)$ for some $\psi \in A_{n+1}$. We thus have an isomorphism

$$\bar{f} : H_n(A^\tau) \xrightarrow{\cong} \pi_{n+1}(\mathcal{MC}(A), \tau).$$

□

## 4.3 Remarks: interpretation of the low dimensional twisting coderivations

In this subsection, we give an interpretation of the differentials $\partial^0, \partial^1, \partial^2$ and $\partial^3$ computed in Lemma 3.6 and Corollary 3.9. This interpretation will be obtained by the study of the first simplices associated to the Maurer-Cartan simplicial set of $\text{Hom}(\Lambda^{-1}\mathcal{A}s^\vee, \text{End}_A)$ for some $A \in \text{dgMod}_k$.

Recall that for every non-symmetric cooperad $\mathcal{C}$ and non-symmetric operad $\mathcal{P}$ such that $\mathcal{C}(0) = \mathcal{P}(0) = 0$, the sequence $\text{Hom}(\mathcal{C}, \mathcal{P})$ is endowed with the structure of an operad such that, for every $f \in \text{Hom}(\mathcal{C}, \mathcal{P})(k)$, $g_1 \in \text{Hom}(\mathcal{C}, \mathcal{P})(i_1)$, $\dots$, $g_k \in \text{Hom}(\mathcal{C}, \mathcal{P})(i_k)$ with $n = i_1 + \dots + i_k$, the composition $\gamma(f \otimes g_1 \otimes \dots \otimes g_k)$ is given by the composite

$$\begin{array}{ccc}
\mathcal{C}(n) \xrightarrow{\Delta} \mathcal{C} \circ \mathcal{C}(n) & \longrightarrow & \mathcal{C}(k) \otimes \mathcal{C}(i_1) \otimes \dots \otimes \mathcal{C}(i_k) \\
\multicolumn{3}{c}{\searrow} \\
\mathcal{P}(k) \otimes \mathcal{P}(i_1) \otimes \dots \otimes \mathcal{P}(i_k) & \longleftarrow & \mathcal{P} \circ \mathcal{P}(n) \xrightarrow{\gamma} \mathcal{P}(n)
\end{array}$$

From [GV95, Proposition 1], we deduce that $\bigoplus_{n \ge 1} \text{Hom}(\mathcal{C}(n), \mathcal{P}(n))$ is endowed with the structure of a brace algebra. The braces are given by

$$f\langle g_1, \dots, g_n \rangle = \sum_{1 \le i_1 < \dots < i_n \le r} \gamma(f \otimes i_d \otimes \dots \otimes g_1 \otimes \dots \otimes g_n \otimes \dots \otimes i_d)$$
where $f \in \operatorname{Hom}(\mathcal{C}(r), \mathcal{P}(r))$, $g_1 \in \operatorname{Hom}(\mathcal{C}(m_1), \mathcal{P}(m_1))$, ..., $g_n \in \operatorname{Hom}(\mathcal{C}(m_n), \mathcal{P}(m_n))$. We immediately see that $\bigoplus_{n \ge 2} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))$ is a sub brace algebra of $\bigoplus_{n \ge 1} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))$. Since $\prod_{n \ge 2} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))$ is the completion of $\bigoplus_{n \ge 2} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))$ under the filtration defined by

$$F_p(\bigoplus_{n \ge 2} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))) := \bigoplus_{n \ge p+1} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n)),$$

we have that the above brace algebra structure on $\bigoplus_{n \ge 1} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))$ induces a complete brace algebra structure on $\prod_{n \ge 2} \operatorname{Hom}(\mathcal{C}(n), \mathcal{P}(n))$.

We now consider the non-symmetric operad $\mathcal{A}s$ such that $\mathcal{A}s(0) = 0$ and $\mathcal{A}s(n) = \mathbb{K}$ for every $n \ge 1$ with trivial operadic compositions. Since $\mathcal{A}s$ is self-dual for Koszul duality (see for instance [LV12, Proposition 9.1.9]), the operad $\mathcal{A}s_\infty = B^c(\Lambda^{-1}\mathcal{A}s^\vee)$ encodes associative algebras up to homotopy. We apply the above analysis with $\mathcal{C} = \Lambda^{-1}\mathcal{A}s^\vee$ and $\mathcal{P} = \operatorname{End}_A$ for some $A \in \operatorname{dgMod}_{\mathbb{K}}$ in order to study morphisms from $\mathcal{A}s_\infty$ to $\operatorname{End}_A$, or equivalently associative up to homotopy algebra structures on $A$. Note that we have an isomorphism of operads

$$\operatorname{Hom}(\Lambda^{-1}\mathcal{A}s^\vee, \operatorname{End}_A) \simeq \operatorname{End}_{\Sigma A}.$$

We set

$$\bar{B}(A) = \bigoplus_{n \ge 1} (\Sigma A)^{\otimes n} ; \quad B_{\ge 2}(A) = \bigoplus_{n \ge 2} (\Sigma A)^{\otimes n}$$

so that $\bar{B}(A) = \Sigma A \oplus B_{\ge 2}(A)$. Let $d$ be the differential of $\bar{B}(A)$ obtained from the internal differential of $A$ by the Leibniz rule. Recall that $\bar{B}(A)$ is a coalgebra with as coproduct

$$\Delta(a_1 \otimes \cdots \otimes a_n) = \sum_{k=1}^{n-1} (a_1 \otimes \cdots \otimes a_k) \otimes (a_{k+1} \otimes \cdots \otimes a_n)$$

for every $n \ge 2$ and $a_1, \dots, a_n \in \Sigma A$. The above isomorphism of operads provides a complete brace algebra structure on $\operatorname{Hom}(B_{\ge 2}(A), \Sigma A) \simeq \prod_{n \ge 2} \operatorname{End}_{\Sigma A}(n)$. Note that we have the isomorphism $\operatorname{Hom}(\bar{B}(A), \Sigma A) \simeq \operatorname{Hom}(\Sigma A, \Sigma A) \oplus \operatorname{Hom}(B_{\ge 2}(A), \Sigma A)$. In the following, we denote by $1 \in \operatorname{Hom}(\Sigma A, \Sigma A)$ the identity morphism so that we have a natural inclusion $\mathbb{K}1 \oplus \operatorname{Hom}(B_{\ge 2}(A), \Sigma A) \subset \operatorname{Hom}(\bar{B}(A), \Sigma A)$.

**Proposition 4.17.** *Giving a Maurer-Cartan element $\phi \in \mathcal{MC}(\operatorname{Hom}(B_{\ge 2}(A), \Sigma A))$ is equivalent to giving a coderivation of coalgebra of the form $d + \partial_\phi$ on $\bar{B}(A)$, where $\partial_\phi$ is the morphism obtained from $\phi$ by the Leibniz rule in the coalgebra $\bar{B}(A)$.*

*Proof.* Let $\phi \in \operatorname{Hom}(B_{\ge 2}(A), \Sigma A)$ be a degree $-1$ morphism. Then $(d + \partial_\phi)^2 = 0$ if and only if $d(\phi) + \phi \partial_\phi = 0$. By definition of $\partial_\phi$, we have, for every $a_1, \dots, a_n \in \Sigma A$,

$$\partial_\phi(a_1 \otimes \cdots \otimes a_n) = \sum_{i=1}^{n} \sum_{j=1}^{n-i} \pm a_1 \otimes \cdots \otimes a_{i-1} \otimes \phi(a_i \otimes \cdots \otimes a_{i+j}) \otimes a_{i+j+1} \otimes \cdots \otimes a_n,$$

which gives $\phi \partial_\phi = \phi \langle \phi \rangle$. We thus have obtained that $d + \partial_\phi$ is a derivation of coalgebra if and only if $\phi \in \mathcal{MC}(\operatorname{Hom}(\bar{B}(A), \Sigma A))$. $\square$

Since giving a morphism of operads $\mathcal{A}s_\infty \to \operatorname{End}_A$ is equivalent to giving a Maurer-Cartan element in $\operatorname{Hom}(B_{\ge 2}(A), \Sigma A)$, we have the following classical definition.
**Definition 4.18.** An associative algebra up to homotopy is a pair $(A, \phi)$ where $A$ is a dg $\mathbb{K}$-module and $\phi \in \mathcal{M}\mathcal{C}(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$.

For every $\phi \in \mathcal{M}\mathcal{C}(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$, we denote by $\bar{B}(A, \phi)$ the dg $\mathbb{K}$-module $\bar{B}(A)$ endowed with the coderivation $d + \partial_{\phi}$.

**Definition 4.19.** Let $(A_1, \phi_1)$ and $(A_2, \phi_2)$ be two associative algebras up to homotopy. An $\infty$-morphism $f : (A_1, \phi_1) \to (A_2, \phi_2)$ is a morphism of coalgebras $f : \bar{B}(A, \phi_1) \to \bar{B}(A, \phi_2)$ which commutes with the derivations.

In the following, we consider the category of associative algebras up to homotopy with set morphisms the $\infty$-morphisms.

**Remark 4.20.** Note that since $B(A_2)$ is cofree, giving a morphism of coalgebras $\bar{B}(A_1) \to \bar{B}(A_2)$ is equivalent to giving a morphism $\bar{B}(A_1) \to \Sigma A_2$.

**Proposition 4.21.** Let $\phi_0, \phi_1 \in \mathcal{M}\mathcal{C}(\mathrm{Hom}(\bar{B}(A), \Sigma A))$. Then giving

$$-\phi_0 \otimes \underline{0}^{\vee} - \phi_1 \otimes \underline{1}^{\vee} - \phi_{01} \otimes \underline{01}^{\vee} \in \mathcal{M}\mathcal{C}_1(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$$

is equivalent to giving a morphism of coalgebras

$$\Phi_{01} : \bar{B}(A, \phi_1) \to \bar{B}(A, \phi_0)$$

which is the identity on $\Sigma A \subset \bar{B}(A)$.

*Proof.* Let $\omega := -\phi_0 \otimes \underline{0}^{\vee} - \phi_1 \otimes \underline{1}^{\vee} - \phi_{01} \otimes \underline{01}^{\vee} \in \mathrm{Hom}(B_{\ge 2}(A), \Sigma A) \otimes \Sigma N^*(\Delta^1)$. Let $\Phi_{01} : \bar{B}(A) \to \bar{B}(A)$ be the unique morphism of coalgebras such that its composite with the projection $\pi_{\Sigma A} : \bar{B}(A) \to \Sigma A$ is $1 + \phi_{01}$. We characterize the equation

$$(d + \partial_{\phi_0})\Phi_{01} = \Phi_{01}(d + \partial_{\phi_1}).$$

Since $\bar{B}(A)$ is cofree, this identity is equivalent to

$$\pi_{\Sigma A}(d + \partial_{\phi_0})\Phi_{01} = \pi_{\Sigma A}\Phi_{01}(d + \partial_{\phi_1}),$$

and then to

$$d(\phi_{01}) = \phi_1 + \phi_{01}\partial_{\phi_1} - \phi_0\Phi_{01}.$$

We precisely have $\phi_{01}\partial_{\phi_1} = \phi_{01}\langle\phi_1\rangle$ and $\phi_0\Phi_{01} = \phi_0 \odot (1 + \phi_{01})$ by definition of $\partial_{\phi_1}$ and $\Phi_{01}$. We thus have obtained that $\Phi_{01}$ commutes with the differentials if and only if

$$d(\phi_{01}) = \phi_1 + \phi_{01}\langle\phi_1\rangle - \phi_0 \odot (1 + \phi_{01}).$$

By Lemma 4.7, this identity is equivalent to ask $\omega \in \mathcal{M}\mathcal{C}_1(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$, which proves the proposition. $\square$

We now characterize elements of $\mathcal{M}\mathcal{C}_2(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$. First, note that for every associative algebra $E$, every $\phi \in \mathcal{M}\mathcal{C}(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$ induces an element in $\mathcal{M}\mathcal{C}(\mathrm{Hom}(B_{\ge 2}(A \otimes E), \Sigma(A \otimes E)))$, which we still denote by $\phi$, and which is defined by applying $\phi$ on the left, and the algebra structure of $E$ on the right. In particular, for every morphism of associative algebras $f : E \to E'$,
we have an $\infty$-morphism $id \otimes f : (A \otimes E, \phi) \to (A \otimes E', \phi)$.

Next, recall that, for every $n \ge 0$, the dg $\mathbb{K}$-module $N^*(\Delta^n)$ is endowed with the structure of an associative algebra. This associative algebra structure is obtained by the dualization of the coassociative coalgebra structure on $N_*(\Delta^n)$ given by the Alexander-Whitney diagonal $AW : N_*(\Delta^n) \to N_*(\Delta^n) \otimes N_*(\Delta^n)$ which is the operation given by the permutation $(12) \in \mathcal{E}(2)_0$. Explicitly, we have

$$AW(\underline{a_0 \cdots a_d}) = \sum_{k=0}^{d} \underline{a_0 \cdots a_k} \otimes \underline{a_k \cdots a_d},$$

for every $0 \le a_0 < \cdots < a_d \le n$.

**Proposition 4.22.** *Giving a Maurer-Cartan element in $\mathcal{MC}_2(\operatorname{Hom}(B_{\ge 2}(A), \Sigma A))$ is equivalent to giving Maurer-Cartan elements $\phi_0, \phi_1, \phi_2 \in \mathcal{MC}(\operatorname{Hom}(B_{\ge 2}(A), \Sigma A))$ and a diagram of the form*

<figure><img src="image_18.png" /></figure>

in the category of $A_\infty$-algebras, where $\Phi_{012} : (A, \phi_2) \to (A \otimes N^*(\Delta^1), \phi_0)$ is a homotopy from $\Phi_{01}\Phi_{12}$ to $\Phi_{02}$.

*Proof.* We consider

$$\omega := -\phi_0 \otimes \underline{0}^\vee - \phi_1 \otimes \underline{1}^\vee - \phi_2 \otimes \underline{2}^\vee - \phi_{01} \otimes \underline{01}^\vee - \phi_{02} \otimes \underline{02}^\vee - \phi_{12} \otimes \underline{02}^\vee - \phi_{012} \otimes \underline{012}^\vee.$$

We characterize the Maurer-Cartan condition on $\omega$. By definition of the $\Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$-algebra structure on $\Sigma\operatorname{Hom}(B_{\ge 2}(A), \Sigma A) \otimes N^*(\Delta^2)$, and by Corollary 3.9, looking at the vertices of $d(\omega) + \sum_{n \ge 1} \omega[\langle\omega\rangle]_n$ gives the Maurer-Cartan condition on $\phi_0, \phi_1, \phi_2 \in \operatorname{Hom}(B_{\ge 2}(A), \Sigma A)$. Looking at the components given by $\underline{01}^\vee$, $\underline{02}^\vee$ and $\underline{12}^\vee$ also give the Maurer-Cartan condition on the elements

$$-\phi_0 \otimes \underline{0}^\vee - \phi_1 \otimes \underline{1}^\vee - \phi_{01} \otimes \underline{01}^\vee,$$

$$-\phi_0 \otimes \underline{0}^\vee - \phi_2 \otimes \underline{1}^\vee - \phi_{02} \otimes \underline{01}^\vee,$$

$$-\phi_1 \otimes \underline{0}^\vee - \phi_2 \otimes \underline{1}^\vee - \phi_{12} \otimes \underline{01}^\vee.$$

In particular, by Proposition 4.21, such datas are equivalent to giving three $\infty$-morphisms $\Phi_{01} : (A, \phi_1) \to (A, \phi_0)$, $\Phi_{02} : (A, \phi_2) \to (A, \phi_0)$ and $\Phi_{12} : (A, \phi_2) \to (A, \phi_1)$ which reduce to the identity on $\Sigma A$. We now analyze the $\underline{012}^\vee$ component of $d(\omega) + \sum_{n \ge 1} \omega[\langle\omega\rangle]_n$. By Corollary 3.9, the Maurer-Cartan condition on $\omega$ gives, when looking at the $\underline{012}^\vee$ component,

$$d(\phi_{012}) - \phi_{01}\bar{\phi}_{12} + \phi_{02} + \phi_{012}\langle\phi_2\rangle + \sum_{i,j \ge 0} \phi_0\langle\underbrace{\phi_{02}, \dots, \phi_{02}}_i, \phi_{012}, \underbrace{\phi_{01}\bar{\phi}_{12}, \dots, \phi_{01}\bar{\phi}_{12}}_j\rangle = 0.$$

Now let $\Phi_{012} : \bar{B}(A) \to \bar{B}(A \otimes N^*(\Delta^1))$ be the unique morphism of coalgebras such that its composite with the projection on $\Sigma A \otimes N^*(\Delta^1)$ is

$$(1 + \phi_{02}) \otimes \underline{0}^\vee + \phi_{012} \otimes \underline{01}^\vee + (1 + \phi_{01}\bar{\phi}_{12}) \otimes \underline{1}^\vee.$$
We characterize the equation

$$ \pi_{\Sigma A \otimes N^*(\Delta^1)}(d + \partial_{\phi_0})\Phi_{012} = \pi_{\Sigma A \otimes N^*(\Delta^1)}\Phi_{012}(d + \partial_{\phi_2}). $$

On one hand, we have

$$ \begin{aligned} \pi_{\Sigma A \otimes N^*(\Delta^1)}(d + \partial_{\phi_0})\Phi_{012} &= (d_{\Sigma A} + d\phi_{02} + \phi_0 \odot (1 + \phi_{02})) \otimes \underline{0}^{\vee} \\ &\quad + (d_{\Sigma A} + d\phi_{01}\overline{\odot}\phi_{12} + \phi_0 \odot (1 + \phi_{01}\overline{\odot}\phi_{12})) \otimes \underline{1}^{\vee} \\ &\quad + (d\phi_{012} - \phi_{01}\overline{\odot}\phi_{12} + \phi_{02} + \sum_{i,j \ge 0} \phi_0 \underbrace{\langle \phi_{02}, \dots, \phi_{02}, \phi_{012}, \phi_{01}\overline{\odot}\phi_{12}, \dots, \phi_{01}\overline{\odot}\phi_{12} \rangle}_{i,j}) \otimes \underline{01}^{\vee}. \end{aligned} $$

On the other hand, we have

$$ \begin{aligned} \pi_{\Sigma A \otimes N^*(\Delta^1)} \Phi_{012}(d + \partial_{\phi_2}) &= (d_{\Sigma A} + \phi_{02}d + \phi_2 + \phi_{02}\langle\phi_2\rangle) \otimes \underline{0}^{\vee} \\ &\quad + (d_{\Sigma A} + \phi_{01}\overline{\odot}\phi_{12}d + (\phi_{01}\overline{\odot}\phi_{12})\langle\phi_2\rangle) \otimes \underline{1}^{\vee} \\ &\quad - (\phi_{012}d + \phi_{012}\langle\phi_2\rangle) \otimes \underline{01}^{\vee}, \end{aligned} $$

which proves the proposition. $\square$

We now characterize $\mathcal{M}\mathcal{C}_3(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$. We first show how to compose homotopies from $(A, \phi)$ to $(A \otimes N^*(\Delta^1), \phi')$ for some Maurer-Cartan elements $\phi, \phi' \in \mathcal{M}\mathcal{C}(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$. Let $f, g, h : (A, \phi) \to (A, \phi')$. Let $H_1 : (A, \phi) \to (A \otimes N^*(\Delta^1), \phi')$ be a homotopy from $f$ to $g$, and $H_2 : (A, \phi) \to (A \otimes N^*(\Delta^1), \phi')$ be a homotopy from $g$ to $h$. We consider the pullback

<figure><img src="image_19.png" /></figure>

where we identify $N^*(\Delta^0)$ with $\mathbb{K}$. Explicitly, we have $N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1) = (N^*(\Delta^1) \times N^*(\Delta^1))/(\underline{1}^{\vee}, 0) \sim (0, \underline{0}^{\vee})$. One can see that the algebra structure of $N^*(\Delta^1) \times N^*(\Delta^1)$ preserves the equivalence relation $\sim$ so that $A \otimes (N^*(\Delta^1) \times N^*(\Delta^1))$ is a path object for $A$ in the category of $A_\infty$-algebras. We thus obtain a homotopy $H := H_2 \times H_1 : (A, \phi) \to (A \otimes (N^*(\Delta^1) \times N^*(\Delta^1)), \phi')$ from $f$ to $h$.

Now let $G_1, G_2 : (A, \phi) \to (A \otimes (N^*(\Delta^1) \times N^*(\Delta^1)), \phi')$ be two homotopies from $f$ to $h$ obtained as above. In the next proposition, we use a particular way to compose $G_1$ with $G_2$. This composition is defined as follows. Let $N^*_{\square}(\square^2) = N^*(\Delta^1) \otimes N^*(\Delta^1)$ and $N^*_{\square}(\partial\square^2) = N^*_{\square}(\square^2)/(\mathbb{K} \cdot \underline{01}^{\vee} \otimes \underline{01}^{\vee})$. We consider the morphisms of algebras $\Gamma_1, \Gamma_2 : N^*_{\square}(\partial\square^2) \to N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1)$ defined by
$$
\begin{array}{ccc}
\Gamma_1: N_{\square}^*(\partial\Box^2) & \to & N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1) \\
\mathbf{0}^{\vee} \otimes \mathbf{0}^{\vee} & \mapsto & (\mathbf{0}^{\vee}, 0) \\
\mathbf{0}^{\vee} \otimes \mathbf{1}^{\vee} & \mapsto & (\mathbf{1}^{\vee}, 0) \\
\mathbf{1}^{\vee} \otimes \mathbf{1}^{\vee} & \mapsto & (0, \mathbf{1}^{\vee}) \\
\mathbf{0}^{\vee} \otimes \mathbf{01}^{\vee} & \mapsto & (\mathbf{01}^{\vee}, 0) \\
\mathbf{01}^{\vee} \otimes \mathbf{1}^{\vee} & \mapsto & (0, \mathbf{01}^{\vee})
\end{array}
$$

and

$$
\begin{array}{ccc}
\Gamma_2: N_{\square}^*(\partial\Box^2) & \to & N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1) \\
\mathbf{0}^{\vee} \otimes \mathbf{0}^{\vee} & \mapsto & (\mathbf{0}^{\vee}, 0) \\
\mathbf{1}^{\vee} \otimes \mathbf{0}^{\vee} & \mapsto & (\mathbf{1}^{\vee}, 0) \\
\mathbf{1}^{\vee} \otimes \mathbf{1}^{\vee} & \mapsto & (0, \mathbf{1}^{\vee}) \\
\mathbf{01}^{\vee} \otimes \mathbf{0}^{\vee} & \mapsto & (\mathbf{01}^{\vee}, 0) \\
\mathbf{1}^{\vee} \otimes \mathbf{01}^{\vee} & \mapsto & (0, \mathbf{01}^{\vee})
\end{array}
$$

From a geometrical point of view, the morphism $\Gamma_1$ allows us to see the product $N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1)$ as the top left corner of $N_{\square}^*(\Box^2)$, while $\Gamma_2$ allows us to see it at the bottom right corner of $N_{\square}^*(\Box^2)$. In particular, one can check that $N_{\square}^*(\partial\Box^2)$ is the pullback of the diagram

$$
\begin{array}{ccc}
N_{\square}^{*}(\partial\Box^2) \xrightarrow{\Gamma_1} & N^{*}(\Delta^1) \times_{\mathbb{K}} N^{*}(\Delta^1) & \\
\Gamma_2 \xrightarrow{\substack{| \\ \searrow}} & & \\
N^{*}(\Delta^1) \times_{\mathbb{K}} N^{*}(\Delta^1) \longrightarrow & \mathbb{K} \cdot (\mathbf{0}^{\vee}, 0) \oplus \mathbb{K}(0, \mathbf{1}^{\vee}) &
\end{array}
$$

Since $G_1$ and $G_2$ are homotopies from $f$ to $h$, their projection on $\Sigma A \otimes \mathbb{K} \cdot (\mathbf{0}^{\vee}, 0)$ (respectively $\Sigma A \otimes \mathbb{K} \cdot (0, \mathbf{1}^{\vee})$) agree and are given by $h$ (respectively $f$). Therefore, the morphisms $G_1$ and $G_2$ induce an $\infty$-morphism $G_1 \square G_2 : (A, \phi) \to (A \otimes N_{\square}^*(\partial\Box^2), \phi')$ given by the following pullback square diagram:

$$
\begin{tikzcd}[column sep=2.8em, row sep=2.8em]
    (A, \phi) \arrow[r, "G_1 \square G_2"] & (A \otimes N_{\square}^*(\partial\Box^2), \phi') \arrow[r, "id \otimes \Gamma_1"] & (A \otimes (N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1))), \phi') \arrow[r, "id \otimes \Gamma_2"] & (A \otimes (N^*(\Delta^1) \times N^*(\Delta^1))), \phi') \arrow[r, "\dashv"] & (A \otimes (\mathbb{K} \cdot (\mathbf{0}^{\vee}, 0) \oplus \mathbb{K} \cdot (0, \mathbf{1}^{\vee})), \phi') \\
    (A, \phi) \arrow[r, "G_2"] & (A \otimes N_{\square}^*(\partial\Box^2), \phi') \arrow[r, "id \otimes \Gamma_1"] & (A \otimes (N^*(\Delta^1) \times_{\mathbb{K}} N^*(\Delta^1))), \phi') \arrow[r, "id \otimes \Gamma_2"] & (A \otimes (N^*(\Delta^1) \times N^*(\Delta^1))), \phi') \arrow[r, "\dashv"] & (A \otimes (\mathbb{K} \cdot (\mathbf{0}^{\vee}, 0) \oplus \mathbb{K} \cdot (0, \mathbf{1}^{\vee})), \phi')
\end{tikzcd}
$$

**Proposition 4.23.** *Giving a Maurer-Cartan element in $\operatorname{Hom}(B_{\ge 2}(A), \Sigma A) \otimes \Sigma N^*(\Delta^3)$ is equivalent to giving $\phi_0, \phi_1, \phi_2, \phi_3 \in \mathcal{MC}(\operatorname{Hom}(B_{\ge 2}(A), \Sigma A))$, two homotopy diagrams of the form*

$$
\begin{tikzcd}[column sep=2.8em, row sep=2.8em]
    (A, \phi_2) \arrow[r, "\Phi_{12}"] & (A, \phi_1) \arrow[d, "\Phi_{123}"] \\
    (A, \phi_3) \arrow[r, "\Phi_{03}"] & (A, \phi_0)
\end{tikzcd}
$$

$$
\begin{tikzcd}[column sep=2.8em, row sep=2.8em]
    (A, \phi_2) \arrow[r, "\Phi_{12}"] & (A, \phi_1) \arrow[d, "\Phi_{012}"] \\
    (A, \phi_3) \arrow[r, "\Phi_{023}"] & (A, \phi_0)
\end{tikzcd}
$$

and a lifting diagram
<figure><img src="image_20.png" /></figure>

where we denote by $H_1, H_2 : (A, \phi_3) \to (A \otimes (N^*(\Delta^1) \times_K N^*(\Delta^1)), \phi_0)$ the homotopies from $\Phi_{01}\Phi_{12}\Phi_{23}$ to $\Phi_{03}$ given by the homotopy diagrams.

*Proof.* Let

$$ \omega := -\phi_0 \otimes \underline{0}^{\vee} - \phi_1 \otimes \underline{1}^{\vee} - \phi_2 \otimes \underline{2}^{\vee} - \phi_3 \otimes \underline{3}^{\vee} \\ - \phi_{01} \otimes \underline{01}^{\vee} - \phi_{02} \otimes \underline{02}^{\vee} - \phi_{12} \otimes \underline{12}^{\vee} - \phi_{03} \otimes \underline{03}^{\vee} - \phi_{13} \otimes \underline{13}^{\vee} - \phi_{23} \otimes \underline{23}^{\vee} \\ - \phi_{012} \otimes \underline{012}^{\vee} - \phi_{013} \otimes \underline{013}^{\vee} - \phi_{023} \otimes \underline{023}^{\vee} - \phi_{123} \otimes \underline{123}^{\vee} - \phi_{0123} \otimes \underline{0123}^{\vee} $$

be an element of $\mathrm{Hom}(B_{\ge 2}(A), \Sigma A) \otimes \Sigma N^*(\Delta^3)$. By Proposition 4.22, the Maurer-Cartan condition on the four faces of $\omega$ is precisely equivalent to giving the first two diagrams given in the assertion of the proposition, since the $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra structure of $\mathrm{Hom}(B_{\ge 2}(A), \Sigma A) \otimes N^*(\Delta^3)$ is compatible with the simplicial structures. From now on, we suppose that $d_0\omega, d_1\omega, d_2\omega$ and $d_3\omega$ are elements of $\mathcal{MC}_2(\mathrm{Hom}(B_{\ge 2}(A), \Sigma A))$. Then the only possibly non-zero component of $d(\omega) + \sum_{n \ge 1} \omega[\omega]^n$ is the $\underline{0123}^{\vee}$ component. By Corollary 3.9, this component is 0 if and only if we have the identity

$$ \begin{aligned} d(\phi_{0123}) + \phi_{023} - \phi_{123} + \phi_{013} - \phi_{012} - \phi_{0123} \langle \phi_3 \rangle &+ \sum_{k \ge 1} \phi_{012} \langle \underbrace{\phi_{23}, \dots, \phi_{23}}_k \rangle - \sum_{i,j \ge 0} \phi_{01} \langle \underbrace{\phi_{13}, \dots, \phi_{13}}_i, \phi_{123}, \underbrace{\phi_{12} \bar{\otimes} \phi_{23}, \dots, \phi_{12} \bar{\otimes} \phi_{23}}_j \rangle \\ &+ \sum_{i,j,k \ge 0} \phi_0 \langle \underbrace{\phi_{03}, \dots, \phi_{03}}_i, \phi_{023}, \underbrace{\phi_{02} \bar{\otimes} \phi_{23}}_j, \phi_{012} \bar{\otimes} (1 + \phi_{23}), \underbrace{\phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}, \dots, \phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}}_m \rangle \\ &+ \sum_{i,j,m \ge 0} \phi_0 \langle \underbrace{\phi_{03}, \dots, \phi_{03}}_i, \phi_{013}, \underbrace{\phi_{01} \bar{\otimes} \phi_{13}, \dots, \phi_{01} \bar{\otimes} \phi_{13}}_j, \phi_{123}, \underbrace{\phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}, \dots, \phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}}_k \rangle \\ &+ \sum_{i,j,k,l,m \ge 0} \phi_0 \langle \underbrace{\phi_{03}, \dots, \phi_{03}}_i, \phi_{013}, \underbrace{\phi_{01} \bar{\otimes} \phi_{13}, \dots, \phi_{01} \bar{\otimes} \phi_{13}}_j, \phi_{123}, \underbrace{\phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}, \dots, \phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}}_m \rangle \\ &+ \sum_{i,j \ge 0} \phi_0 \langle \underbrace{\phi_{03}, \dots, \phi_{03}}_i, \phi_{0123}, \underbrace{\phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}, \dots, \phi_{01} \bar{\otimes} \phi_{12} \bar{\otimes} \phi_{23}}_j \rangle = 0. \end{aligned} $$

We let $\Phi_{0123} : \bar{B}(A) \to \bar{B}(A \otimes N^*_{\square}(\square^2))$ to be the morphism of coalgebras whose composite with
the projection on $\Sigma A \otimes N_{\square}^*(\square^2)$ is

$$ (1 + \phi_{03}) \otimes \underline{0}^{\vee} \otimes \underline{0}^{\vee} \\ + (1 + \phi_{01}) \odot (1 + \phi_{13}) \otimes \underline{0}^{\vee} \otimes \underline{1}^{\vee} \\ + (1 + \phi_{02}) \odot (1 + \phi_{23}) \otimes \underline{1}^{\vee} \otimes \underline{0}^{\vee} \\ (1 + \phi_{01}) \odot (1 + \phi_{12}) \odot (1 + \phi_{23}) \otimes \underline{1}^{\vee} \otimes \underline{1}^{\vee} \\ + \phi_{023} \otimes \underline{01}^{\vee} \otimes \underline{0}^{\vee} \\ + \phi_{013} \otimes \underline{0}^{\vee} \otimes \underline{01}^{\vee} \\ + \phi_{012} \odot (1 + \phi_{23}) \otimes \underline{1}^{\vee} \otimes \underline{01}^{\vee} \\ + \left( \phi_{123} + \sum_{i,j \ge 0} \phi_{01} \langle \underbrace{\phi_{12}, \dots, \phi_{12}}_i, \phi_{123}, \underbrace{\phi_{12} \overline{\circ} \phi_{23}, \dots, \phi_{12} \overline{\circ} \phi_{23}}_j \right) \otimes \underline{01}^{\vee} \otimes \underline{1}^{\vee} \\ + \phi_{0123} \otimes \underline{01}^{\vee} \otimes \underline{01}^{\vee}. $$

We check that $\omega$ is a Maurer-Cartan element if and only if $\Phi_{0123}$ commutes with the differentials. The latter condition is expressed by the identity

$$ \pi_{\Sigma A \otimes N_{\square}^*(\square^2)} \Phi_{0123}(d + \partial_{\phi_3}) - \pi_{\Sigma A \otimes N_{\square}^*(\square^2)}(d + \partial_{\phi_0}) \Phi_{0123} = 0. $$

Since the morphisms $\Phi_{03}, \Phi_{01}\Phi_{13}, \Phi_{02}\Phi_{23}$ and $\Phi_{01}\Phi_{12}\Phi_{23}$ commute with the differentials, the components given by $\underline{0}^{\vee} \otimes \underline{0}^{\vee}, \underline{0}^{\vee} \otimes \underline{1}^{\vee}, \underline{1}^{\vee} \otimes \underline{0}^{\vee}$ and $\underline{1}^{\vee} \otimes \underline{1}^{\vee}$ are indeed 0. Since $\Phi_{023}$ and $\Phi_{013}$ commute with the differentials, the components given by $\underline{01}^{\vee} \otimes \underline{0}^{\vee}$ and $\underline{0}^{\vee} \otimes \underline{01}^{\vee}$ are also 0. We now look at the component given by $\underline{1}^{\vee} \otimes \underline{01}^{\vee}$. Since the algebra structure of $N_{\square}^*(\square^2)$ is compatible with its underlying simplicial structure, it is equivalent to check that the element

$$ (1 + \phi_{02} \overline{\circ} \phi_{23}) \otimes \underline{0}^{\vee} + \phi_{012} \odot (1 + \phi_{23}) \otimes \underline{01}^{\vee} + (1 + \phi_{01} \overline{\circ} \phi_{12} \overline{\circ} \phi_{23}) \otimes \underline{1}^{\vee} $$

is a Maurer-Cartan element in $\operatorname{Hom}(B_{\ge 2}(A), A) \otimes N^*(\Delta^1)$. From Proposition 4.22, one can see that it is equivalent to check that the composite $\Phi_{012}\Phi_{23} : (\overline{B}(A), \phi_3) \to (\overline{B}(A \otimes N^*(\Delta^1), \phi_0)$ commutes with the differentials, which is the case since $\Phi_{012}$ and $\Phi_{23}$ commute with the differentials. Analogously, we $\underline{01}^{\vee} \otimes \underline{1}^{\vee}$ is also 0, since the composite $\Phi_{01}\Phi_{123}$ commutes with the differentials.

We now look at the $\underline{01}^{\vee} \otimes \underline{01}^{\vee}$ component. The composite $\pi_{\Sigma A \otimes N_{\square}^*(\square^2)} \Phi_{0123}(d + \partial_{\phi_3})$ gives

$$ (\phi_{0123} d + \phi_{0123} \langle \phi_3 \rangle) \otimes \underline{01}^{\vee} \otimes \underline{01}^{\vee} $$

as $\underline{01}^{\vee} \otimes \underline{01}^{\vee}$ component. We now compute the $\underline{01}^{\vee} \otimes \underline{01}^{\vee}$ component given by the composite $\pi_{\Sigma A \otimes N_{\square}^*(\square^2)}(d + \partial_{\phi_0})\Phi_{0123}$. Computing $\pi_{\Sigma A \otimes N_{\square}^*(\square^2)}d\Phi_{0123}$ gives the terms

$$ (d\phi_{0123} - \phi_{013} + \phi_{023} + \phi_{012} \odot (1 + \phi_{23}) \\ - \phi_{123} - \sum_{i,j \ge 0} \phi_{01} \langle \underbrace{\phi_{12}, \dots, \phi_{12}}_i, \phi_{123}, \underbrace{\phi_{12} \overline{\circ} \phi_{23}, \dots, \phi_{12} \overline{\circ} \phi_{23}}_j \rangle) \otimes \underline{01}^{\vee} \otimes \underline{01}^{\vee}. $$

We now compute $\pi_{\Sigma A \otimes N_{\square}^*(\square^2)}\partial_{\phi_0}\Phi_{0123}$. Note that the only way to write $\underline{01}^{\vee} \otimes \underline{01}^{\vee}$ as a product in $N^*(\Delta^1) \otimes N^*(\Delta^1)$ are given by one of the three following products:

$$ (\underline{0}^{\vee} \otimes \underline{0}^{\vee}) \cdot \cdot \cdot (\underline{0}^{\vee} \otimes \underline{0}^{\vee}) \cdot (\underline{01}^{\vee} \otimes \underline{0}^{\vee}) \cdot (\underline{1}^{\vee} \otimes \underline{0}^{\vee}) \cdot \cdot \cdot (\underline{1}^{\vee} \otimes \underline{01}^{\vee}) \cdot (\underline{1}^{\vee} \otimes \underline{1}^{\vee}) \cdot \cdot \cdot (\underline{1}^{\vee} \otimes \underline{1}^{\vee}); $$
$$ (\underline{0}^{\vee} \otimes \underline{0}^{\vee}) \cdot^{i} \cdot (\underline{0}^{\vee} \otimes \underline{0}^{\vee}) \cdot (\underline{0}^{\vee} \otimes \underline{01}^{\vee}) \cdot (\underline{0}^{\vee} \otimes \underline{1}^{\vee}) \cdot^{j} \cdot (\underline{0}^{\vee} \otimes \underline{1}^{\vee}) \cdot (\underline{01}^{\vee} \otimes \underline{1}^{\vee}) \cdot (\underline{1}^{\vee} \otimes \underline{1}^{\vee}) \cdot^{k} \cdot (\underline{1}^{\vee} \otimes \underline{1}^{\vee}); \\ (\underline{0}^{\vee} \otimes \underline{0}^{\vee}) \cdot^{i} \cdot (\underline{0}^{\vee} \otimes \underline{0}^{\vee}) \cdot (\underline{01}^{\vee} \otimes \underline{01}^{\vee}) \cdot (\underline{1}^{\vee} \otimes \underline{1}^{\vee}) \cdot^{j} \cdot (\underline{1}^{\vee} \otimes \underline{1}^{\vee}). $$

for every $i, j, k \ge 0$. These type of products give respectively

$$ - \sum_{i,j,k \ge 0} \underbrace{\phi_0 \langle \phi_{03}, \dots, \phi_{03}, \phi_{023}, \phi_{02} \overline{\otimes} \phi_{23}, \dots, \phi_{02} \overline{\otimes} \phi_{23}, \phi_{012} \otimes (1 + \phi_{23}),}_{j} \underbrace{\phi_{01} \overline{\otimes} \phi_{12} \overline{\otimes} \phi_{23}, \dots, \phi_{01} \overline{\otimes} \phi_{12} \overline{\otimes} \phi_{23}}_{k} \otimes \underline{01}^{\vee} \otimes \underline{01}^{\vee}; $$

$$ \sum_{i,j,k,l,m \ge 0} \underbrace{\phi_0 \langle \phi_{03}, \dots, \phi_{03}, \phi_{013}, \phi_{01} \overline{\otimes} \phi_{13}, \dots, \phi_{01} \overline{\otimes} \phi_{13},}_{j} \underbrace{\phi_{123} + \phi_{01} \langle \phi_{12}, \dots, \phi_{12}, \phi_{123}, \phi_{12} \overline{\otimes} \phi_{23}, \dots, \phi_{12} \overline{\otimes} \phi_{23} \rangle,}_{l} \underbrace{\phi_{01} \overline{\otimes} \phi_{12} \overline{\otimes} \phi_{23}, \dots, \phi_{01} \overline{\otimes} \phi_{12} \overline{\otimes} \phi_{23}}_{m} \otimes \underline{01}^{\vee} \otimes \underline{01}^{\vee}; $$

$$ \sum_{i,j \ge 0} \phi_0 \langle \phi_{03}, \dots, \phi_{03}, \phi_{0123}, \phi_{01} \overline{\otimes} \phi_{12} \overline{\otimes} \phi_{23}, \dots, \phi_{01} \overline{\otimes} \phi_{12} \overline{\otimes} \phi_{23} \rangle \otimes \underline{01}^{\vee} \otimes \underline{01}^{\vee}, $$

as $\underline{01}^{\vee} \otimes \underline{01}^{\vee}$. We thus have obtained that $\Phi_{0123}$ commutes with the differentials if and only if $\omega$ is a Maurer-Cartan element, which proves the lemma. $\square$

## 4.4 A Goldman-Millson theorem

Our next goal is to prove an extension of the classical Goldman-Millson theorem for Lie-algebras (see [GM88, §2.4]). The proof of our analogue will be adapted from the proof given in [MR23, §6] in the setting of associative algebras up to homotopy.

We first prove that the category $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$ admits finite products.

**Lemma 4.24.** Let $(V_1, Q_{V_1})$, $(V_2, Q_{V_2}) \in \Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$. Then there exists a $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra structure on $V_1 \times V_2$ such that the morphisms $\pi_{V_1}: V_1 \times V_2 \to V_1$ and $\pi_{V_2}: V_1 \times V_2 \to V_2$ are strict morphisms of $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebras.

Moreover, for every $\phi_1: W \rightsquigarrow V_1$ and $\phi_2: W \rightsquigarrow V_2$, there exists a unique $\infty$-morphism denoted by $\phi_1 \times \phi_2: W \rightsquigarrow V_1 \times V_2$ such that $\pi_{V_1}(\phi_1 \times \phi_2) = \phi_1$ and $\pi_{V_2}(\phi_1 \times \phi_2) = \phi_2$.

*Proof.* We let $Q_{V_1 \times V_2}$ be the coderivation produced by the morphism

$$ Q_{V_1 \times V_2}^0 : \Gamma \mathrm{Perm}^c(V_1 \times V_2) \longrightarrow \Gamma \mathrm{Perm}^c(V_1) \times \Gamma \mathrm{Perm}^c(V_2) \xrightarrow{Q_{V_1}^0 \times Q_{V_2}^0} V_1 \times V_2. $$

Recall that the coderivation $Q_{V_1 \times V_2}$ is obtained from $Q_{V_1 \times V_2}^0$ by $Q_{V_1 \times V_2} = \tilde{\Psi}_1(Q_{V_1 \times V_2}^0) + \tilde{\Psi}_2(Q_{V_1 \times V_2}^0)$ (see the proof of Proposition 2.15). We check that $Q_{V_1 \times V_2} Q_{V_1 \times V_2} = 0$. By definition of $\tilde{\Psi}_1$, we
have the following commutative diagram:

$$
\begin{tikzcd}
(V_1 \times V_2) \otimes \Gamma(V_1 \times V_2) \arrow[r, "\tilde{\Psi}_1(Q_{V_1 \times V_2}^0)"] \arrow[d] & (V_1 \times V_2) \otimes \Gamma(V_1 \times V_2) \arrow[r, "Q_{V_1 \times V_2}^0"] \arrow[d] & V_1 \times V_2 \arrow[d, "\uparrow Q_{V_1}^0 \times Q_{V_2}^0"'] \\
(V_1 \otimes \Gamma(V_1)) \times (V_2 \otimes \Gamma(V_2)) \arrow[r, "\tilde{\Psi}_1(Q_{V_1}^0) \times \tilde{\Psi}_1(Q_{V_2}^0)"] & (V_1 \otimes \Gamma(V_1)) \times (V_2 \otimes \Gamma(V_2)) &
\end{tikzcd}
$$

We also have the commutative diagram, by definition of $\tilde{\Psi}_2$:

$$
\begin{tikzcd}
(V_1 \times V_2) \otimes \Gamma(V_1 \times V_2) \arrow[r, "\tilde{\Psi}_2(Q_{V_1 \times V_2}^0)"] \arrow[d] & (V_1 \times V_2) \otimes \Gamma(V_1 \times V_2) \arrow[r, "Q_{V_1 \times V_2}^0"] \arrow[d] & V_1 \times V_2 \arrow[d, "\uparrow Q_{V_1}^0 \times Q_{V_2}^0"'] \\
(V_1 \otimes \Gamma(V_1)) \times (V_2 \otimes \Gamma(V_2)) \arrow[r, "\tilde{\Psi}_2(Q_{V_1}^0) \times \tilde{\Psi}_2(Q_{V_2}^0)"] & (V_1 \otimes \Gamma(V_1)) \times (V_2 \otimes \Gamma(V_2)) &
\end{tikzcd}
$$

Finally, we have proved that $Q_{V_1 \times V_2}^0 \tilde{\Psi}(Q_{V_1 \times V_2}^0)$ fits in the following commutative diagram

$$
\begin{tikzcd}
(V_1 \times V_2) \otimes \Gamma(V_1 \times V_2) \arrow[r, "\tilde{\Psi}(Q_{V_1 \times V_2}^0)"] \arrow[d] & (V_1 \times V_2) \otimes \Gamma(V_1 \times V_2) \arrow[r, "Q_{V_1 \times V_2}^0"] \arrow[d] & V_1 \times V_2 \arrow[d, "\uparrow Q_{V_1}^0 \times Q_{V_2}^0"'] \\
(V_1 \otimes \Gamma(V_1)) \times (V_2 \otimes \Gamma(V_2)) \arrow[r, "\tilde{\Psi}(Q_{V_1}^0) \times \tilde{\Psi}(Q_{V_2}^0)"] & (V_1 \otimes \Gamma(V_1)) \times (V_2 \otimes \Gamma(V_2)) &
\end{tikzcd}
$$

which proves that $(V_1 \times V_2, Q_{V_1 \times V_2}) \in \Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$. Now let $\phi_1 : W \rightsquigarrow V_1$ and $\phi_2 : W \rightsquigarrow V_2$ be two $\infty$-morphisms. We define $\phi : W \rightsquigarrow V_1 \times V_2$ by $\phi^0 = \phi_1^0 \times \phi_2^0$. We prove that this gives an $\infty$-morphism i.e. $\phi^0 Q = (Q_{V_1}^0 \times Q_{V_2}^0) \phi$. This can be proven with the following commutative diagram:

$$
\begin{tikzcd}
\Gamma\operatorname{Perm}^c(W) \arrow[r, "Q"] & \Gamma\operatorname{Perm}^c(W) \arrow[d, "\phi"] \\
\Gamma\operatorname{Perm}^c(V_1 \times V_2) \arrow[r, "\pi_{V_1} \times \pi_{V_2}"] & \Gamma\operatorname{Perm}^c(V_1) \times \Gamma\operatorname{Perm}^c(V_2) \arrow[r, "Q_{V_1}^0 \times Q_{V_2}^0"] \arrow[d, "\phi_1^0 \times \phi_2^0"] & V_1 \times V_2 \arrow[d, "Q_{V_1 \times V_2}^0"]
\end{tikzcd}
$$

The identities $\pi_{V_1}(\phi_1 \times \phi_2) = \phi_1$ and $\pi_{V_2}(\phi_1 \times \phi_2) = \phi_2$ follow by immediate computations. $\square$

**Remark 4.25.** By an immediate check, the above definitions extend to the category $\Gamma\Lambda\tilde{\mathcal{P}}\mathcal{L}_\infty$ of complete $\Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$-algebras. Explicitly, if $(V_1, Q_{V_1})$ and $(V_2, Q_{V_2})$ are complete with respect to some filtrations, then $(V_1 \times V_2, Q_{V_1 \times V_2})$ is also complete with the filtration

$$F_n(V_1 \times V_2) = F_n V_1 \times F_n V_2.$$

In this setting, we deduce immediately from the definition of $Q_{V_1 \times V_2}^0$ that we have a bijection

$$\mathcal{M}\mathcal{C}(V_1 \times V_2) \simeq \mathcal{M}\mathcal{C}(V_1) \times \mathcal{M}\mathcal{C}(V_2).$$
We give an analogue of [MR23, Proposition 5.2].

**Lemma 4.26.** Let $A, B \in \operatorname{dg\widehat{Mod}}_{\mathbb{K}}$ be such that $A \otimes \Sigma N^*(\Delta^\bullet)$ and $B \otimes \Sigma N^*(\Delta^\bullet)$ are endowed with the structure of simplicial $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras. Let $\Theta : A \to B$ be a morphism in $\operatorname{dg\widehat{Mod}}_{\mathbb{K}}$ such that $\Theta \otimes \mathrm{id}$ is a strict morphism of $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras for every $n \ge 0$. Suppose that $\Theta$ is an acyclic fibration of $\operatorname{dg\widehat{K}}$-modules.

Then the map

$$ \Theta \otimes \mathrm{id} : \mathcal{M}\mathcal{C}(A \otimes \Sigma N^*(\Delta^\bullet)) \to \mathcal{M}\mathcal{C}(B \otimes \Sigma N^*(\Delta^\bullet)) $$

is a weak equivalence of simplicial sets.

*Proof.* Since the two simplicial sets $\mathcal{M}\mathcal{C}(A \otimes \Sigma N^*(\Delta^\bullet))$ and $\mathcal{M}\mathcal{C}(B \otimes \Sigma N^*(\Delta^\bullet))$ are Kan complexes by Theorem 4.5, it suffices to show that $\Theta \otimes \mathrm{id}$ induces a bijection on the sets of connected components and an isomorphism on every homotopy groups. Let $\tau : B \to A$ and $h : A \to A$ be such that

$$ \Theta\tau = \mathrm{id}; \mathrm{id} - \tau\Theta = dh + hd. $$

We endow $\mathrm{Ker}(\Theta)$ with the brace algebra structure defined by $a\langle b_1, \dots, b_r \rangle = 0$ for every $r \ge 1$ and $a, b_1, \dots, b_r \in \mathrm{Ker}(\Theta)$. Our first goal is to define a morphism $\Psi_\bullet : A \otimes \Sigma N^*(\Delta^\bullet) \rightsquigarrow \mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^\bullet)$ of simplicial $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras. For every $n \ge 0$, let $(\Psi_n)_0^0 = (id - \tau\Theta) \otimes \mathrm{id}$. We set, for every $k \ne 0$,

$$ (\Psi_n)_k^0 = (\Psi_n)_0^0(h \otimes \mathrm{id})Q_k^0, $$

where we denote by $Q$ the coderivation on $\Gamma\mathrm{Perm}^c(A \otimes \Sigma N^*(\Delta^n))$ given by the $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebra structure on $A \otimes \Sigma N^*(\Delta^n)$. We check that $\Psi_n : A \otimes \Sigma N^*(\Delta^n) \rightsquigarrow \mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^n)$:

$$
\begin{align*}
d(\Psi_n)_k^0 &= d(\Psi_n)_0^0(h \otimes \mathrm{id})Q_k^0 \\
&= (\Psi_n)_0^0 d(h \otimes \mathrm{id})Q_k^0 \\
&= (\Psi_n)_0^0 Q_k^0 - (\Psi_n)_0^0 (h \otimes \mathrm{id})dQ_k^0 \\
&= (\Psi_n)_0^0 Q_k^0 + \sum_{i=1}^k (\Psi_n)_0^0 (h \otimes \mathrm{id})Q_i^0 Q_k^i \\
&= (\Psi_n)_0^0 Q_k^0 + \sum_{i=1}^k (\Psi_n)_i^0 Q_k^i \\
&= \sum_{i=0}^k (\Psi_n)_i^0 Q_k^i,
\end{align*}
$$

which proves that $\Psi_n : A \otimes \Sigma N^*(\Delta^n) \rightsquigarrow \mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^n)$. Since $\Psi_\bullet$ is defined in terms of morphisms which are compatible with the simplicial structure, the morphism $\Psi_\bullet$ is a morphism of simplicial $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras. Consider now, for every $n \ge 0$, the $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebra structure on $(B \times \mathrm{Ker}(\Theta)) \otimes \Sigma N^*(\Delta^\bullet)$ given by the isomorphism

$$ (B \times \mathrm{Ker}(\Theta)) \otimes \Sigma N^*(\Delta^\bullet) \simeq (B \otimes \Sigma N^*(\Delta^\bullet)) \times (\mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^\bullet)). $$

Let $g_n = (\Theta \otimes \mathrm{id}) \times \Psi_n : A \otimes \Sigma N^*(\Delta^n) \rightsquigarrow (B \times \mathrm{Ker}(\Theta)) \otimes \Sigma N^*(\Delta^n)$. Then $g_\bullet$ is a morphism of simplicial $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras. We also have that $\mathcal{M}\mathcal{C}(g_\bullet)$ is an isomorphism of simplicial sets. Indeed, we have that $(g_\bullet)_0^0$ is an isomorphism of simplicial sets, with as inverse $(b, k) \otimes \underline{x} \mapsto (\tau(b) + k) \otimes \underline{x}$,
for every $b \in B$, $k \in \mathrm{Ker}(\Theta)$ and $\underline{x} \in \Sigma N^*(\Delta^\bullet)$. By looking at the Maurer-Cartan spaces degree wise, we obtain the following commutative diagram:

$$
\begin{tikzcd}
\mathcal{M}\mathcal{C}(A \otimes \Sigma N^*(\Delta^\bullet)) \arrow[r, "\mathcal{M}\mathcal{C}(g_\bullet)"'] & \mathcal{M}\mathcal{C}((B \times \mathrm{Ker}(\Theta)) \otimes \Sigma N^*(\Delta^\bullet)) \arrow[d, "\mathcal{M}\mathcal{C}(\Theta \otimes \mathrm{id})"] \\
\mathcal{M}\mathcal{C}(B \otimes \Sigma N^*(\Delta^\bullet)) \arrow[u, "\mathcal{M}\mathcal{C}(\Theta \otimes \Delta^\bullet)"'] & \mathcal{M}\mathcal{C}(B \otimes \Sigma N^*(\Delta^\bullet)) \times \mathcal{M}\mathcal{C}(\mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^\bullet)) \arrow[u, "\cong"']
\end{tikzcd}
$$

It is then sufficient to prove that the projection $\mathcal{M}\mathcal{C}(B \otimes \Sigma N^*(\Delta^\bullet)) \times \mathcal{M}\mathcal{C}(\mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^\bullet)) \to \mathcal{M}\mathcal{C}(B \otimes \Sigma N^*(\Delta^\bullet))$ is a weak equivalence of simplicial sets, which is true because $\mathcal{M}\mathcal{C}(\mathrm{Ker}(\Theta) \otimes \Sigma N^*(\Delta^\bullet)) = \mathcal{M}\mathcal{C}_\bullet(\mathrm{Ker}(\Theta))$, and this simplicial set has trivial $\pi_0$ and homotopy groups according to the computations of the connected components and the homotopy groups made in §4.2. We then have the result. $\square$

The next lemma is an analogue of [MR23, Proposition 5.5].

**Lemma 4.27.** Let $A, B, C$ be $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras. Let $\Theta : A \rightsquigarrow C$ and $\Phi : B \rightsquigarrow C$ be two $\infty$-morphisms of $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebras. We suppose that $\Phi$ is strict, and that $\Phi_0^0$ is surjective. Then there exists a $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebra structure on $A \times \mathrm{Ker}(\Phi_0^0)$ and $H : A \times \mathrm{Ker}(\Phi_0^0) \rightsquigarrow A \times B$ such that the following diagram is a pullback square diagram in $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$:

$$
\begin{tikzcd}
A \times \mathrm{Ker}(\Phi_0^0) \arrow[r, "\pi_B H"] & B \arrow[d, "\pi_A"] \\
\phantom{A} \arrow[u, "\Phi"] & \phantom{B} C \arrow[u, "\pi_A H"] \\
A \arrow[r, "\pi_A"] & B \arrow[u, "\pi_A"] \arrow[d, "\pi_B"] \\
\phantom{A} \arrow[u, "\Phi"] & \phantom{B} C \arrow[u, "\pi_B"]
\end{tikzcd}
$$

*Proof.* We follow the proof of [MR23, Proposition 5.5]. Let $\sigma : C \to B$ be a morphism of $\mathbb{K}$-modules such that $\Phi_0^0\sigma = \mathrm{id}$. We define two morphisms $J_0^0, H_0^0 : A \times B \to A \times B$ by $J_0^0(a, b) = (a, b - \sigma\Theta_0^0(a))$ and $H_0^0(a, b) = (a, \sigma\Theta_0^0(a) + b)$. We set

$$
\begin{align*}
H_n^0 &= (0, \sigma\Theta_n^0\pi_A); \\
J_n^0 &= (0, -\sigma\Theta_n^0\pi_A),
\end{align*}
$$

An immediate computation gives $HJ = JH = \mathrm{id}$. Therefore, if we denote by $\tilde{Q}$ the $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebra structure on $A \times B$, then $\tilde{Q} = JQH$ is a degree $-1$ coderivation on $\Gamma\mathrm{Perm}^c(A \times B)$. We note that $\tilde{Q}$ preserves $\Gamma\mathrm{Perm}^c(A \times \mathrm{Ker}(\Phi_0^0))$ and the filtrations, so that $A \times \mathrm{Ker}(\Phi_0^0)$ is a $\Gamma\Lambda\widehat{P\mathcal{L}}_\infty$-algebra such that $H : A \times \mathrm{Ker}(\Phi_0^0) \rightsquigarrow A \times B$. Consider now a diagram of $\infty$-morphisms:

$$
\begin{tikzcd}
D \arrow[r, "\Psi_2"] & B \arrow[d, "\Psi_1"] \\
A \arrow[u, "\Theta"'] & C \arrow[u, "\Psi_2"]
\end{tikzcd}
$$
Then, we have the commutative diagram

<figure><img src="image_21.png" /></figure>

which proves the result.

<figure><img src="image_22.png" /></figure>

We now prove Theorem C.

**Theorem 4.28.** Let $\Theta : A \to B$ be a morphism of complete brace algebras such that $\Theta$ is a weak equivalence in $\mathrm{dgMod}_k$. Then $\mathcal{MC}_\bullet(\Theta) : \mathcal{MC}_\bullet(A) \to \mathcal{MC}_\bullet(B)$ is a weak equivalence.

Before giving the proof, note that if $B$ is a complete brace algebra, and if we set $B^I = B \otimes N^*(\Delta^1)$, then we have the following decomposition of the diagonal map in the category of $\Gamma\widehat{\Lambda\mathcal{P}\mathcal{L}}_\infty$-algebras

$$ B \otimes \Sigma N^*(\Delta^n) \xrightarrow[\Delta]{\Delta} \underset{s_0 \otimes id}{\overset{B^I \otimes \Sigma N^*(\Delta^n)}{\rightleftarrows}} \underset{(d_0 \times d_1) \otimes id}{\overset{\Sigma N^*(\Delta^n)}{\rightleftarrows}} (B \times B) \otimes \Sigma N^*(\Delta^n) $$

for every $n \ge 0$. This decomposition comes from Proposition 1.33 with $\mathcal{P} = \Lambda\mathcal{B}race$ and $R = B \otimes \Sigma N^*(\Delta^n)$. The map $s_0 : B \to B^I$ is given by the simplicial map $s_0 : N^*(\Delta^0) \to N^*(\Delta^1)$ and the maps $d_0, d_1 : B^I \to B$ are given by $d_0, d_1 : N^*(\Delta^1) \to N^*(\Delta^0)$. In particular, the morphisms $s_0, d_0$ and $d_1$ induce strict morphisms of $\Gamma\widehat{\Lambda\mathcal{P}\mathcal{L}}_\infty$-algebras, since the action of $\mathcal{E}$ on $N^*(\Delta^n)$ is compatible with its underlying simplicial structure.

*Proof.* Lemma 4.26 proves the theorem in the case of an acyclic fibration $\Theta$. Consider now the general case. Since $d_0 \otimes id : B^I \otimes \Sigma N^*(\Delta^n) \to B \otimes \Sigma N^*(\Delta^n)$ is a strict morphism and surjective, we can apply Lemma 4.27:

<figure><img src="image_23.png" /></figure>

where $H_n : (A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^n) \rightsquigarrow (A \times B^I) \otimes \Sigma N^*(\Delta^n)$ is given by Lemma 4.27, and $\Psi_n : A \otimes \Sigma N^*(\Delta^n) \rightsquigarrow (A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^n)$ is the unique $\infty$-morphism which makes the
previous diagram commutative.

We recall the $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra structure on $(A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^n)$. Let $Q_n$ be the cod derivation on $\Gamma\mathrm{Perm}^c((A \times B^I) \otimes \Sigma N^*(\Delta^n))$ given by the product

$$ (A \times B^I) \otimes \Sigma N^*(\Delta^n) \simeq (A \otimes \Sigma N^*(\Delta^n)) \times (B^I \otimes \Sigma N^*(\Delta^n)). $$

Consider the morphisms $H_n, J_n : (A \times B^I) \otimes \Sigma N^*(\Delta^n) \rightsquigarrow (A \times B^I) \otimes \Sigma N^*(\Delta^n)$ defined in the proof of Lemma 4.27. We note that these morphisms are strict, as the morphism $\Theta \otimes \mathrm{id} : A \otimes \Sigma N^*(\Delta^n) \to B \otimes \Sigma N^*(\Delta^n)$ is strict, and are defined by

$$ (H_n)_0^0((a, b) \otimes \underline{x}) = (a, \sigma\Theta(a) + b) \otimes \underline{x}; $$

$$ (J_n)_0^0((a, b) \otimes \underline{x}) = (a, b - \sigma\Theta(a)) \otimes \underline{x}, $$

for every $a \in A, b \in B$ and $\underline{x} \in \Sigma N^*(\Delta^n)$, and where $\sigma : B \to B^I$ is a splitting of $d_0 : B^I \to B$. Then, the $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra structure on $(A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^n)$ is given by

$$ (\tilde{Q}_n)_p^q = (J_n)_q^q(Q_n)_p^q(H_n)_p^p. $$

We thus see that the $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra structures on $(A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^n)$ for all $n \ge 0$ endow the simplicial set $(A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^\bullet)$ with the structure of a strict simplicial object in $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$. Moreover, the map $\pi_A : A \times \mathrm{Ker}(d_0) \to A$ is an acyclic fibration, and a simple computation shows that $\pi_A \otimes \mathrm{id}$ is a strict morphism. By Lemma 4.26, we deduce that $\mathcal{MC}(\pi_A \otimes \mathrm{id})$ is a weak equivalence. By the 2 out of 3 axiom in sSet, we also have that $\mathcal{MC}(\Psi_\bullet)$ is a weak equivalence of simplicial sets.

Let $h : A \times B^I \to A \times B^I$ be the morphism such that $(H_n)_0^0 = h \otimes \mathrm{id}$. For every $n \ge 0$, we set

$$ P_n = d_1 \pi_{B^I} h \otimes \mathrm{id} : (A \times \mathrm{Ker}(d_0)) \otimes \Sigma N^*(\Delta^n) \to B \otimes \Sigma N^*(\Delta^n) $$

We show that $P_n$ is a strict acyclic fibration. First, for every $n \ge 0$, the morphism $P_n$ is strict as it is the composite of strict morphisms. Moreover, we have the identity $\Theta \otimes \mathrm{id} = \Psi_\bullet P_\bullet$, which shows that $P_n$ is acyclic for every $n \ge 0$. We now prove that $P_n$ is surjective. For every $b \in B$ and $\underline{x} \in \Sigma N^*(\Delta^n)$, we have $P_n(0, b \otimes \underline{0}^\vee \otimes \underline{x}) = b \otimes \underline{x}$ which proves that $P_n$ is surjective for every $n \ge 0$. By Lemma 4.26, we have that $\mathcal{MC}(P_\bullet)$ is a weak equivalence. Finally, since we have $\Theta \otimes \mathrm{id} = \Psi_\bullet P_\bullet$, it follows that $\mathcal{MC}(\Theta \otimes \mathrm{id})$ is also a weak equivalence, which proves the theorem. $\square$

## 4.5 Comparison with the deformation theory of shifted $\mathcal{L}ie_\infty$-algebras

Let $\mathcal{L}ie_\infty$ be an operad which encodes Lie algebras up to homotopy, for instance $\mathcal{L}ie_\infty = B^c(\Lambda^{-1}\mathcal{C}om)$. We call a $\Lambda\mathcal{L}e_\infty$-algebra any algebra over the operad $\Lambda\mathcal{L}ie_\infty$. These algebras have been widely studied in the literature. Recall (for instance from [DR15a, §2], or [Ber15] for the non-shifted analogue) that giving a $\Lambda\mathcal{L}e_\infty$-algebra structure on a graded $\mathbb{K}$-module $V$ is equivalent to giving degree $-1$ brackets

$$ \underbrace{[\dots, \dots, \dots]}_n : (V^{\otimes n})_{\Sigma_n} \to V $$
In the following, we distinguish the $\Lambda\mathcal{P}\mathcal{L}_\infty$-algebra structure with the $\Lambda\mathcal{L}_\infty$-algebra structure.
More precisely, for every complete $\Lambda\mathcal{P}\mathcal{L}_\infty$-algebra $V$, we set:

$$ \mathcal{M}\mathcal{C}^{\Lambda\mathcal{P}\mathcal{L}_{\infty}}(V) = \mathcal{M}\mathcal{C}(V); \mathcal{M}\mathcal{C}^{\Lambda\mathcal{L}_{\infty}}(V) = \mathcal{M}\mathcal{C}(L(V)). $$

We also set, for every complete brace algebra $A$,

$$ \mathcal{M}\mathcal{C}_{\bullet}^{\Lambda\mathcal{P}\mathcal{L}_{\infty}}(A) = \mathcal{M}\mathcal{C}_{\bullet}(A); \mathcal{M}\mathcal{C}_{\bullet}^{\Lambda\mathcal{L}_{\infty}}(A) = \mathcal{M}\mathcal{C}_{\bullet}(L(A)), $$

where $L(A)$ is the Lie algebra endowed with the bracket $[x, y] = x\langle y \rangle - (-1)^{|x||y|}y\langle x \rangle$.

**Lemma 4.30.** Let $V$ be a $\widehat{\Lambda\mathcal{P}\mathcal{L}}_{\infty}$-algebra. Then

$$ \mathcal{M}\mathcal{C}^{\Lambda\mathcal{L}_{\infty}}(V) = \mathcal{M}\mathcal{C}^{\Lambda\mathcal{P}\mathcal{L}_{\infty}}(V). $$

*Proof.* By Proposition 4.29, we have, for every $x_1, \dots, x_n \in V$,

$$ [x_1, \dots, x_n] = \sum_{k=1}^{n} \pm x_k \langle [x_1, \dots, \widehat{x_k}, \dots, x_n] \rangle. $$

Then, the Maurer-Cartan equation

$$ d(x) + \sum_{n \ge 1} \frac{1}{n!} x \underbrace{[x, \dots, x]}_{n} = 0 $$

is equivalent to the equation

$$ d(x) + \sum_{n \ge 2} \frac{1}{n!} [x, \dots, x] = 0, $$

which is precisely the Maurer-Cartan equation in complete $\Lambda\mathcal{L}_{\infty}$-algebras. $\square$

Let $A$ be a complete brace algebra, $B$ a dg commutative and associative algebra, and $E$ be a $\mathcal{E}$-algebra. Then the tensor products $(A \hat{\otimes} B) \hat{\otimes} \Sigma E$ and $(A \hat{\otimes} E) \hat{\otimes} \Sigma B$ are endowed with a complete $\Lambda\mathcal{L}_\infty$-algebra structure. Indeed, the first one is induced by the composite

$$ \Lambda\mathcal{L}ie_{\infty} \longrightarrow \Lambda\mathcal{B}race_{\mathbb{H}} \otimes \mathcal{E} \xrightarrow{\simeq} (\mathcal{B}race_{\mathbb{H}} \otimes \mathcal{Com})_{\mathbb{H}} \otimes \Lambda\mathcal{E}, $$

while the second one is induced by the composite

$$ \Lambda\mathcal{L}ie_{\infty} \longrightarrow \Lambda\mathcal{B}race_{\mathbb{H}} \otimes \mathcal{E} \xrightarrow{\simeq} (\mathcal{B}race_{\mathbb{H}} \otimes \mathcal{E})_{\mathbb{H}} \otimes \Lambda\mathcal{Com}. $$

**Lemma 4.31.** The isomorphism

$$ (A \hat{\otimes} B) \hat{\otimes} \Sigma E \xrightarrow{\simeq} (A \hat{\otimes} E) \hat{\otimes} \Sigma B $$

which exchanges $E$ and $B$ is an isomorphism of complete $\Lambda\mathcal{L}_\infty$-algebras.
Proof. Straightforward computations.

□

We thus obtain the following theorem.

**Theorem 4.32.** Let A be a complete brace algebra. Then there exists a simplicial set $S_{\bullet}^A$ and a zig-zag of weak equivalences in simplicial sets:

$$MC_{\bullet}^{\Lambda P L_{\infty}}(A) \xrightarrow{\sim} S_{\bullet}^{A} \leftarrow \sim MC_{\bullet}^{\Lambda L_{\infty}}(A).$$

One major consequence of this theorem is that the homotopy groups that we have computed are isomorphic to the one's found in [Ber15] if the field is of characteristic 0.

Proof. We first remark that, for any $n \ge 0$, we have a morphism of complete brace algebras which is a weak equivalence:

$$A \xrightarrow{\sim} A \hat{\otimes} \Omega^*(\Delta^n).$$

By Theorem 4.28, we obtain a weak equivalence

$$MC_{\bullet}^{\Lambda P L_{\infty}}(A) \xrightarrow{\sim} MC_{\bullet}^{\Lambda P L_{\infty}}(A \hat{\otimes} \Omega^*(\Delta^n)) = MC_{\bullet}^{\Lambda P L_{\infty}}((A \hat{\otimes} \Omega^*(\Delta^n)) \hat{\otimes} \Sigma N^*(\Delta^*))$$

We now apply [GJ09, Chapter IV, Proposition 1.9]. Recall that the *diagonal* of a bisimplicial set X (see [GJ09, Chapter IV, §1]) is the simplicial set Diag(X) defined by

$$Diag(X)_n = X_{nn}$$

for every $n \ge 0$. Since we have a point-wise weak equivalence, this extends to the following weak-equivalence of simplicial sets

$$MC_{\bullet}^{\Lambda P L_{\infty}}(A) \xrightarrow{\sim} Diag(MC^{\Lambda P L_{\infty}}((A \hat{\otimes} \Omega^*(\Delta^*)) \hat{\otimes} \Sigma N^*(\Delta^*))$$

Similarly, by [DR15b, Theorem 1.1], we have a weak equivalence

$$Diag(MC^{\Lambda L_{\infty}}((A \otimes N^*(\Delta^*)) \otimes \Sigma \Omega^*(\Delta^*)) \xrightarrow{\sim} MC_{\bullet}^{\Lambda L_{\infty}}(A).$$

By combining the above weak equivalences with the two previous lemmas, we obtain the following diagram:

$$\begin{aligned} MC_{\bullet}^{\Lambda P L_{\infty}}(A) &\xrightarrow{\sim} Diag(MC^{\Lambda P L_{\infty}}((A \otimes \Omega^*(\Delta^*)) \otimes \Sigma N^*(\Delta^*)) \\ &\qquad \qquad \qquad \qquad \qquad \qquad = \\ &\qquad Diag(MC^{\Lambda L_{\infty}}((A \otimes \Omega^*(\Delta^*)) \otimes \Sigma N^*(\Delta^*)) \\ &\qquad \qquad \qquad \qquad \qquad \qquad \sim \\ &\qquad Diag(MC^{\Lambda L_{\infty}}((A \otimes N^*(\Delta^*)) \otimes \Sigma \Omega^*(\Delta^*)) \xrightarrow{\sim} MC_{\bullet}^{\Lambda L_{\infty}}(L(A)) \end{aligned}$$

which proves the theorem.

□
## 5 A mapping space in the category of non-symmetric operads

In this section, we give an explicit construction of a mapping space $Map_{\mathcal{O}_p}(B^c(C), \mathcal{P})$ in the category of non symmetric operads in terms of $\Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$ operations. Explicitly, we give a construction of a mapping space as the simplicial Maurer-Cartan set associated to the complete brace algebra $\text{Hom}_{\text{Seq}_K}(\bar{C}, \bar{P})$.

In §5.1, we make recollections on the construction of the free operad functor and on the model structure used for operads in this memoir. In this memoir, we use an explicit description of the free operad functor in terms of trees with inputs, which we define in this section.

In §5.2, we give an explicit construction of a cosimplicial frame associated to the cobar construction $B^c(C)$ of a coaugmented non symmetric cooperad as a sequence.

In §5.3, we finally prove Theorem D, which gives a description of a mapping space $Map_{\mathcal{O}_p}(B^c(C), \mathcal{P})$ in the category of non symmetric operads as the simplicial Maurer-Cartan set associated to the brace algebra $\text{Hom}_{\text{Seq}_K}(\bar{C}, \bar{P})$. This gives a computation of the connected components and the homotopy groups of $Map_{\mathcal{O}_p}(B^c(C), \mathcal{P})$ by using Theorem B.

### 5.1 The free operad functor and the model structure on $\mathcal{O}_p$

We first recall the definition of the free operad functor and the model structure on operads. We will mostly follow conventions of [Mur11]. Let $\text{Seq}_K$ be the category of sequences in $\text{dgMod}_K$. Recall that we have an obvious model structure on $\text{Seq}_K$ which is defined in the standard model structure on $\text{dgMod}_K$.

The model structure on the category of non symmetric operads $\mathcal{O}_p$ is obtained by transferring the model structure of $\text{Seq}_K$ from an adjunction

$$ \mathcal{F}: \text{Seq}_K \stackrel{\leftarrow}{\longrightarrow} \mathcal{O}_p : \omega, $$

where $\omega: \mathcal{O}_p \longrightarrow \text{Seq}_K$ is the functor which forgets the operad structure. The left adjoint $\mathcal{F}: \text{Seq}_K \longrightarrow \mathcal{O}_p$ is the *free operad* functor, for which we recall the construction.

We define the notion of tree with inputs, which is analogue to the notion of "planted planar tree with inputs" given in [Mur11, Definition 3.4].

**Definition 5.1.** Let $n \ge 0$. A (planar) tree with inputs is the data of a tree $T \in \mathcal{PT}(n)$ and, for each vertex of $T$, an integer which represents the number of ingoing arrows, which may includes some edges of $T$. We also add an outgoing arrow on the root of $T$. The ingoing arrows with only one vertex of $T$ are called the inputs of the tree.
<figure><img src="image_24.png" /></figure>

We usually denote by $\overline{T}$ any tree with inputs with underlying tree $T \in \mathcal{PT}$. We call $T$ the shape of $\overline{T}$, and set $\text{Shape}(\overline{T}) = T$. We also set $V_{\overline{T}} = V_T$. For every vertex $v \in V_{\overline{T}}$, we denote by $\text{val}_{\overline{T}}(v)$ the number of ingoing arrows which go to $v$. We denote by $\overline{\mathcal{PT}}_k(n)$ the set of trees with $n$ vertices and $k$ inputs and $\overline{\mathcal{Tree}}_k(n) = \mathbb{K}[\overline{\mathcal{PT}}_k(n)]$.

As in Definition 1.20, we can consider trees with inputs $\overline{T} \in \overline{\mathcal{PT}}_k(a_1 < \dots < a_n)$ in a general totally ordered finite set $a_1 < \dots < a_n$. We say that $\overline{T}$ is *canonical* (or in the canonical order) if its shape $\text{Shape}(\overline{T}) \in \overline{\mathcal{PT}}(a_1 < \dots < a_n)$ is canonical.

For every tree $\overline{T} \in \overline{\mathcal{PT}}_k(n)$, we endow the inputs with the canonical labeling from 1 to $k$ obtained by following the canonical order of $T$. For instance, the tree given in the definition is seen as

<figure><img src="image_25.png" /></figure>

As for trees in $\mathcal{PT}$, we have the following definitions.

**Definition 5.2.** Let $\overline{T}$ be a tree with inputs and with underlying shape $T \in \mathcal{PT}$.

* A subtree of $\overline{T}$ is the data of a subtree $S$ of $T$, endowed with the unique choice of arrows such that, for every $v \in V_S \subset V_T$, we have $\text{val}_S(v) = \text{val}_{\overline{T}}(v)$.

* If $\overline{S}$ is a subtree of $\overline{T}$, we denote by $\overline{T}/\overline{S}$ the tree of shape $T/S$ obtained by contracting the tree $\overline{S}$ on the tree with only one vertex, denoted by $S$, with the same number of inputs as $\overline{S}$.

For instance, if we consider the above tree with inputs $\overline{T}$, then the following tree with inputs

<figure><img src="image_26.png" /></figure>
is a subtree of $\mathcal{T}$ such that

<figure><img src="image_27.png" /></figure>

Let $p, q, n, m \ge 0, 1 \le i \le p$ and $\mathcal{U} \in \mathcal{T}ree_p(n), \mathcal{V} \in \mathcal{T}ree_q(m)$. We let $\mathcal{U} \circ_i \mathcal{V}$ to be the tree in $\mathcal{T}ree_{p+q-1}(n+m)$ given by the attachment of the unique outgoing arrow of $\mathcal{V}$ to the $i$-th ingoing arrow of $\mathcal{U}$. This defines a morphism

$$ \circ_i : \mathcal{T}ree_p(n) \otimes \mathcal{T}ree_q(m) \to \mathcal{T}ree_{p+q-1}(n+m). $$

**Lemma 5.3.** Let $\mathcal{T}ree$ be the sequence defined by

$$ \mathcal{T}ree(k) = \bigoplus_{n \ge 0} \mathcal{T}ree_k(n). $$

Then the morphisms $\circ_i : \mathcal{T}ree_p \otimes \mathcal{T}ree_q \to \mathcal{T}ree_{p+q-1}$ endow the sequence $\mathcal{T}ree$ with the structure of an operad.

Using this notion of tree, we set

$$ \mathcal{F}(M)(k) = \bigoplus_{n \ge 0} \left( \bigoplus_{\mathcal{T} \in \mathcal{P}\mathcal{R}\mathcal{T}_k(n)} \mathcal{T} \otimes \bigotimes_{i=1}^n M(\mathrm{val}_{\mathcal{T}}(i)) \right)_{\Sigma_n} $$

where, in the sum, we consider the action of $\Sigma_n$ on $\mathcal{P}\mathcal{R}\mathcal{T}_k(n)$ by the permutation of the labels of the vertices, and the action of $\Sigma_n$ on $\otimes_{i=1}^n M(\mathrm{val}_{\mathcal{T}}(i))$ by permutations. The operadic structure of $\mathcal{F}(M)$ is given by the operadic structure of $\mathcal{T}ree$ and the concatenation of the elements in $M$. We denote by $\mathcal{F}_{(\mathcal{T})}(M) = \otimes_{i=1}^n M(\mathrm{val}_{\mathcal{T}}(i))$ the $\mathcal{T}$-component of $\mathcal{F}(M)$ associated to some tree $\mathcal{T} \in \mathcal{P}\mathcal{R}\mathcal{T}_k(n)$.

We can check that the functor $\mathcal{F} : \mathrm{Seq}_{\mathbb{K}} \to \mathcal{Op}$ is left adjoint to the forgetful functor $\omega : \mathcal{Op} \to \mathrm{Seq}_{\mathbb{K}}$ which forgets the operad structure:

$$ \mathcal{F} : \mathrm{Seq}_{\mathbb{K}} \stackrel{\leftarrow}{\rightleftharpoons} \mathcal{Op} : \omega. $$

This adjunction implies the following result.

**Proposition 5.4** (see [Mur11, Theorem 1.1]). The category $\mathcal{Op}$ is endowed with a cofibrantly generated model structure such that the forgetful functor $\omega : \mathcal{Op} \to \mathrm{Seq}_{\mathbb{K}}$ creates weak-equivalences and fibrations. Cofibrations are given by the left lifting property.

**Remark 5.5.** In the following subsections, we also use the notion of cofree cooperad generated by a sequence $M$ such that $M(0) = 0$. For every $k \ge 0$ and $n \ge 1$, let $\mathcal{P}\mathcal{R}\mathcal{T}_k^0(n)$ be the subset of
$\underline{\mathcal{PRT}}_k(n)$ given by trees $\underline{T}$ such that $\mathrm{val}_{\underline{T}}(v) \neq 0$ for every $v \in V_{\underline{T}}$. Let $\underline{\mathcal{Tree}}_k^0(n) := \mathbb{K}[\underline{\mathcal{PRT}}_k^0(n)]$. Then the sequence $\underline{\mathcal{Tree}}^0$ defined by

$$ \underline{\mathcal{Tree}}^0(n) = \bigoplus_{k \ge 0} \underline{\mathcal{Tree}}_k^0(n) $$

is a suboperad of $\underline{\mathcal{Tree}}$ such that, for every $n \ge 0$, the $\mathbb{K}$-module $\underline{\mathcal{Tree}}^0(n)$ is finite dimensional. By Remark 1.7, the dual symmetric sequence $(\underline{\mathcal{Tree}}^0)^{\vee}$ is endowed with the structure of a cooperad. We then define

$$ \mathcal{F}^c(M)(k) = \bigoplus_{n \ge 1} \left( \bigoplus_{\underline{T} \in \mathcal{PRT}_k^0(n)} \underline{T}^{\vee} \otimes \bigotimes_{i=1}^n M(\mathrm{val}_{\underline{T}}(i)) \right)^{\Sigma_n} $$

where we consider the action of $\Sigma_n$ on $\underline{T}^{\vee}$ by permutation of the vertices, and the action of $\Sigma_n$ on $\bigotimes_{i=1}^n M(\mathrm{val}_{\underline{T}}(i))$ by permutations. We endow $\mathcal{F}^c(M)$ with the cooperad structure given by the cooperadic structure of $(\underline{\mathcal{Tree}}^0)^{\vee}$, and by the deconcatenation coproduct in the tensor coalgebra of $\bigoplus_{n \ge 1} M(n)$. As for operads, we have an adjunction

$$ \omega : \mathcal{O}p^c \stackrel{\leftarrow}{\longrightarrow} \mathrm{Seq}_{\mathbb{K}} : \mathcal{F}^c $$

where $\omega : \mathcal{O}p^c \to \mathrm{Seq}_{\mathbb{K}}$ is the functor which forgets the cooperad structure.

We will need to consider operadic compositions (resp. cooperadic cocompositions) shaped on trees with inputs. This can be formalized as follows. Let $\mathcal{P}$ be an augmented operad $\mathcal{P} \simeq I \oplus \bar{\mathcal{P}}$ and $\mathcal{C}$ be a coaugmented cooperad $\mathcal{C} \simeq I \oplus \bar{\mathcal{C}}$ such that $\mathcal{P}(0) = \mathcal{C}(0) = 0$ and $\mathcal{P}(1) = \mathcal{C}(1) = \mathbb{K}$. By the universal property satisfied by $\mathcal{F}$, we have a unique operad morphism $\mathcal{F}(\mathcal{P}) \to \mathcal{P}$ which reduces to the identity on $\mathcal{P} \subset \mathcal{F}(\mathcal{P})$. Analogously, we have a unique cooperad morphism $\mathcal{C} \to \mathcal{F}^c(\mathcal{C})$ whose projection on $\mathcal{C}$ is given by the identity on $\mathcal{C}$.

**Definition 5.6.** Let $k \ge 1$ and $\underline{T} \in \mathcal{PRT}_k^0$. We define $\gamma_{(\underline{T})} : \mathcal{F}_{(\underline{T})}(\bar{\mathcal{P}}) \to \bar{\mathcal{P}}$ and $\Delta_{(\underline{T})} : \bar{\mathcal{C}} \to \mathcal{F}_{(\underline{T})}^c(\bar{\mathcal{C}})$ by the composites

$$ \gamma_{(\underline{T})} : \mathcal{F}_{(\underline{T})}(\bar{\mathcal{P}}) \longrightarrow \mathcal{F}_{(\underline{T})}(\mathcal{P}) \xrightarrow{\gamma} \mathcal{P} \dashrightarrow \bar{\mathcal{P}} ; $$

$$ \Delta_{(\underline{T})} : \bar{\mathcal{C}} \longrightarrow \mathcal{C} \xrightarrow{\Delta} \mathcal{F}_{(\underline{T})}^c(\mathcal{C}) \dashrightarrow \mathcal{F}_{(\underline{T})}^c(\bar{\mathcal{C}}) . $$

For every $p, q, n, m \ge 0$ and $1 \le i \le p$, we define a morphism

$$ \bullet_i : \underline{\mathcal{Tree}}_p(n) \otimes \underline{\mathcal{Tree}}_q(m) \to \underline{\mathcal{Tree}}_p(n+m-1) $$

by the following. Let $\underline{U} \in \mathcal{PRT}_p(n)$ and $\underline{V} \in \mathcal{PRT}_q(m)$. If the number of arrows on the $i$-th vertex of $\underline{U}$ is not $q$, then $\underline{U} \bullet_i \underline{V} = 0$. Else, we define $\underline{U} \bullet_i \underline{V}$ as the unique tree obtained by putting $\underline{V}$ in the $i$-th vertex of $\underline{U}$, and attaching the ingoing arrows of the $i$-th vertex of $\underline{U}$ into the inputs of $\underline{V}$.

**Lemma 5.7.** Let $\underline{T}$ be a tree with inputs and $\underline{S}$ be a subtree of $\underline{T}$. Then $\underline{T}/\underline{S} \bullet_S \underline{S} = \underline{T}$.

*Proof.* It is an immediate consequence of the definitions. $\square$

The morphisms defined in Definition 5.6 also behave well with the compositions $\circ_i$ and $\bullet_i$.
**Lemma 5.8.** Let $k \ge 1$, $\underline{T} \in \mathcal{PT}\underline{T}_k^0(n)$ and $\underline{S} \subset \underline{T}$. Then

$$ \gamma(\underline{T}/\underline{S}) \circ_{\underline{S}} \gamma(\underline{S}) = \gamma(\underline{T}) ; \quad \Delta(\underline{T}/\underline{S}) \circ_{\underline{S}} \Delta(\underline{S}) = \Delta(\underline{T}) $$

in the endomorphism operad $\text{End}_{\oplus_{n \ge 2} \mathcal{P}(n)}$ and in the coendomorphism operad $\text{CoEnd}_{\oplus_{n \ge 2} \mathcal{C}(n)}$ respectively.

*Proof.* These are direct consequences of the (co)associativity axioms in a (co)operad. $\square$

## 5.2 A cosimplicial frame for $\mathcal{B}^c(\mathcal{C})$

Let $\mathcal{C} \simeq I \oplus \bar{\mathcal{C}}$ be a coaugmented non-symmetric cooperad with $\mathcal{C}(0) = 0$ and $\mathcal{C}(1) = \mathbb{K}$. The goal of this subsection is to construct a cosimplicial frame $\mathcal{B}^c(\mathcal{C}) \otimes \Delta^n$ associated to $\mathcal{B}^c(\mathcal{C})$. We will explicitly define $\mathcal{B}^c(\mathcal{C}) \otimes \Delta^n$ as the free operad induced by a cooperad up to homotopy that will be given by $\bar{\mathcal{C}} \otimes N_*(\Delta^n)$.

Let $E$ be a $\mathcal{E}$-coalgebra. We endow the operad $\mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$ with a general twisting morphism such that if $E = N_*(\Delta^0) \simeq \mathbb{K}$, then $\mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E) \simeq \mathcal{B}^c(\mathcal{C})$. Explicitly, we construct $\beta^E : \bar{\mathcal{C}} \otimes \Sigma^{-1}E \to \mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$ such that the morphism $\partial_{\beta^E} : \mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E) \to \mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$ obtained from $\beta^E$ by the Leibniz rule is a twisting morphism. If we denote by $d$ the differential induced by $\bar{\mathcal{C}} \otimes \Sigma^{-1}E$ on $\mathcal{F}^c(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$, then the morphism $\beta^E$ shall needs (see [LV12] or [Fre09b] for instance) to be such that

$$ d(\beta^E) + \partial_{\beta^E} \beta^E = 0. $$

Let $k \ge 1$ and $\underline{T} \in \mathcal{PT}\underline{T}_k^0$ be a canonical tree with inputs with shape $T \in \mathcal{PT}$. We define $\beta_{(T)}^E : \bar{\mathcal{C}} \otimes \Sigma^{-1}E \to \mathcal{F}(\bar{\mathcal{C}})(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$ by

$$ \beta_{(T)}^E = \Delta_{(T)} \tilde{\otimes} \Lambda \mu_T^E $$

where, for every $\mu \in \mathcal{E}(n)$, we denote by $\mu^E$ the morphism in $\text{Hom}(E, E^{\otimes n})$ given by the $\mathcal{E}$-coalgebra structure $E$.

This gives a well defined morphism of sequences $\beta^E : \bar{\mathcal{C}} \otimes \Sigma^{-1}E \to \mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$ by summing over all canonical trees $T$. Note that such a sum of morphisms is will defined on $\bar{\mathcal{C}} \otimes \Sigma^{-1}E$ since $\bar{\mathcal{C}}(0) = \bar{\mathcal{C}}(1) = 0$. It is also natural in $E$ by definition.

**Proposition 5.9.** The morphism $\beta^E$ defined above satisfies

$$ d(\beta^E) + \partial_{\beta^E} \beta^E = 0. $$

We thus have a derivation of operads $d + \partial_{\beta^E}$ on $\mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1}E)$.

*Proof.* It is sufficient to prove the formula on $\bar{\mathcal{C}} \otimes \Sigma^{-1}E$. Let $\underline{T}$ be a tree with inputs with shape a canonical tree $T \in \mathcal{PT}$. We show that the $\underline{T}$-component of the morphism $d(\beta^E) + \partial_{\beta^E} \beta^E$ is 0. First, we have

$$ d(\beta_{(T)}^E) = \Delta_{(T)} \tilde{\otimes} d(\Lambda \mu_T)^{\Sigma E}, $$

since the cooperadic structure on $\mathcal{C}$ is preserved by its differential. Next, we have by definition

$$ (\partial_{\beta^E} \beta^E)_{(T)} = \sum_{\underline{S} \subset \underline{T}} (\Delta_{(\underline{T}/\underline{S})} \circ_{\text{Shape}(\underline{S})} \Delta_{(\underline{S})}) \tilde{\otimes} (\Lambda \mu_{\underline{T}/\text{Shape}(\underline{S})} \circ_{\text{Shape}(\underline{S})} \Lambda \mu_{\text{Shape}(\underline{S})})^{\Sigma E}. $$
Note that taking a subtree $\mathcal{S}$ of $\mathcal{T}$ is equivalent to taking a subtree $\mathcal{S}$ of $\mathcal{T}$. We thus have

$$ (\partial_{\beta^E} \beta^E)_{(\mathcal{T})} = \sum_{S \subset \mathcal{T}} \Delta_{(\mathcal{T})} \tilde{\otimes} (\Lambda \mu_{T/S} \circ_S \Lambda \mu_S)^{\Sigma E}. $$

The proposition follows by Theorem 3.3. $\square$

We can now construct a cosimplicial frame for $B^c(\mathcal{C})$. Recall that the normalized chain complex $N_*(\Delta^n)$ admits a structure of a $\mathcal{E}$-coalgebra.

**Definition 5.10.** Let $n \ge 0$. We set

$$ B^c(\mathcal{C}) \otimes \Delta^n = (\mathcal{F}(\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^n)), \partial_{\beta_{N_*(\Delta^n)}}). $$

We immediately see that $B^c(\mathcal{C}) \otimes \Delta^\bullet$ defines a cosimplicial object in the category of non symmetric operads. By Corollary 3.9, we also have that $B^c(\mathcal{C}) \otimes \Delta^0 = B^c(\mathcal{C})$.

Recall from [Fre17b, §3.2.2–§3.2.3] that a cosimplicial frame associated to $B^c(\mathcal{C})$ is a cosimplicial set $B^c(\mathcal{C}) \otimes \Delta^\bullet$ such that, for every $n \ge 0$, the morphism $B^c(\mathcal{C}) \otimes \Delta^n \to B^c(\mathcal{C}) \otimes \Delta^0$ is a weak equivalence and the morphism $B^c(\mathcal{C}) \otimes \partial \Delta^n \to B^c(\mathcal{C}) \otimes \Delta^n$ is a cofibration for every $n \ge 0$.

**Theorem 5.11.** The cosimplicial object $B^c(\mathcal{C}) \otimes \Delta^\bullet$ defines a cosimplicial frame for $B^c(\mathcal{C})$ in the category of operads.

*Proof.* Since the morphisms $N_*(\partial \Delta^n) \to N_*(\Delta^n)$ are cofibrations for every $n \ge 0$, the morphisms $\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\partial \Delta^n) \to \bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^n)$ are cofibrations (see [Fre09b, Proposition 1.4.13]). We now prove that $B^c(\mathcal{C}) \otimes \Delta^n \to B^c(\mathcal{C})$ is a weak equivalence. We first note that $B^c(\mathcal{C}) \otimes \Delta^n$ admits a natural filtration $(F_p(B^c(\mathcal{C}) \otimes \Delta^n))_p$ defined by

$$ F_p(B^c(\mathcal{C}) \otimes \Delta^n) = \bigoplus_{k \ge 1} \bigoplus_{\substack{\mathcal{T} \in \mathcal{PRT}_k^0 \\ \mathcal{T} \text{ canonical} \\ |\mathcal{T}| \ge p+1}} \underline{\mathcal{T}} \otimes \mathcal{F}_{(\mathcal{T})}(\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^n)). $$

By definition, the differential $\partial^n$ preserves this filtration. We thus have a spectral sequence which is convergent dimension-wise:

$$ E_q^0 \Rightarrow H_*(B^c(\mathcal{C}) \otimes \Delta^n) $$

where we have set

$$ E_q^0 = \bigoplus_{k \ge 1} \bigoplus_{\substack{\mathcal{T} \in \mathcal{PRT}_k^0 \\ \mathcal{T} \text{ canonical} \\ |\mathcal{T}|=q+1}} \underline{\mathcal{T}} \otimes \mathcal{F}_{(\mathcal{T})}(\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^n)). $$

Because the twisting part $\partial_{N_*(\Delta^n)}$ increases the number of vertices, the differential is reduced to the internal differential $d$ on $E_q^0$. Because $\bar{\mathcal{C}} \otimes N_*(\Delta^n) \to \bar{\mathcal{C}}$ is a weak equivalence, we have that the morphism $N_*(\Delta^n) \to N_*(\Delta^0)$ induces a weak equivalence on $E_p^0$ for all $p$. It then induces a weak equivalence from $B^c(\mathcal{C}) \otimes \Delta^n$ to $B^c(\mathcal{C})$.

We then have the result. $\square$
## 5.3 Computation of $Map_{Op}(B^c(C), P)$

In his last subsection, we give an explicit description of a mapping space $Map_{Op}(B^c(C), P)$. We know from Theorem 5.11 that we can set

$$ \mathrm{Map}_{Op}(B^c(C), P)_n = \mathrm{Mor}_{Op}(B^c(C) \otimes \Delta^n, P) $$

for every $n \ge 0$. The goal of this subsection is to link this object with some $\Gamma\Lambda P\mathcal{L}_\infty$-algebra structure on $\mathrm{Hom}_{\mathrm{Seq}_k}(\bar{C} \otimes \Sigma^{-1}N_*(\Delta^n), \bar{P})$. We consider the dg $\mathbb{K}$-module

$$ \mathcal{L}(\mathrm{Hom}(\bar{C} \otimes N_*(\Delta^n), \bar{P})) = \bigoplus_{k \ge 2} \mathrm{Hom}(\bar{C}(k) \otimes N_*(\Delta^n), \bar{P}(k)). $$

This dg $\mathbb{K}$-module is endowed with a filtration defined by

$$ F_p(\mathcal{L}(\mathrm{Hom}(\bar{C} \otimes N_*(\Delta^n), \bar{P}))) = \bigoplus_{k \ge p+1} \mathrm{Hom}(\bar{C}(k) \otimes N_*(\Delta^n), \bar{P}(k)), $$

so that $\mathrm{Hom}_{\mathrm{Seq}_k}(\bar{C} \otimes N_*(\Delta^n), \bar{P})$ is the completion of $\mathcal{L}(\mathrm{Hom}(\bar{C} \otimes N_*(\Delta^n), \bar{P}))$ with respect to this filtration.

**Lemma 5.12.** The dg $\mathbb{K}$-module $\mathcal{L}(\mathrm{Hom}(\bar{C} \otimes N_*(\Delta^n), \bar{P}))$ is endowed with the structure of a *Brace* $\otimes_H \mathcal{E}$-algebra defined by

$$ (T \otimes \mu)(f_1, \dots, f_k) = \pm \sum_{\underline{T} \in \mathrm{Shape}^{-1}(T)} \gamma_{(\underline{T})} \circ (f_1 \otimes \dots \otimes f_k) \circ (\Delta_{(\underline{T})} \tilde{\otimes} \mu^{N_*(\Delta^n)}), $$

for any $T \in \mathcal{PRT}(k)$, $\mu \in \mathcal{E}(k)$ and $f_1, \dots, f_k \in \mathcal{L}(\mathrm{Hom}(\bar{C} \otimes N_*(\Delta^n), \bar{P}))$ homogeneous, where we consider the tensor product $\tilde{\otimes}$ (see Definition 1.1). The sign is yielded by the commutation of $\mu$ with the $f_i$'s. Note that the sum is finite point-wise since we have supposed that $\mathcal{C}(0) = 0$.

*Proof.* Let $\sigma \in \Sigma_n$. By definition of $\gamma_{(\underline{T})}$ and $\Delta_{(\underline{T})}$ for every $\underline{T} \in \mathcal{PRT}$, we have

$$ (\sigma \cdot T \otimes \sigma \cdot \mu)(f_1, \dots, f_k) = \pm (T \otimes \mu)(f_{\sigma^{-1}(1)}, \dots, f_{\sigma^{-1}(k)}), $$

where we consider the action of $\sigma$ on $(\bar{C} \otimes N_*(\Delta^n))^{\otimes k}$ by permutation of the tensors.

We now prove the compatibility with the operadic structure. Let $p, q \ge 0$ and $U \in \mathcal{PRT}(p), V \in \mathcal{PRT}(q), \mu \in \mathcal{E}(p), \nu \in \mathcal{E}(q)$ and $1 \le i \le p$. By Lemma 5.8, we have

$$ \begin{aligned} (U \otimes \mu)(f_1, \dots, f_{i-1}, (V \otimes \nu)(f_i, \dots, f_{i+q-1}), f_{i+q}, \dots, f_{p+q-1}) \\ = \pm \sum_{\substack{\underline{U} \in \mathrm{Shape}^{-1}(U) \\ \underline{V} \in \mathrm{Shape}^{-1}(V)}} \gamma_{(\underline{U} \cdot_i \underline{V})} \circ (f_1 \otimes \dots \otimes f_{p+q-1}) \circ (\Delta_{(\underline{U} \cdot_i \underline{V})} \tilde{\otimes} (\mu \circ_i \nu)^{N_*(\Delta^n)}). \end{aligned} $$

Now, write $U \circ_i V = T_1 + \dots + T_m$ for some $T_1, \dots, T_m \in \mathcal{PRT}$. By definition of the composition product in *Brace*, for every $\underline{U} \in \mathrm{Shape}^{-1}(U)$ and $\underline{V} \in \mathrm{Shape}^{-1}(V)$, the tree $\underline{U} \cdot_i \underline{V}$ has shape $T_j$ for some $1 \le j \le m$ given by the particular choice of attachments forced by $\underline{V}$. In the converse
direction, for every tree $\underline{T}$ with shape $T_j$, there exists a unique subtree $\underline{V} \subset \underline{T}$ with shape $V$ such that $\underline{U} := \underline{T}/\underline{V}$ has shape $U$. This then proves that

$$ (U \otimes \mu)(f_1, \dots, f_{i-1}, (V \otimes \nu)(f_i, \dots, f_{i+q-1}), f_{i+q}, \dots, f_{p+q-1}) = \pm (U \circ_i V \otimes \mu \circ_i \nu)(f_1 \otimes \dots \otimes f_{p+q-1}). \quad \square $$

**Proposition 5.13.** The dg $\mathbb{K}$-module $\text{Hom}_{\text{Seq}_k}(\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}})$ is endowed with the structure of a $\Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$-algebra.

*Proof.* By Lemma 5.12 and Theorem 3.11, the dg $\mathbb{K}$-module $\Sigma\mathcal{L}(\text{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}}))$ is endowed with the structure of a $\Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$-algebra. By taking the completion, we obtain that $\text{Hom}(\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}})$ is a $\Gamma\Lambda\mathcal{P}\mathcal{L}_\infty$-algebra. $\square$

From the definition of the *Brace* $\otimes_{\mathbb{H}}^{\mathcal{E}}$-algebra structure on $\text{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}})$, we deduce a first computation of $\text{Map}_{\mathcal{O}_p}(B^c(\mathcal{C}), \mathcal{P})$.

**Corollary 5.14.** We have the isomorphism of simplicial sets

$$ \text{Map}_{\mathcal{O}_p}(B^c(\mathcal{C}), \mathcal{P}) \simeq \mathcal{M}\mathcal{C}(\text{Hom}_{\text{Seq}_k}(\bar{\mathcal{C}} \otimes \Sigma^{-1} N_*(\Delta^\bullet), \bar{\mathcal{P}})). $$

Our goal is now to link this computation with the simplicial Maurer-Cartan set of $\text{Hom}_{\text{Seq}_k}(\bar{\mathcal{C}}, \bar{\mathcal{P}})$. Recall that $N_*(\Delta^n)$ has a basis given by increasing sequence of integers $0 \le a_0 < \dots < a_r \le n$ which we denote by $\underline{a}_0 \dots \underline{a}_r$. We let $\mathcal{B}_n$ to be this basis.

**Lemma 5.15.** Let $n \ge 0$. We set

$$ \phi_n : \mathcal{L}(\text{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}})) \xrightarrow{f} \mathcal{L}(\text{Hom}(\bar{\mathcal{C}}, \bar{\mathcal{P}})) \otimes N^*(\Delta^n) \xrightarrow{\sum_{\underline{x} \in \mathcal{B}_n} f^{\underline{x}} \otimes \underline{x}^{\vee}} \mathcal{B}_n $$

where, for every $\underline{x} \in N_*(\Delta^n)$ and $f \in \mathcal{L}(\text{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}}))$, we denote by $f^{\underline{x}} \in \mathcal{L}(\text{Hom}(\bar{\mathcal{C}}, \bar{\mathcal{P}}))$ the map defined by $f^{\underline{x}}(c) = (-1)^{|c|} f(c)$ for every $c \in \bar{\mathcal{C}}$.

Then $\phi_n$ is an isomorphism of *Brace* $\otimes_{\mathbb{H}}^{\mathcal{E}}$-algebras.

Moreover, the sequence of isomorphisms $(\phi_n)_{n \ge 0}$ is compatible with the simplicial structures.

*Proof.* We first prove that $\phi_n$ commutes with the differentials. Let $f \in \mathcal{L}(\text{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}}))$. Then

$$ d(\phi_n(f)) = \sum_{\underline{x} \in \mathcal{B}_n} d(f^{\underline{x}}) \otimes \underline{x}^{\vee} + \sum_{\underline{x} \in \mathcal{B}_n} (-1)^{|f|+|\underline{x}|} f^{\underline{x}} \otimes d(\underline{x}^{\vee}). $$

For every $c \in \bar{\mathcal{C}}$, we have that

$$ d(f^{\underline{x}})(c) = (-1)^{|c|} d(f(c \otimes \underline{x})) - (-1)^{|f|+|\underline{x}|(|c|+1)} f(d(c) \otimes \underline{x}) $$

which gives

$$ d(f^{\underline{x}}) = d(f)^{\underline{x}} + (-1)^{|f|} f^{d(\underline{x})}. $$

We thus obtain

$$ d(\phi_n(f)) = \sum_{\underline{x} \in \mathcal{B}_n} d(f)^{\underline{x}} \otimes \underline{x}^{\vee} + \sum_{\underline{x} \in \mathcal{B}_n} (-1)^{|f|} \left( f^{d(\underline{x})} \otimes \underline{x}^{\vee} + (-1)^{|\underline{x}|} f^{\underline{x}} \otimes d(\underline{x}^{\vee}) \right). $$
It remains to prove that

$$ \sum_{\underline{x} \in \mathcal{B}_n} f^{d(\underline{x})} \otimes \underline{x}^{\vee} = - \sum_{\underline{x} \in \mathcal{B}_n} (-1)^{|\underline{x}|} f^{\underline{x}} \otimes d(\underline{x}^{\vee}). $$

For every $\underline{x} \in \mathcal{B}_n$, we write $d(\underline{x}) = \sum_{\underline{x} \in \mathcal{B}_n} \lambda_{\underline{x}}^{\underline{y}} \underline{y}$ where $\lambda_{\underline{x}}^{\underline{y}} \in \{-1; 0; 1\}$. We thus have, for every $\underline{y} \in \mathcal{B}_n$,

$$ d(\underline{y}^{\vee}) = -(-1)^{|\underline{y}|} \sum_{\underline{x} \in \mathcal{B}_n} \lambda_{\underline{x}}^{\underline{y}} \underline{x}^{\vee}. $$

We thus have

$$ \sum_{\underline{x} \in \mathcal{B}_n} f^{d(\underline{x})} \otimes \underline{x}^{\vee} = \sum_{\underline{x}, \underline{y} \in \mathcal{B}_n} \lambda_{\underline{x}}^{\underline{y}} f^{\underline{y}} \otimes \underline{x}^{\vee} \\ = - \sum_{\underline{y} \in \mathcal{B}_n} (-1)^{|\underline{y}|} f^{\underline{y}} \otimes d(\underline{y}^{\vee}). $$

At the end, we have obtained that

$$ d(\phi_n(f)) = \sum_{\underline{x} \in \mathcal{B}_n} d(f)^{\underline{x}} \otimes \underline{x}^{\vee} = \phi_n(d(f)) $$

so that $\phi_n$ commutes with the differentials.

Now, let $f_1, \dots, f_r \in \mathcal{L}(\mathrm{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}}))$ be homogeneous elements. Let $T \in \mathcal{PRT}$ be a canonical tree with $r$ vertices and $\mu \in \mathcal{E}(r)$. We have

$$ (T \otimes \mu)(\phi_n(f_1), \dots, \phi_n(f_r)) = \sum_{\underline{T} \in \mathrm{Shape}^{-1}(T)} \sum_{\underline{x}_1, \dots, \underline{x}_r \in \mathcal{B}_n} \pm (\gamma_{\underline{T}}) \circ (f_1^{\underline{x}_1} \otimes \cdots \otimes f_r^{\underline{x}_r}) \circ \Delta_{\underline{T}}) \otimes \mu^{N^*(\Delta^n)}(\underline{x}_1^{\vee}, \dots, \underline{x}_r^{\vee}), $$

where the sign is given by

$$ \prod_{i<j} (-1)^{|\underline{x}_i|(|f_j|+|\underline{x}_j|)} \times \prod_{j=1}^{r} (-1)^{|\mu|(|f_j|+|\underline{x}_j|)}. $$

Now, for every $\underline{x} \in \mathcal{B}_n$, we write

$$ \mu^{N^*(\Delta^n)}(\underline{x}) = \sum_{\underline{x}_1, \dots, \underline{x}_r \in \mathcal{B}_n} \lambda_{\underline{x}}^{\underline{x}_1, \dots, \underline{x}_r} \underline{x}_1 \otimes \cdots \otimes \underline{x}_r $$

where $\lambda_{\underline{x}}^{\underline{x}_1, \dots, \underline{x}_r} \in \{-1; 0; 1\}$ by definition of the interval cuts operations. This gives, for every $\underline{x}_1, \dots, \underline{x}_r \in \mathcal{B}_n$,

$$ \mu^{N^*(\Delta^n)}(\underline{x}_1^{\vee}, \dots, \underline{x}_r^{\vee}) = \sum_{\underline{x} \in \mathcal{B}_n} \pm \underline{x}^{\vee} $$

where the sign is given by

$$ \lambda_{\underline{x}}^{\underline{x}_1, \dots, \underline{x}_r} \prod_{i<j} (-1)^{|\underline{x}_i||\underline{x}_j|} \prod_{j=1}^{r} (-1)^{|\mu||\underline{x}_j|}. $$
We thus have

$$ (T \otimes \mu)(\phi_n(f_1), \dots, \phi_n(f_r)) = \sum_{\underline{T} \in \text{Shape}^{-1}(T)} \sum_{\underline{x}, \underline{x}_1, \dots, \underline{x}_r \in \mathcal{B}_n} \pm (\gamma_{(\underline{T})} \circ (f_1^{\underline{x}_1} \otimes \cdots \otimes f_r^{\underline{x}_r}) \circ \Delta_{(\underline{T})}) \otimes \underline{x}^{\vee} $$

where the sign is

$$ \lambda_{\underline{x}}^{\underline{x}_1, \dots, \underline{x}_r} \prod_{i<j} (-1)^{|\underline{x}_i||f_j|} \prod_{j=1}^{r} (-1)^{|\mu||f_j|}. $$

We now use that, for every $c \in \bar{\mathcal{C}}$,

$$ (f_1^{\underline{x}_1} \otimes \cdots \otimes f_r^{\underline{x}_r})(\Delta_{(\underline{T})}(c)) = \pm (f_1 \otimes \cdots \otimes f_r) \circ (\Delta_{(\underline{T})}(c) \tilde{\otimes} (\underline{x}_1 \otimes \cdots \otimes \underline{x}_r)) $$

where the sign is given by

$$ \prod_{i<j} (-1)^{|\underline{x}_i||f_j|} \times \prod_{j=1}^{r} (-1)^{|c||\underline{x}_j|}. $$

We deduce

$$ (T \otimes \mu)(\phi_n(f_1), \dots, \phi_n(f_r))(c) = \sum_{\underline{T} \in \text{Shape}^{-1}(T)} \sum_{\underline{x} \in \mathcal{B}_n} \pm \gamma_{(\underline{T})}((f_1 \otimes \cdots \otimes f_r)(\Delta_{(\underline{T})}(c) \tilde{\otimes} \mu^{N_*(\Delta^n)}(\underline{x}))) $$

where the sign is

$$ \prod_{j=1}^{r} (-1)^{|\mu||f_j|} \times (-1)^{|c|(|\underline{x}|+|\mu|)}, $$

since, for every $\underline{x}, \underline{x}_1, \dots, \underline{x}_r \in \mathcal{B}_n$ such that $\lambda_{\underline{x}}^{\underline{x}_1, \dots, \underline{x}_r} \neq 0$, we have $|\mu| - \sum_{i=1}^r |\underline{x}_i| = -|\underline{x}|$. We finally obtain

$$ (T \otimes \mu)(\phi_n(f_1), \dots, \phi_n(f_r)) = \sum_{\underline{x} \in \mathcal{B}_n} ((T \otimes \mu)(f_1, \dots, f_r))^{\underline{x}} \otimes \underline{x}^{\vee} = \phi_n((T \otimes \mu)(f_1, \dots, f_r)). $$

We thus have proved that $\phi_n : \mathcal{L}(\text{Hom}(\bar{\mathcal{C}} \otimes N_*(\Delta^n), \bar{\mathcal{P}))) \to \mathcal{L}(\text{Hom}(\bar{\mathcal{C}}, \bar{\mathcal{P}))) \otimes N^*(\Delta^n)$ is a morphism of complete $\mathcal{B}race \otimes_{\mathbb{H}} \mathcal{E}$-algebras, and $\phi$ is obviously a bijection, with as inverse

$$ \phi_n^{-1}(f \otimes \underline{x}^{\vee}) = (c \otimes \underline{x} \mapsto (-1)^{|c||\underline{x}|} f(c)) $$

for every $f \in \mathcal{L}(\text{Hom}(\bar{\mathcal{C}}, \bar{\mathcal{P}}))$ and $\underline{x} \in \mathcal{B}_n$.

The compatibility of the sequence $(\phi_n)_{n \ge 0}$ with the simplicial structures follows directly from the definition of the $\phi_n$'s. $\square$

In particular, the map $\phi_n$ induces an isomorphism of $\Gamma(\mathcal{P}re\mathcal{L}ie_\infty, -)$-algebras. We thus obtain Theorem D.

**Theorem 5.16.** We have the identity

$$ \mathrm{Map}_{\mathcal{O}_p}(\mathcal{B}^c(\mathcal{C}), \mathcal{P}) = \mathcal{M}\mathcal{C}_{\bullet}(\mathrm{Hom}_{\mathrm{Seq}_k}(\bar{\mathcal{C}}, \bar{\mathcal{P}})). $$
Proof. For every $n \ge 0$, since the morphism $\phi_n$ given in Lemma 5.15 preserves the filtrations, taking the completions gives an isomorphism

$$ \mathrm{Hom}_{\mathrm{Seq}_{\mathbb{K}}} (\bar{C} \otimes \Sigma^{-1} N_{*}(\Delta^{n}), \bar{P}) \xrightarrow{\simeq} \Sigma \mathrm{Hom}_{\mathrm{Seq}_{\mathbb{K}}} (\bar{C}, \bar{P}) \otimes N^{*}(\Delta^{n}) $$

of $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebras for every $n \ge 0$. Since this isomorphism preserves the simplicial structures, we obtain the theorem. $\square$

# 6 A mapping space in the category of symmetric connected operads

In this last section, we show that we can describe a mapping space in the category of symmetric and connected operads as the degree-wise Maurer-Cartan set of some complete $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra.

In §6.1, we recall the construction of the free operad functor in the category of symmetric connected operads and the model structure on the latter category.

In §6.2, we use the surjection cooperad $\mathbf{Sur}_{\mathbb{K}}$ to obtain a $\Sigma_{*}$-cofibrant replacement $B^{c}(\mathcal{C} \underset{\mathbb{H}}{\otimes} \mathbf{Sur}_{\mathbb{K}})$ of the cobar construction $B^{c}(\mathcal{C})$ associated to a symmetric cooperad $\mathcal{C}$ such that $\mathcal{C}(0) = 0$. We construct an explicit cosimplicial frame associated to $B^{c}(\mathcal{C} \underset{\mathbb{H}}{\otimes} \mathbf{Sur}_{\mathbb{K}})$.

In §6.3, we finally deduce Theorem E which gives a computation of the mapping spaces in the category of symmetric connected operads in terms of a degree-wise simplicial Maurer-Cartan set of some $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebras.

## 6.1 The free symmetric operad functor and the model structure on $\Sigma\mathcal{O}p^0$

In this subsection, we recall the construction of the free operad functor in the category $\Sigma\mathcal{O}p$ and recall the model structure on the category of symmetric connected operads $\Sigma\mathcal{O}p^0$.

We have a functor $- \otimes \Sigma : \mathrm{Seq}_{\mathbb{K}} \to \Sigma\mathrm{Seq}_{\mathbb{K}}$ defined, for every $M \in \mathrm{Seq}_{\mathbb{K}}$, by

$$ (M \otimes \Sigma)(n) = M(n) \otimes \mathbb{K}[\Sigma_n], $$

where $M(n) \otimes \mathbb{K}[\Sigma_n]$ is endowed with the $\Sigma_n$ action defined, for every $m \in M(n)$ and $\sigma, \tau \in \Sigma_n$, by

$$ \sigma \cdot (m \otimes \tau) = m \otimes \sigma\tau. $$

The functor $- \otimes \Sigma$ fits in an adjunction

$$ - \otimes \Sigma : \mathrm{Seq}_{\mathbb{K}} \underset{\leftarrow}{\longrightarrow} \Sigma\mathrm{Seq}_{\mathbb{K}} : \omega, $$

where $\omega : \Sigma\mathrm{Seq}_{\mathbb{K}} \to \mathrm{Seq}_{\mathbb{K}}$ is the functor which forgets the symmetric groups actions. We have the following result.
**Proposition 6.1** (see [Fre09a, Proposition 11.4.A]). The category $\Sigma\mathrm{Seq}_{\mathbb{K}}$ is endowed with a cofibrantly generated model category structure such that the forgetful functor $\omega: \Sigma\mathrm{Seq}_{\mathbb{K}} \to \mathrm{Seq}_{\mathbb{K}}$ creates weak-equivalences and fibrations. Cofibrations are given by the left lifting property.

In fact, the category $\Sigma\mathrm{Op}$ does not have a model structure but rather a semi-model structure (see [Spi01, Theorem 3]). We instead consider the subcategory $\Sigma\mathrm{Op}^0$ of operads $\mathcal{P}$ such that $\mathcal{P}(0) = 0$. Such an operad is said to be connected. We also denote by $\Sigma\mathrm{Seq}_{\mathbb{K}}^0$ the subcategory of $\Sigma\mathrm{Seq}_{\mathbb{K}}$ given by symmetric sequences $M$ such that $M(0) = 0$. As for the non symmetric context, we are searching for a convenient adjunction

$$ \mathcal{F}: \Sigma\mathrm{Seq}_{\mathbb{K}}^0 \mathrel{\substack{\longleftarrow \\ \longrightarrow}} \Sigma\mathrm{Op}^0 : \omega $$

where $\omega: \Sigma\mathrm{Op}^0 \to \Sigma\mathrm{Seq}_{\mathbb{K}}^0$ is the functor which forgets the operad structure. To achieve this, recall that the functor $\otimes \Sigma: \mathrm{Seq}_{\mathbb{K}} \to \Sigma\mathrm{Seq}_{\mathbb{K}}$ restricts to a functor $\otimes \Sigma: \mathrm{Op} \to \Sigma\mathrm{Op}$ where, for every $\mathcal{P} \in \mathrm{Op}$, the operad structure on $\mathcal{P} \otimes \Sigma$ is defined by

$$ \gamma((f \otimes \sigma) \otimes (g_1 \otimes \tau_1) \otimes \cdots \otimes (g_n \otimes \tau_n)) = \pm \gamma(f \otimes g_{\sigma^{-1}(1)} \otimes \cdots \otimes g_{\sigma^{-1}(n)}) \otimes \sigma(\tau_1, \dots, \tau_n), $$

where we consider the composite of $\sigma$ with $\tau_1, \dots, \tau_n$ (see after Lemma 1.34).

**Definition 6.2.** Consider the operad $\mathcal{Tree}$ defined in Lemma 5.3. We set

$$ \Sigma\mathcal{Tree} = \mathcal{Tree} \otimes \Sigma. $$

For every $\mathcal{T} \in \Sigma\mathcal{Tree}$, we denote by $V_{\mathcal{T}}$ the set of vertices, and by $\mathrm{val}_{\mathcal{T}}(v)$ the number of ingoing arrows on a vertex $v \in V_{\mathcal{T}}$.

The elements of $\Sigma\mathcal{Tree}(n)$ can then be seen as trees with inputs endowed with a choice of labeling on the inputs. We identify such a choice with a permutation in $\Sigma_n$. For instance, if we consider the tree with inputs $\mathcal{T} \in \mathcal{Tree}$ given in Definition 5.1, then

<figure><img src="image_28.png" /></figure>

For every $M \in \Sigma\mathrm{Seq}_{\mathbb{K}}$, we consider the sequence

$$ k \mapsto \bigoplus_{n \ge 0} \bigoplus_{\mathcal{T} \in \mathcal{P}\mathcal{R}\mathcal{T}_k(n)} \bigoplus_{\sigma \in \Sigma_k} (\mathcal{T} \otimes \sigma) \otimes \bigotimes_{i=1}^{n} M(\mathrm{val}_{\mathcal{T}}(i)). $$

This sequence is endowed with the structure of a non-symmetric operad given by the operadic structure of $\Sigma\mathcal{Tree}$ and the concatenation of elements in the tensor algebra of $\bigoplus_{n \ge 0} M(n)$. In order to endow this sequence with the structure of a symmetric operad, we need to identify some
where we denote by $(\Sigma \mathcal{O} p^c)^0$ the subcategory of $\Sigma \mathcal{O} p^c$ given by connected cooperads, and where $\omega : (\Sigma \mathcal{O} p^c)^0 \to \Sigma \mathrm{Seq}_{\mathbb{K}}^0$ is the functor which forgets the cooperad structure.

As for the non symmetric context, we consider operadic (resp. cooperadic) compositions (resp. cocompositions) shaped on trees with inputs. Let $\mathcal{P}$ be an augmented operad $\mathcal{P} \simeq I \oplus \bar{\mathcal{P}}$ and $\mathcal{C}$ be a coaugmented cooperad $\mathcal{C} \simeq I \oplus \bar{\mathcal{C}}$ such that $\mathcal{P}(0) = \mathcal{C}(0) = 0$ and $\mathcal{P}(1) = \mathcal{C}(1) = \mathbb{K}$. By the universal property satisfied by $\mathcal{F}$, we have a unique operad morphism $\mathcal{F}(\mathcal{P}) \to \mathcal{P}$ which reduces to the identity on $\mathcal{P} \subset \mathcal{F}(\mathcal{P})$. Analogously, we have a unique cooperad morphism $\mathcal{C} \to \mathcal{F}^c(\mathcal{C})$ whose projection on $\mathcal{C}$ is given by the identity on $\mathcal{C}$.

**Definition 6.7.** Let $\underline{T}$ be a tree with inputs. We define $\gamma_{(\underline{T})} : \mathcal{F}_{(\underline{T})}(\bar{\mathcal{P}}) \to \bar{\mathcal{P}}$ and $\Delta_{(\underline{T})} : \bar{\mathcal{C}} \to \mathcal{F}^c(\bar{\mathcal{C}})$ by the composites

$$ \gamma_{(\underline{T})} : \mathcal{F}_{(\underline{T})}(\bar{\mathcal{P}}) \xrightarrow{\gamma} \mathcal{F}_{(\underline{T})}(\mathcal{P}) \xrightarrow{\mathcal{P}} \bar{\mathcal{P}}; $$

$$ \Delta_{(\underline{T})} : \bar{\mathcal{C}} \xrightarrow{\bar{\mathcal{C}}} \mathcal{C} \xrightarrow{\Delta} \mathcal{F}_{(\underline{T})}^c(\mathcal{C}) \xrightarrow{\mathcal{F}_{(\underline{T})}^c(\bar{\mathcal{C}})} \mathcal{F}_{(\underline{T})}^c(\bar{\mathcal{C}}). $$

For every $p, q, n, m \ge 0$ and $1 \le i \le p$, as for the non symmetric context, we define a morphism

$$ \bullet_i : \Sigma \underline{\mathcal{T}} \mathit{ree}_p(n) \otimes \Sigma \underline{\mathcal{T}} \mathit{ree}_q(m) \to \Sigma \underline{\mathcal{T}} \mathit{ree}_p(n+m-1) $$

defined as follows. Let $\underline{U} \in \Sigma \underline{\mathcal{T}} \mathit{ree}_p(n)$ and $\underline{V} \in \Sigma \underline{\mathcal{T}} \mathit{ree}_q(m)$. If $\mathrm{val}_{\underline{U}}(i) \neq q$, we set $\underline{U} \bullet_i \underline{V} = 0$. Else, we define $\underline{U} \bullet_i \underline{V}$ as the tree obtained by changing the $i$-th vertex of $\underline{U}$ into the tree $\underline{V}$. The attachment of the $q$ arrows on the $i$-th vertex of $\underline{U}$ on the tree $\underline{V}$ are given following the order of the labeling in $\underline{V}$.

As for the non symmetric context, we have the two following lemmas.

**Lemma 6.8.** Let $\underline{T}$ be a tree with inputs and $\underline{S}$ be a subtree of $\underline{T}$. Then $\underline{T}/\underline{S} \bullet_S \underline{S} = \underline{T}$.

**Lemma 6.9.** Let $\underline{T} \in \underline{\mathcal{P}}\underline{\mathcal{T}}\underline{T}(n)$ and $\underline{S} \subset \underline{T}$. Then

$$ \gamma_{(\underline{T}/\underline{S})} \circ_S \gamma_{(\underline{S})} = \gamma_{(\underline{T})} ; \quad \Delta_{(\underline{T}/\underline{S})} \circ_S \Delta_{(\underline{S})} = \Delta_{(\underline{T})} $$

in the endomorphism operad $\mathrm{End}_{\oplus_{n \ge 2} \mathcal{P}(n)}$ and in the coendomorphism operad $\mathrm{CoEnd}_{\oplus_{n \ge 2} \mathcal{C}(n)}$ respectively.

## 6.2 A cosimplicial frame for $\mathcal{B}^c(\mathcal{C} \otimes \mathbf{Sur}_{\mathbb{K}})$

Let $\mathbf{Sur}_{\mathbb{K}}$ be the surjection cooperad defined in [BCN23, Theorem A.1]. This cooperad is actually equal, as a symmetric sequence, to the surjection operad $\chi$ recalled in §1.4. Note however that the cooperad structure on $\mathbf{Sur}_{\mathbb{K}}$ is not the cooperad structure obtained by dualizing the operad structure on $\chi$. We have a weak-equivalence $\mathcal{B}^c(\mathcal{C}) \xrightarrow{\sim} \mathcal{B}^c(\mathcal{C} \otimes \mathbf{Sur}_{\mathbb{K}})$, which provides a $\Sigma_*$-cofibrant replacement of $\mathcal{B}^c(\mathcal{C})$.

In this section, we construct an explicit cosimplicial frame associated to $\mathcal{B}^c(\mathcal{C} \otimes \mathbf{Sur}_{\mathbb{K}})$ for every symmetric coaugmented cooperad $\mathcal{C}$. To be more precise, we construct a twisted differential
$\partial^n : \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n)) \to \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$ by an inductive process analogue
to the one given in Theorem 3.7.

For every $k \ge 1$, we define $\Phi_n^0, H_n^0 : (\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))^{\otimes k} \to (\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))^{\otimes k}$
by

$$
\Phi_n^0 = id_{\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}}} \otimes \tilde{\phi}_n^0;
$$

$$
H_n^0 = id_{\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}}} \otimes \tilde{h}_n^0,
$$

where we use the tensor product $\tilde{\otimes}$ defined in Definition 1.1, and the morphisms $\phi_n^0, h_n^0 : (\Sigma^{-1}N_*(\Delta^n))^{\otimes k} \to (\Sigma^{-1}N_*(\Delta^n))^{\otimes k}$ defined after Lemma 3.6. We extend $\Phi_n^0$ and $H_n^0$ on $\mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1}N_*(\Delta^n))$ by using the identification given in Proposition 6.5. Note however that the morphism $H_n^0$ does not preserve the action of the symmetric groups on $\mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1}N_*(\Delta^n))$.

Since the action of the symmetric groups on $\mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}}$ is free, we can chose an explicit choice of representatives for the orbits. For every $n \ge 1$, we let $\mathbf{Sur}_{\mathbb{K}}^{id}(n)$ to be the dg $\mathbb{K}$-module generated by surjections $u \in \mathbf{Sur}_{\mathbb{K}}(n)$ of the form

$$
u = \left|
\begin{array}{cccc}
u_0(1) & \cdots & u_0(r_0 - 1) & u_0(r_0) \\
\vdots & & \vdots & \\
u_{d-1}(1) & \cdots & u_{d-1}(r_{d-1} - 1) & u_{d-1}(r_{d-1}) \\
u_d(1) & \cdots & u_d(r_d - 1) & u_d(r_d)
\end{array}
\right|
$$

with

$$
u_0(1) \cdots u_0(r_0 - 1) \cdots u_{d-1}(1) \cdots u_{d-1}(r_{d-1} - 1) u_d(1) \cdots u_d(r_d) = 1 \cdots n.
$$

We thus have an isomorphism of graded symmetric sequences $\mathbf{Sur}_{\mathbb{K}} \simeq \mathbf{Sur}_{\mathbb{K}}^{id} \otimes \Sigma$. This gives an
isomorphism

$$
\mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}} \simeq (\mathcal{C} \otimes \mathbf{Sur}_{\mathbb{K}}^{id}) \otimes \Sigma
$$

defined by sending $c \otimes (u \otimes \sigma) \in \mathcal{C} \otimes_{\mathbb{H}} (\mathbf{Sur}_{\mathbb{K}}^{id} \otimes \Sigma)$ to $(\sigma^{-1} \cdot c \otimes u) \otimes \sigma \in (\mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}}^{id}) \otimes \Sigma$. Note that
the differential $d_{\mathcal{C}}$ preserves such a decomposition, but not the differential $d_{\mathbf{Sur}_{\mathbb{K}}}$, since it does not
preserve $\mathbf{Sur}_{\mathbb{K}}^{id}$. For every $l \ge 0$, we let $F_l \mathbf{Sur}_{\mathbb{K}}^{id}$ to be the sequence given by surjections of degree
equal or less than $l$ in $\mathbf{Sur}_{\mathbb{K}}^{id}$ and we set $F_l(\mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}}^{id}) = \mathcal{C} \otimes_{\mathbb{H}} F_l \mathbf{Sur}_{\mathbb{K}}^{id}$.

For every $n \ge 0$, we aim to define a derivation of operads on $\mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$ which reduces to the internal differential of $\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n)$ on trees with only one vertex. We denote by $d_{\mathcal{C}}$, $d_{\mathbf{Sur}_{\mathbb{K}}}$ and $d_{\Sigma^{-1}N_*(\Delta^n)}$ the corresponding differentials on $\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n)$. Let $\mathcal{C}_{ns} := \mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}}^{id} \otimes \mathrm{id}$. We construct $\beta^n : \overline{\mathcal{C}}_{ns} \otimes \Sigma^{-1} N_*(\Delta^n) \to \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$ which reduces to $d_{\mathbf{Sur}_{\mathbb{K}}} + d_{\Sigma^{-1}N_*(\Delta^n)}$ on trees with one vertex and which is such that

$$
d_{\mathcal{C}}(\beta^n) + \partial^n \beta^n = 0,
$$
where $\partial^n : \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n)) \to \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$ is the morphism obtained from $\beta^n$ by applying the Leibniz rule in $\mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$.

In the following, we endow the sequence of operads $\mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^\bullet))$ with the structure of a cosimplicial set with as coface maps (resp. codegeneracy maps) the coface maps (resp. codegeneracy maps) of the cosimplicial set $\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^\bullet)$ taken tensor-wise. Recall that the cosimplicial relations are given by the following:

* If $i < j$, then $d^j d^i = d^i d^{j-1}$;

* If $i < j$, then $s^j d^i = d^i s^{j-1}$;

* $s^j d^j = s^j d^{j+1} = id$;

* If $i > j + 1$, then $s^j d^i = d^{i-1} s^j$;

* If $i \le j$, then $s^j s^i = s^i s^{j+1}$.

Note that we have an extra codegeneracy $s^{-1} : N_*(\Delta^n) \to N_*(\Delta^{n-1})$ defined for every $0 \le a_0 < \dots < a_r \le n$ by $s^{-1}(a_0 \cdots a_r) = (a_0 - 1) \cdots (a_r - 1)$, with the convention $s^{-1}(a_0 \cdots a_r) = 0$ if $a_0 = 0$. One can easily check that the above relations are still satisfied with the addition of this degeneracy.

**Construction 6.10.** We define a sequence of degree $-1$ morphisms $\beta^n : \overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n) \to \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$ by induction on $n \ge 0$. Let $\partial^n : \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n)) \to \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n))$ be the morphism obtained from $\beta^n$ by the Leibniz rule.

We let $\beta^0$ to be such that $d_{\overline{\mathcal{C}}} + \partial^0$ is the differential of the cobar construction $B^c(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}}) \simeq \mathcal{F}(\overline{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^0))$. We set $\beta_{(0)}^n = d_{\overline{\mathbf{Sur}}_{\mathbb{K}}} + d_{\Sigma^{-1} N_*(\Delta^n)}$. For every $k \ge 1$, we define the component $\beta_{(k)}^n$ of $\beta^n$ given by trees with $k+1$ vertices by induction on $k$ and $n$. More precisely, we define $\beta_{(k)}^n$ on $F_l \overline{\mathcal{C}}_{ns} \otimes \Sigma^{-1} N_*(\Delta^n)$ by induction on $l \ge 0$. Let $c \in F_l \overline{\mathcal{C}}_{ns}$ and $\underline{x} \in N_*(\Delta^n)$ be a basis element.

* If $\underline{x} \ne 0 \cdots n$, let $0 \le i \le n$ be such that $\underline{x} = 0 \cdots (i-1)a_i \cdots a_r$ with $i < a_i < \cdots < a_r \le n$. We set

$$ \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) = d^i \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} s^{i-1} \underline{x}); $$

* If $\underline{x} = 0 \cdots n$, we set

$$ \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = (-1)^{|c|} H_n^0 d^0 \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{0 \cdots (n-1)}) \\ - H_n^0 \beta_{(k)}^n (d_{\overline{\mathbf{Sur}}_{\mathbb{K}}}(c) \otimes \Sigma^{-1} \underline{0 \cdots n}) - \sum_{\substack{p+q=k \\ p,q \ne 0}} H_n^0 \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}). $$

The morphism $\beta_{(k)}^n$ is then extended on $F_l \overline{\mathcal{C}}_{ns} \otimes \Sigma \otimes \Sigma^{-1} N_*(\Delta^n)$ by symmetry.
**Lemma 6.11.** For every $n, k \ge 0$, we have

$$ \forall 0 \le j \le n-1, d^j \beta_{(k)}^{n-1} = \beta_{(k)}^n d^j; $$

$$ \forall 0 \le j \le n, s^j \beta_{(k)}^n = \beta_{(k)}^{n-1} s^j, $$

where we consider the morphisms $\beta_{(k)}^n$ defined in Construction 6.10.

*Proof.* Since the coface maps and codegeneracy maps preserve the action of the symmetric groups on $\bar{\mathcal{C}} \otimes \bar{\mathbf{Sur}}_k \otimes \Sigma^{-1} N_*(\Delta^n)$, it is sufficient to prove it on $\bar{\mathcal{C}}_{ns} \otimes \Sigma^{-1} N_*(\Delta^n)$. Let $c \in \bar{\mathcal{C}}_{ns}$ and let $\underline{x} \in N_*(\Delta^n)$ be a basis element. We prove the formulas by induction on $n, k, l \ge 0$. The assertion is obviously true for $n=0$, and for $n \ge 1$ and $k=0$. We now suppose that $n, k \ge 1$.

We prove the first line of the lemma. Let $0 \le j \le n-1$. If $\underline{x} = 0 \cdots (n-1)$, then we indeed have $d^j \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}\underline{x}) = \beta_{(k)}^n d^j(c \otimes \Sigma^{-1}\underline{x})$ by definition of $\beta_{(k)}^n$. Suppose now that $\underline{x} \ne 0 \cdots (n-1)$. Then there exists $0 \le i \le n-1$ such that $\underline{x} = 0 \cdots (i-1)a_i \cdots a_r$ with $i < a_i < \cdots < a_r \le n-1$. If $j=i$, then

$$ \begin{aligned} \beta_{(k)}^n d^j(c \otimes \Sigma^{-1}\underline{x}) &= \beta_{(k)}^n(c \otimes \Sigma^{-1}d^i\underline{x}) \\ &= d^i \beta_{(k)}^n(c \otimes \Sigma^{-1}s^{i-1}d^i\underline{x}) \\ &= d^j \beta_{(k)}^n(c \otimes \Sigma^{-1}\underline{x}), \end{aligned} $$

since $s^{i-1}d^i = \text{id}$. If $j > i$, then

$$ \begin{aligned} \beta_{(k)}^n d^j(c \otimes \Sigma^{-1}\underline{x}) &= \beta_{(k)}^n(c \otimes \Sigma^{-1}d^j\underline{x}) \\ &= d^i \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}s^{i-1}d^j\underline{x}) \\ &= d^i \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}d^{j-1}s^{i-1}\underline{x}), \end{aligned} $$

since $s^{i-1}d^j = d^{j-1}s^{i-1}$. By induction hypothesis on $n-1$, we deduce

$$ \begin{aligned} \beta_{(k)}^n d^j(c \otimes \Sigma^{-1}\underline{x}) &= d^i d^{j-1} \beta_{(k)}^{n-2}(c \otimes \Sigma^{-1}s^{i-1}\underline{x}) \\ &= d^j d^i \beta_{(k)}^{n-2}(c \otimes \Sigma^{-1}s^{i-1}\underline{x}) \\ &= d^j \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}\underline{x}). \end{aligned} $$

If $j < i$, then

$$ \begin{aligned} \beta_{(k)}^n d^j(c \otimes \Sigma^{-1}\underline{x}) &= d^j \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}s^{j-1}d^j\underline{x}) \\ &= d^j \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}\underline{x}), \end{aligned} $$

since $s^{j-1}d^j = \text{id}$. We thus have proved that $d^j \beta_{(k)}^{n-1} = \beta_{(k)}^n d^j$.

We now prove the second line of the lemma. Let $\underline{x} \in N_*(\Delta^n)$. We first consider $\underline{x} = 0 \cdots n$. Then, by definition of $\beta_{(k)}^n$,

$$ \begin{aligned} s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) &= (-1)^{|c|} s^j H_n^0 d^0 \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{0 \cdots (n-1)}) \\ &\quad - s^j H_n^0 \beta_{(k)}^n (d_{\bar{\mathbf{Sur}}_k}(c) \otimes \Sigma^{-1} \underline{0 \cdots n}) - \sum_{\substack{p+q=k \\ p,q \ne 0}} s^j H_n^0 \partial_{(p)} \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}). \end{aligned} $$
Since $s^j H_n^0 = H_{n-1}^0 s^j$, we have

$$s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = (-1)^{|c|} H_{n-1}^0 s^j d^0 \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{0 \cdots (n-1)}) \\ - H_{n-1}^0 s^j \beta_{(k)}^n (d_{\mathbf{Sur}_k}(c) \otimes \Sigma^{-1} \underline{0 \cdots n}) - \sum_{\substack{p+q=k \ p,q \neq 0}} H_{n-1}^0 s^j \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}).$$

By induction hypothesis, we have that $s^j$ commutes with $\partial_{(p)}^n \beta_{(q)}^n$ for every $p, q \neq 0$ such that $p+q=n$. Since $s^j(\underline{0 \cdots n}) = 0$, we have that the sum in the above identity is 0. Analogously, by induction hypothesis on $l \ge 0$, we have that $s^j \beta_{(k)}^n (d_{\mathbf{Sur}_k}(c) \otimes \Sigma^{-1} \underline{0 \cdots n}) = 0$. If $j > 0$, we have $s^j d^0 = d^0 s^{j-1}$ so that the first term is also 0. If $j=0$, then $s^j d^0 = id$. We thus have

$$s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = (-1)^{|c|} H_{n-1}^0 \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{0 \cdots (n-1)}).$$

By definition of $\beta_{(k)}^{n-1}$, the term $\beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{0 \cdots (n-1)})$ is in the image of $H_{n-1}^0$. Since $H_{n-1}^0 H_{n-1}^0 = 0$, we obtain $s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = 0$. We thus have proved that $s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = \beta_{(k)}^n s^j (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = 0$. Suppose now that $\underline{x} \neq \underline{0 \cdots n}$. Then there exists $0 \le i \le n$ such that $\underline{x} = \underline{0 \cdots (i-1)a_i \cdots a_r}$ with $i < a_i < \cdots < a_r \le n$. We thus have

$$s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) = s^j d^i \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{s^{i-1} x}).$$

If $i < j$, then $s^j d^i = d^i s^{j-1}$ so that

$$s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) = d^i s^{j-1} \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{s^{i-1} x}).$$

By induction hypothesis on $n-1$, we obtain

$$\begin{align*}
s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) &= d^i \beta_{(k)}^{n-2} (c \otimes \Sigma^{-1} \underline{s^{j-1} s^{i-1} x}) \\
&= d^i \beta_{(k)}^{n-2} (c \otimes \Sigma^{-1} \underline{s^{i-1} s^j x}) \\
&= \beta_{(k)}^n s^j (c \otimes \Sigma^{-1} \underline{x}).
\end{align*}$$

If $i = j, j+1$, then

$$s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) = \beta_{(k)}^{n-1} (c \otimes \Sigma^{-1} \underline{s^{i-1} x}).$$

Since we have $s^{i-1} \underline{x} = s^i \underline{x}$, in any case, this gives

$$s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) = \beta_{(k)}^{n-1} s^j (c \otimes \Sigma^{-1} \underline{x}).$$

If $i > j + 1$, then, by induction hypothesis on $n-1$,

$$\begin{align*}
s^j \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{x}) &= d^{i-1} \beta_{(k)}^{n-2} (c \otimes \Sigma^{-1} \underline{s^j s^{i-1} x}) \\
&= d^{i-1} \beta_{(k)}^{n-2} (c \otimes \Sigma^{-1} \underline{s^{i-2} s^j x}) \\
&= 0 \\
&= \beta_{(k)}^n s^j (c \otimes \Sigma^{-1} \underline{x}),
\end{align*}$$

since $s^j \underline{x} = 0$. At the end, we have proved that $s^j \beta_{(k)}^n = \beta_{(k)}^{n-1} s^j$ and thus the lemma. $\square$
**Remark 6.12.** In particular, this lemma implies that

$$ \beta_{(k)}^{n}(c \otimes \Sigma^{-1}\underline{0 \cdots n}) = - \sum_{\substack{p+q=k \\ p \neq 0}} H_{n}^{0} \partial_{(p)}^{n} \beta_{(q)}^{n}(c \otimes \Sigma^{-1}\underline{0 \cdots n}) $$

for every $n, k \ge 0$ and $c \in \overline{C}_{ns}$. Indeed, we have

$$ \begin{aligned} \partial_{(k)}^{n} \beta_{(0)}^{n}(c \otimes \Sigma^{-1}\underline{0 \cdots n}) &= -(-1)^{|c|} \sum_{i=0}^{n} (-1)^{i} \beta_{(k)}^{n}(c \otimes \Sigma^{-1}\underline{0 \cdots \hat{i} \cdots n}) \\ &\quad + \beta_{(k)}^{n}(d\mathbf{Sur}_{\mathcal{K}}(c) \otimes \Sigma^{-1}\underline{0 \cdots n}) \\ &= -(-1)^{|c|} \sum_{i=0}^{n} (-1)^{i} d^{i} \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}\underline{0 \cdots (n-1)}) \\ &\quad + \beta_{(k)}^{n}(d\mathbf{Sur}_{\mathcal{K}}(c) \otimes \Sigma^{-1}\underline{0 \cdots n}). \end{aligned} $$

By an immediate computation, if $i \neq 0$, then $H_{n}^{0}d^{i} = H_{n-1}^{0}d^{i}$. Since $\beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}\underline{0 \cdots (n-1)})$ is in the image of $H_{n-1}^{0}$ by construction, and that $H_{n-1}^{0}H_{n-1}^{0} = 0$, we obtain

$$ H_{n}^{0} \partial_{(k)}^{n} \beta_{(0)}^{n}(c \otimes \Sigma^{-1}\underline{0 \cdots n}) = -(-1)^{|c|} H_{n}^{0} d^{0} \beta_{(k)}^{n-1}(c \otimes \Sigma^{-1}\underline{0 \cdots (n-1)}) + H_{n}^{0} \beta_{(k)}^{n}(d\mathbf{Sur}_{\mathcal{K}}(c) \otimes \Sigma^{-1}\underline{0 \cdots n}) $$

which proves the above formula.

**Theorem 6.13.** Let $n \ge 0$. The morphism $\partial^n : \mathcal{F}(\overline{\mathcal{C}} \otimes \overline{\mathbf{Sur}}_{\mathcal{K}} \otimes \Sigma^{-1}N_*(\Delta^n)) \to \mathcal{F}(\overline{\mathcal{C}} \otimes \overline{\mathbf{Sur}}_{\mathcal{K}} \otimes \Sigma^{-1}N_*(\Delta^n))$ is such that $d_C + \partial^n$ is a derivation of operads. Moreover, the sequence $\partial^\bullet : \mathcal{F}(\overline{\mathcal{C}} \otimes \overline{\mathbf{Sur}}_{\mathcal{K}} \otimes \Sigma^{-1}N_*(\Delta^\bullet)) \to \mathcal{F}(\overline{\mathcal{C}} \otimes \overline{\mathbf{Sur}}_{\mathcal{K}} \otimes \Sigma^{-1}N_*(\Delta^\bullet))$ is a morphism of cosimplicial sets.

*Proof.* By Lemma 6.11, the morphisms $\partial^\bullet$ preserve the cosimplicial structure of $\mathcal{F}(\overline{\mathcal{C}} \otimes \overline{\mathbf{Sur}}_{\mathcal{K}} \otimes \Sigma^{-1}N_*(\Delta^\bullet))$. We need to prove that $d_C + \partial^n$ is a differential of operads. This is equivalent to prove that

$$ d_C(\beta^n) + \partial^n \beta^n = 0 $$

for every $n \ge 0$. By an immediate induction, we have $d_C(\beta^n) = 0$. It remains to prove that $\partial^n \beta^n = 0$. This is equivalent to prove that

$$ \sum_{p+q=k} \partial_{(p)}^{n} \beta_{(q)}^{n} = 0 $$

for every $k \ge 0$. We prove it on $F_l \overline{C}_{ns} \otimes \Sigma^{-1}N_*(\Delta^n)$ by induction on $n, k, l \ge 0$. Since all the maps are symmetric by construction, we can prove it on $\overline{C}_{ns} \otimes \Sigma^{-1}N_*(\Delta^n)$. It is true for $n = 0$, and for $n \ge 1$ and $k = 0$. We now suppose that $n \ge 1$ and $k \ge 1$. Let $c \in F_l \overline{C}_{ns}$ and $\underline{x} \in N_*(\Delta^n)$ be a basis element. If $\underline{x} \neq \underline{0} \cdots \underline{n}$, then there exists $0 \le i \le n-1$ and $\underline{y} \in N_*(\Delta^{n-1})$ such that $d^i \underline{y} = \underline{x}$. Since the morphisms $d_C + \partial^\bullet$ are compatible with the cosimplicial structure of $\mathcal{F}(\overline{\mathcal{C}} \otimes \overline{\mathbf{Sur}}_{\mathcal{K}} \otimes \Sigma^{-1}N_*(\Delta^n))$ by Lemma 6.11, we have

$$ \sum_{p+q=k} \partial_{(p)}^{n} \beta_{(q)}^{n}(c \otimes \Sigma^{-1}\underline{x}) = \sum_{p+q=k} d^{i} \partial_{(p)}^{n-1} \beta_{(q)}^{n-1}(c \otimes \Sigma^{-1}\underline{y}) $$
which is 0 by induction hypothesis on $n-1$. Suppose now that $\underline{x} = \underline{0\cdots n}$. By using that

$$ \partial_{(0)}^n H_n^0 + H_n^0 \partial_{(0)}^n = id - \Phi_n^0, $$

we have

$$ \begin{aligned} \partial_{(0)}^n \beta_{(k)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = & \sum_{\substack{p+q=k \\ p \neq 0}} H_n^0 \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) \\ & - \sum_{\substack{p+q=k \\ p \neq 0}} \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) + \sum_{\substack{p+q=k \\ p \neq 0}} \Phi_n^0 \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}). \end{aligned} $$

By Lemma 6.11, the morphism $\Phi_n^0$ commutes with the $\partial^n$'s. We thus have that the last sum is 0, since $\phi_n^0(\underline{0\cdots n}) = 0$ because $n \ge 1$. Now, we claim that

$$ \sum_{\substack{p+q=k \\ p \neq 0}} \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = 0. $$

We write

$$ \sum_{\substack{p+q=k \\ p \neq 0}} \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = \partial_{(0)}^n \partial_{(k)}^n \beta_{(0)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) + \sum_{\substack{p+q=k \\ p,q \neq 0}} \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}). $$

We first deal with the sum at the right hand-side. Since $p < k$, we can use our induction hypothesis on $p$ to obtain

$$ \sum_{\substack{p+q=k \\ p,q \neq 0}} \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = - \sum_{\substack{s+t=k \\ p,q \neq 0}} \sum_{\substack{s+t=p \\ s \neq 0}} \partial_{(s)}^n \partial_{(t)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}). $$

By a variable substitution, this gives

$$ \sum_{\substack{p+q=k \\ p,q \neq 0}} \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = - \sum_{\substack{s+t=k \\ s,t \neq 0}} \partial_{(s)}^n \left( \sum_{\substack{p+q=t \\ q \neq 0}} \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) \right). $$

Now, since applying $\beta_{(0)}^n$ on $c \otimes \Sigma^{-1} \underline{0\cdots n}$ allows us to apply our induction hypothesis, we have

$$ \partial_{(0)}^n \partial_{(k)}^n \beta_{(0)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = - \sum_{\substack{s+t=k \\ s,t \neq 0}} \partial_{(s)}^n \partial_{(t)}^n \beta_{(0)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}). $$

At the end, we obtain that

$$ \sum_{\substack{p+q=k \\ p,q \neq 0}} \partial_{(0)}^n \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = - \sum_{\substack{s+t=k \\ s,t \neq 0}} \partial_{(s)}^n \left( \sum_{\substack{p+q=t \\ p,q \neq 0}} \partial_{(p)}^n \beta_{(q)}^n (c \otimes \Sigma^{-1} \underline{0 \cdots n}) \right), $$
and this last sum is 0 by induction hypothesis on $t < k$. We thus have proved that

$$ \partial_{(0)}^{n} \beta_{(k)}^{n} (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = - \sum_{\substack{p+q=k \\ p \neq 0}} \partial_{(p)}^{n} \beta_{(q)}^{n} (c \otimes \Sigma^{-1} \underline{0 \cdots n}), $$

which is equivalent to

$$ \sum_{p+q=k} \partial_{(p)}^{n} \beta_{(q)}^{n} (c \otimes \Sigma^{-1} \underline{0 \cdots n}) = 0. $$

The theorem is proved. □

Under the assumption of the existence of $C_{ns}$, and provided that $\mathcal{C}$ is a $\Sigma_*$-cofibrant cooperad, then we can construct a cosimplicial frame associated to $B^c(\mathcal{C})$ as in Theorem 5.11.

**Theorem 6.14.** Let $\mathcal{C}$ be a symmetric cooperad. Then

$$ B^c(\mathcal{C} \otimes \mathrm{Sur}_{\mathbb{K}}) \otimes \Delta^\bullet := (\mathcal{F}(\bar{\mathcal{C}} \otimes \overline{\mathrm{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^\bullet)), \partial^\bullet) $$

where $\partial^\bullet$ is the twisting derivation constructed in Theorem 6.13 is a cosimplicial frame associated to $B^c(\mathcal{C})$.

*Proof.* The proof uses the same arguments as Theorem 5.11, with the differentials constructed in Theorem 6.13. □

## 6.3 Computation of $\mathrm{Map}_{\Sigma\mathcal{O}p^0}(B^c(\mathcal{C}), \mathcal{P})$

We now describe a mapping space $\mathrm{Map}_{\Sigma\mathcal{O}p^0}(B^c(\mathcal{C} \otimes \mathrm{Sur}_{\mathbb{K}}), \mathcal{P})$ for some coaugmented connected cooperad $\mathcal{C}$ and for some augmented connected operad $\mathcal{P}$. We recall the following definition.

**Definition 6.15.** Let $M \in \Sigma\mathrm{Seq}_{\mathbb{K}}^0$ be an augmented sequence $M \simeq I \oplus \overline{M}$ with differential $d$. The sequence $M$ is an operad up to homotopy if there exists a derivation of cooperads of the form $d + \partial$ on $\mathcal{F}^c(\Sigma\overline{M})$ with $\partial|_{\Sigma\overline{M}} = 0$.

In this situation, we say that $\partial$ is a twisting morphism, and that $d + \partial$ is a twisted derivation. Recall that giving such a differential is equivalent to giving a morphism $\beta : \mathcal{F}^c(\Sigma\overline{M}) \to \Sigma\overline{M}$ such that $\beta|_{\Sigma\overline{M}} = 0$ and, if we denote by $\partial$ the morphism obtained from $\beta$ by the Leibniz rule on $\mathcal{F}^c(\Sigma\overline{M})$, then

$$ d(\beta) + \beta\partial = 0. $$

**Proposition 6.16.** Let $M \in \Sigma\mathrm{Seq}_{\mathbb{K}}^0$ be an operad up to homotopy. Then $\mathcal{L}(M) := \bigoplus_{n \ge 2} M(n)^{\Sigma_n}$ is endowed with the structure of a $\Gamma(\mathrm{PreLie}_\infty, -)$-algebra.

*Proof.* Let $\partial$ be the twisting part of the differential on $\mathcal{F}^c(\Sigma\overline{M})$. We denote by $\beta$ its composite with the projection on $\Sigma\overline{M}$. Let $x, y_1, \dots, y_n \in \mathcal{L}(M)$ be elements with homogeneous degrees and arities, and $r_1, \dots, r_n \ge 0$. We let $E$ to be the symmetric sequence spanned by abstract invariant variables $Y_1, \dots, Y_n, dY_1, \dots, dY_n$ of the same arities and degrees as $y_1, \dots, y_n, d(y_1), \dots, d(y_n)$. Let $\psi : E \to \Sigma\overline{M}$ be the morphism which sends the $Y_i$'s to the $y_i$'s. This gives a morphism of
coalgebras $\psi : \Gamma(\mathcal{L}(E)) \to \Gamma(\Sigma\mathcal{L}(M))$. Let $r_1, \dots, r_n \ge 0$ be such that $r := r_1 + \dots + r_n \ne 0$, and $r_i = 1$ if $Y_i$ has an odd degree. We set $x[\|\|] = d(x)$ and

$$x[\|y_1, \dots, y_n\|]_{r_1, \dots, r_n} = \sum_{\underline{T} \in \mathcal{T}\mathcal{M}(r+1)} \beta(\underline{T}^{\vee} \otimes x \otimes \psi\mathcal{O}(Y_1^{r_1} \cdots Y_n^{r_n})),$$

where we consider the orbit map $\mathcal{O}$ defined in Proposition 2.20 and where, in this sum, we identify every tensor $\underline{T}^{\vee} \otimes z$ such that $z \notin \otimes_{i=1}^{r+1} \Sigma\bar{M}(\text{val}_{\underline{T}}(i))$ with 0. In particular, the above sum is finite.

We first note that these operations preserve $\mathcal{L}(M)$. Indeed, the symmetry relations in the cooperad $\mathcal{F}^c(\Sigma\bar{M})$ will only make involve either actions of symmetric groups elements on $x$ and the $Y_i$'s, which are invariant, or actions on the tensors given by $\mathcal{O}(Y_1^{\otimes r_1} \cdots Y_n^{\otimes r_n})$, which is invariant under the action of $\Sigma_r$. It remains to prove formulas of Theorem 2.22. It is an immediate check that the operations $-[\|-\dots, -\|]_{r_1, \dots, r_n}$ satisfy the relations $(i) - (v)$. We now check relation $(vi)$. First, we have

$$\sum_{\underline{T} \in \mathcal{T}\mathcal{M}(r+1)} d(\beta(\underline{T}^{\vee} \otimes x \otimes \psi\mathcal{O}(Y_1^{r_1} \cdots Y_n^{r_n}))) = x[\|y_1, \dots, y_n\|]_{r_1, \dots, r_n}[\|\|];$$

$$\sum_{\underline{T} \in \mathcal{T}\mathcal{M}(r+1)} \beta(\underline{T}^{\vee} \otimes d(x) \otimes \psi\mathcal{O}(Y_1^{r_1} \cdots Y_n^{r_n})) = x[\|\|][\|y_1, \dots, y_n\|]_{r_1, \dots, r_n};$$

and

$$\sum_{\underline{T} \in \mathcal{T}\mathcal{M}(r+1)} \beta(\underline{T}^{\vee} \otimes x \otimes \psi\mathcal{O}(dY_k \cdot Y_1^{r_1} \cdots Y_k^{r_k-1} \cdots Y_n^{r_n})) \\ = x[\|y_k[\|\|], y_1, \dots, y_n\|]_{1, r_1, \dots, r_k-1, \dots, r_n}.$$

for every $1 \le k \le n$. This gives

$$\sum_{\underline{T} \in \mathcal{T}\mathcal{M}(r+1)} d(\beta)(\underline{T}^{\vee} \otimes x \otimes \psi\mathcal{O}(Y_1^{r_1} \cdots Y_n^{r_n})) \\ = x[\|y_1, \dots, y_n\|]_{r_1, \dots, r_n}[\|\|] + x[\|\|][\|y_1, \dots, y_n\|]_{r_1, \dots, r_n} \\ + \sum_{k=1}^{n} \pm x[\|y_k[\|\|], y_1, \dots, y_n\|]_{1, r_1, \dots, r_k-1, \dots, r_n}.$$

Let $\Delta : \mathcal{F}^c(\Sigma\bar{M}) \to \mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M}))$ be the cooperad morphism induced by the cooperad structure of $\mathcal{F}^c(\Sigma\bar{M})$. Let $\pi_{\Sigma\bar{M}} : \mathcal{F}^c(\Sigma\bar{M}) \to \Sigma\bar{M}$ be the projection on $\Sigma\bar{M}$. We keep the notation $\partial$ for the morphism defined on $\mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M}))$ obtained from $\partial$ by the Leibniz rule. Since $\partial$ is compatible with the cooperad structure on $\mathcal{F}^c(\Sigma\bar{M})$, we have the following commutative diagram:
$$
\begin{tikzcd}
& & \mathcal{F}^c(\Sigma\bar{M}) \arrow[r, "\Delta"] & \mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M})) \\
\mathcal{F}^c(\Sigma\bar{M}) \arrow[r, "\partial"] & \mathcal{F}^c(\Sigma\bar{M}) \arrow[r, "\Delta"] & & \\
& \mathcal{F}^c(\Sigma\bar{M}) \arrow[r, "\Delta"'] & \mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M})) \arrow[r, "\pi_{\Sigma\bar{M}}"] & \\
& & \mathcal{F}^c(\Sigma\bar{M})
\end{tikzcd}
$$

Let $\mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M})$ be the sub symmetric sequence of $\mathcal{F}^c(\Sigma\bar{M})$ given by trees with at least 2 vertices, so that $\mathcal{F}^c(\Sigma\bar{M}) \simeq \Sigma\bar{M} \oplus \mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M})$ as a symmetric sequence. We denote by $\mathcal{F}^c(\Sigma\bar{M}; \mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M}))$ the sub symmetric sequence of $\mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M}))$ given by trees with only one vertex in $\mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M})$, and the other in $\Sigma\bar{M}$. Then we have the following commutative diagram:

$$
\begin{tikzcd}
& & \mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M})) \arrow[r, "\Delta_{(1)}"] & \mathcal{F}^c(\Sigma\bar{M}; \mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M})) \\
& \mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M})) \arrow[r, "\pi_{\Sigma\bar{M}}"] & \mathcal{F}^c(\Sigma\bar{M}; \mathcal{F}^c(\Sigma\bar{M})) \arrow[r, "\pi_{\Sigma\bar{M}}"] & \\
& & \mathcal{F}^c(\Sigma\bar{M}) \arrow[r, "\Delta_{(2)}"] & \mathcal{F}^c(\Sigma\bar{M}; \mathcal{F}^c(\Sigma\bar{M}))
\end{tikzcd}
$$

The above commutative diagrams prove that $\partial : \mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M}) \to \mathcal{F}^c(\Sigma\bar{M})$ is given by the composite:

$$
\partial : \mathcal{F}^c_{(\ge 2)}(\Sigma \bar{M}) \xrightarrow{\Delta_{(1)}} \mathcal{F}^c(\Sigma \bar{M}; \mathcal{F}^c_{(\ge 2)}(\Sigma \bar{M})) \xrightarrow{\mathcal{F}^c(\Sigma \bar{M}; \beta)} \mathcal{F}^c(\Sigma \bar{M}; \Sigma \bar{M}) \xrightarrow{\mathcal{F}^c(id_{\Sigma \bar{M}} \oplus id_{\Sigma \bar{M}})} \mathcal{F}^c(\Sigma \bar{M})
$$

where $\Delta_{(1)}$ is the composite of $\Delta : \mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M}) \to \mathcal{F}^c(\mathcal{F}^c(\Sigma\bar{M}))$ with the projection on trees with only one vertex in $\mathcal{F}^c_{(\ge 2)}(\Sigma\bar{M})$. By definition, for every $\underline{T} \in \mathcal{T}\mathcal{M}(r+1)$, we have

$$
\begin{align*}
& \Delta_{(1)}(\underline{T}^{\vee} \otimes x \otimes \psi \mathcal{O}(Y_1^{r_1} \cdots Y_n^{r_n})) \\
&= \sum_{\substack{p+q=r \\ p \neq 0}} \sum_{\substack{U \in \mathcal{T}\mathcal{M}(q+1) \\ \underline{V} \in \mathcal{T}\mathcal{M}(p+1)}} \sum_{\substack{p_i+q_i=r_i \\ q_1+\cdots+p_n=p \\ \underline{U} \cdot \underline{1}, \underline{V}=\underline{T} \\ q_1+\cdots+q_n=q}} \pm \underline{U}^{\vee} \otimes (\underline{V}^{\vee} \otimes x \otimes \psi \mathcal{O}(Y_1^{p_1} \cdots Y_n^{p_n})) \otimes \psi \mathcal{O}(Y_1^{q_1} \cdots Y_n^{q_n}) \\
&\quad + \sum_{\substack{p+q=r-1 \\ p \neq 0}} \sum_{\substack{k=0 \\ \underline{U} \in \mathcal{T}\mathcal{M}(q+1) \\ \underline{V} \in \mathcal{T}\mathcal{M}(p+1)}} \sum_{\substack{s_i+p'_i+t_i=r_i \\ q_1+\cdots+p'_n=p+1 \\ \underline{U} \cdot \underline{k}, \underline{V}=\underline{T} \\ s_1+\cdots+s_n=k \\ t_1+\cdots+t_n=q-k}} \pm \underline{U}^{\vee} \otimes x \otimes \psi \mathcal{O}(Y_1^{s_1} \cdots Y_n^{s_n}) \\
&\quad \qquad \otimes (\underline{V}^{\vee} \otimes \psi \mathcal{O}(Y_1^{p'_1} \cdots Y_n^{p'_n})) \otimes \psi \mathcal{O}(Y_1^{t_1} \cdots Y_n^{t_n}).
\end{align*}
$$
By some variable substitutions, summing over $\underline{T} \in \mathcal{T}\mathcal{M}(r+1)$ gives

$$ \sum_{\substack{p+q=r \\ p \neq 0}} \sum_{\substack{p_i+q_i=r_i \\ p_1+\cdots+p_n=p \\ q_1+\cdots+q_n=q}} \sum_{\underline{U} \in \mathcal{T}\mathcal{M}(q+1)} \pm \underline{U}^{\vee} \otimes \left( \sum_{\underline{V} \in \mathcal{T}\mathcal{M}(p+1)} \underline{V}^{\vee} \otimes x \otimes \psi \mathcal{O}(Y_1^{p_1} \cdots Y_n^{p_n}) \right) \otimes \psi \mathcal{O}(Y_1^{q_1} \cdots Y_n^{q_n}) \\ + \sum_{\substack{p+q=r-1 \\ p \neq 0}} \sum_{\substack{p'_i+q_i=r_i \\ p'_1+\cdots+p'_n=p+1 \\ q_1+\cdots+q_n=q}} \sum_{\underline{U} \in \mathcal{T}\mathcal{M}(q+1)} \pm \underline{U}^{\vee} \otimes x \\ \otimes \mathrm{Sh} \left( \left( \sum_{\underline{V} \in \mathcal{T}\mathcal{M}(p+1)} \underline{V}^{\vee} \otimes \psi \mathcal{O}(Y_1^{p'_1} \cdots Y_n^{p'_n}) \right) ; \psi \mathcal{O}(Y_1^{q_1} \cdots Y_n^{q_n}) \right), $$

where Sh is defined analogously as in Definition 2.14. We now use that

$$ \psi \mathcal{O}(Y_1^{p'_1} \cdots Y_n^{p'_n}) = \sum_{k=1}^{n} \pm y_k \otimes \psi \mathcal{O}(Y_1^{p'_1} \cdots Y_k^{p'_k-1} \cdots Y_n^{p'_n}) $$

for every $p'_1, \dots, p'_n \ge 0$ (if $p'_k = 0$ for some $k$, we just remove the corresponding term). For a chosen $1 \le k \le n$, we set $p_i = p'_i$ for every $i \ne k$ and $p_k = p'_k - 1$. Then, applying $\mathcal{F}^c(\Sigma \bar{M}; \beta)$, gives

$$ \sum_{\substack{p+q=r \\ p \neq 0}} \sum_{\substack{p_i+q_i=r_i \\ p_1+\cdots+p_n=p \\ q_1+\cdots+q_n=q}} \sum_{\underline{U} \in \mathcal{T}\mathcal{M}(q+1)} \pm \underline{U}^{\vee} \otimes (x[\{y_1, \dots, y_n\}]_{p_1, \dots, p_n}) \otimes \psi \mathcal{O}(Y_1^{q_1} \cdots Y_n^{q_n}) \\ + \sum_{\substack{p+q=r-1 \\ p \neq 0}} \sum_{\substack{k=1 \\ p_i+q_i=r_i, i \neq k \\ p_k+q_k=r_k-1 \\ p_1+\cdots+p_n=p \\ q_1+\cdots+q_n=q}} \sum_{\underline{U} \in \mathcal{T}\mathcal{M}(q+1)} \pm \underline{U}^{\vee} \otimes x \otimes \mathrm{Sh}(y_k[\{y_1, \dots, y_n\}]_{p_1, \dots, p_n}; \psi \mathcal{O}(Y_1^{q_1} \cdots Y_n^{q_n})). $$

Applying $\beta$ again gives

$$ \sum_{\substack{p_i+q_i=r_i \\ p_1+\cdots+p_n \neq 0 \\ q_1+\cdots+q_n \neq 0}} \pm x[\{y_1, \dots, y_n\}]_{p_1, \dots, p_n}[\{y_1, \dots, y_n\}]_{q_1, \dots, q_n} \\ + \sum_{\substack{k=1 \\ p_i+q_i=r_i, i \neq k \\ p_k+q_k=r_k-1 \\ p_1+\cdots+p_n \neq 0}}^{n} \sum_{\underline{U} \in \mathcal{T}\mathcal{M}(q+1)} \pm x[\{y_k[\{y_1, \dots, y_n\}]_{p_1, \dots, p_n}, y_1, \dots, y_n\}]_{1, q_1, \dots, q_n} $$
Finally, the equation $d(\beta) + \beta\partial = 0$ applied on $\sum_{\underline{T} \in \mathcal{T}\mathcal{M}(r+1)} \underline{T} \otimes x \otimes \psi\mathcal{O}(Y_1^{r_1} \cdots Y_n^{r_n})$ gives

$$ \sum_{p_i+q_i=r_i} \pm x \llbracket y_1, \dots, y_n \rrbracket_{p_1, \dots, p_n} \llbracket y_1, \dots, y_n \rrbracket_{q_1, \dots, q_n} \\ + \sum_{k=1}^{n} \sum_{\substack{p_i+q_i=r_i, i \neq k \\ p_k+q_k=r_k-1}} \pm x \llbracket y_k \llbracket y_1, \dots, y_n \rrbracket_{p_1, \dots, p_n}, y_1, \dots, y_n \rrbracket_{1, q_1, \dots, q_n} = 0 $$

as desired.

□

**Corollary 6.17.** Let $M \in \Sigma\mathrm{Seq}_{\mathbb{K}}^0$ be an operad up to homotopy. Then the completion of $\mathcal{L}(\Sigma M)$, which is $\prod_{n \ge 1} \Sigma M(n)^{\Sigma_n}$, is endowed with the structure of a complete $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra.

*Proof.* It is the same proof as for [Ver23, Corollary 2.18].

□

We now apply this proposition to $M = \mathrm{Hom}(\mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}} \otimes N_*(\Delta^n), \mathcal{P})$ for every $n \ge 0$, which will give Theorem E.

**Theorem 6.18.** Let $\mathcal{C}$ be a symmetric cooperad and $\mathcal{P}$ be a symmetric augmented operad such that $\mathcal{P}(0) = \mathcal{C}(0) = 0$ and $\mathcal{P}(1) = \mathcal{C}(1) = \mathbb{K}$. Then, for every $n \ge 0$, the symmetric sequence $\mathrm{Hom}(\mathcal{C} \otimes_{\mathbb{H}} \mathbf{Sur}_{\mathbb{K}} \otimes N_*(\Delta^n), \mathcal{P})$ is an operad up to homotopy such that the underlying $\Gamma\Lambda\mathcal{P}\mathcal{L}_{\infty}$-algebra structure on $\Sigma\mathrm{Hom}_{\Sigma\mathrm{Seq}_{\mathbb{K}}}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes N_*(\Delta^n), \bar{\mathcal{P}})$ satisfies

$$ \mathrm{Map}_{\Sigma\mathcal{O}p^0}^h(B^c(\mathcal{C}), \mathcal{P}) \simeq \mathcal{M}\mathcal{C}(\Sigma\mathrm{Hom}_{\Sigma\mathrm{Seq}_{\mathbb{K}}}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes N_*(\Delta^\bullet), \bar{\mathcal{P}})), $$

where we have set $\mathrm{Map}_{\Sigma\mathcal{O}p^0}^h(B^c(\mathcal{C}), \mathcal{P}) = \mathrm{Map}_{\Sigma\mathcal{O}p^0}(B^c(\mathcal{C} \otimes \mathbf{Sur}_{\mathbb{K}}), \mathcal{P})$

*Proof.* Let $n \ge 0$. We first note that we have an isomorphism

$$ \Sigma\mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes N_*(\Delta^n), \mathcal{P}) \simeq \mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1}N_*(\Delta^n), \mathcal{P}). $$

We thus need to construct a morphism

$$ \beta : \mathcal{F}_{(\ge 2)}^c(\mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}})) \to \mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}}) $$

such that, if we denote by $d$ the differential induced by the internal differential of $\mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}})$, and if we denote by $\partial : \mathcal{F}_{(\ge 2)}^c(\mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}})) \to \mathcal{F}^c(\mathrm{Hom}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}}))$ the morphism obtained from $\beta$ by the Leibniz rule, then $d(\beta) + \beta\partial = 0$.

We set, for every $f_1, \dots, f_m \in \mathrm{Hom}_{\Sigma\mathrm{Seq}_{\mathbb{K}}}(\bar{\mathcal{C}} \otimes_{\mathbb{H}} \overline{\mathbf{Sur}}_{\mathbb{K}} \otimes \Sigma^{-1} N_*(\Delta^n), \bar{\mathcal{P}})$ and $\underline{T} \in \mathcal{T}\mathcal{M}(m)$,

$$ \beta(\underline{T}^\vee \otimes f_1 \otimes \cdots \otimes f_m) = \gamma(\underline{T}) \circ (f_1 \otimes \cdots \otimes f_m) \circ \beta_{(\underline{T})}^n $$
where $\beta_{(T)}^n$ is the composite of the morphism $\beta^n$ defined in Construction 6.10 with the projection on the $T$-component. We first note that $d(\beta) = 0$, since $d(\beta^n) = 0$. Now, note that the $T$-component of $\beta\partial((T^\vee \otimes f_1 \otimes \cdots \otimes f_m))$ is

$$ \sum_{S \subset T} \gamma_{(T/S)} \circ \left( f_1 \otimes \cdots \otimes f_{r(S)-1} \otimes \left( \gamma_{(S)} \circ \left( \bigotimes_{i \in V_S} f_i \right) \circ \beta_{(S)}^n \right) \otimes \bigotimes_{i \in V_{T/S} \setminus \{1, \dots, r(S)-1\}} f_i \right) \circ \beta_{(T/S)}^n. $$

By Lemma 6.9, this is equal to

$$ -\gamma_{(T)} \circ (f_1 \otimes \cdots \otimes f_m) \circ \left( \sum_{S \subset T} \beta_{(T/S)}^n \circ_S \beta_{(S)}^n \right). $$

This terms is 0, since the sum $\sum_{S \subset T} \beta_{(T/S)}^n \circ_S \beta_{(S)}^n$ is precisely the $T$-component of $\partial^n \beta^n$, which is 0 by the proof of Theorem 6.13.

The computation of $\mathrm{Map}^h_{\Sigma O_p^0}(B^c(\mathcal{C}), \mathcal{P})$ comes from the construction of $B^c(\mathcal{C} \otimes \mathrm{Sur}_\mathcal{H}) \otimes \Delta^\bullet$ given by Construction 6.10 and from the complete $\Gamma(\mathrm{PreLie}_\infty, -)$-algebra structure given by Corollary 6.17. $\square$

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UNIV. LILLE, CNRS, UMR 8524 - LABORATOIRE PAUL PAINLEVÉ, F-59000 LILLE, FRANCE

E-mail address: marvin.verstraete@univ-lille.fr