Let $(\diamond N(U), \diamond a, \diamond b, \diamond h)$ be as in Lemma \ref{lemma-represent-homotopy} and its proof. By that lemma there exists a morphism $\diamond N(U) \to N(V)$ representing the triple $(N(a), N(b), \{N_n\})$. We will show there exists a morphism $\psi : N(U \times \Delta[1]) \to \diamond{N(U)}$ such that $\diamond a = \psi \circ N(e_0)$, and $\diamond b = \psi \circ N(e_1)$. Moreover, we will show that the homotopy between $N(e_0), N(e_1) : N(U) \to N(U \times \Delta[1])$ coming from (\ref{equation-homotopy-to-homotopy}) and Lemma \ref{lemma-homotopy-s-N} with $h = \text{id}_{U \times \Delta[1]}$ is mapped via $\psi$ to the canonical homotopy $\diamond h$ between the two maps $\diamond a, \diamond b : N(U) \to \diamond{N(U)}$. Certainly this will imply the lemma. \medskip\noindent Note that $N : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})$ as a functor is a direct summand of the functor $s : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})$. Also, the functor $\diamond$ is compatible with direct sums. Thus it suffices instead to construct a morphism $\Psi : s(U \times \Delta[1]) \to \diamond{s(U)}$ with the corresponding properties. This is what we do below. \medskip\noindent By Definition \ref{definition-homotopy} the morphisms $e_0 : U \to U \times \Delta[1]$ and $e_1 : U \to U \times \Delta[1]$ are homotopic with homotopy $\text{id}_{U \times \Delta[1]}$. By Lemma \ref{lemma-homotopy-s-N} we get an explicit homotopy $\{h_n : s(U)_n \to s(U \times \Delta[1])_{n + 1}\}$ between the morphisms of chain complexes $s(e_0) : s(U) \to s(U \times \Delta[1])$ and $s(e_1) : s(U) \to s(U \times \Delta[1])$. By Lemma \ref{lemma-map-into-diamond} above we get a corresponding morphism $$ \Phi : \diamond{s(U)} \to s(U \times \Delta[1]) $$ According to the construction, $\Phi_n$ restricted to the summand $s(U)[-1]_n = s(U)_{n - 1}$ of $\diamond{s(U)}_n$ is equal to $h_{n - 1}$. And $$ h_{n - 1} = \sum\nolimits_{i = 0}^{n - 1} (-1)^{i + 1} s^n_i \cdot \alpha^n_{i + 1} : U_{n - 1} \to \bigoplus\nolimits_j U_n \cdot \alpha^n_j. $$ with obvious notation. \medskip\noindent On the other hand, the morphisms $e_i : U \to U \times \Delta[1]$ induce a morphism $(e_0, e_1) : U \oplus U \to U \times \Delta[1]$. Denote $W$ the cokernel. Note that, if we write $(U \times \Delta[1])_n = \bigoplus_{\alpha : [n] \to [1]} U_n \cdot \alpha$, then we may identify $W_n = \bigoplus_{i = 1}^n U_n \cdot \alpha^n_i$ with $\alpha^n_i$ as in Section \ref{section-homotopy}. We have a commutative diagram $$ \xymatrix{ 0 \ar[r] & U \oplus U \ar[rd]_{(1, 1)} \ar[r] & U \times \Delta[1] \ar[d]^\pi \ar[r] & W \ar[r] & 0 \\ & & U & & } $$ This implies we have a similar commutative diagram after applying the functor $s$. Next, we choose the splittings $\sigma_n : s(W)_n \to s(U \times \Delta[1])_n$ by mapping the summand $U_n \cdot \alpha^n_i \subset W_n$ via $(-1, 1)$ to the summands $U_n \cdot \alpha^n_0 \oplus U_n \cdot \alpha^n_i \subset (U \times \Delta[1])_n$. Note that $s(\pi)_n \circ \sigma_n = 0$. It follows that $(1, 1) \circ \delta(\sigma)_n = 0$. Hence $\delta(\sigma)$ factors as in Lemma \ref{lemma-map-into-diamond}. By that lemma we obtain a canonical morphism $\Psi : s(U \times \Delta[1]) \to \diamond{s(U)}$. \medskip\noindent To compute $\Psi$ we first compute the morphism $\delta(\sigma) : s(W) \to s(U)[-1] \oplus s(U)[-1]$. According to Homology, Lemma \ref{homology-lemma-ses-termwise-split} and its proof, to do this we have compute $$ d_{s(U \times \delta[1]), n} \circ \sigma_n - \sigma_{n - 1} \circ d_{s(W), n} $$ and write it as a morphism into $U_{n - 1} \cdot \alpha^{n - 1}_0 \oplus U_{n - 1} \cdot \alpha^{n - 1}_n$. We only do this in case $\mathcal{A}$ is the category of abelian groups. We use the short hand notation $x_{\alpha}$ for $x \in U_n$ to denote the element $x$ in the summand $U_n \cdot \alpha$ of $(U \times \Delta[1])_n$. Recall that $$ d_{s(U \times \delta[1]), n} = \sum\nolimits_{i = 0}^n (-1)^i d^n_i $$ where $d^n_i$ maps the summand $U_n \cdot \alpha$ to the summand $U_{n - 1} \cdot (\alpha \circ \delta^n_i)$ via the morphism $d^n_i$ of the simplicial object $U$. In terms of the notation above this means $$ d_{s(U \times \delta[1]), n}(x_\alpha) = \sum\nolimits_{i = 0}^n (-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} $$ Starting with $x_\alpha \in W_n$, in other words $\alpha = \alpha^n_j$ for some $j \in \{1, \ldots, n\}$, we see that $\sigma_n(x_\alpha) = x_\alpha - x_{\alpha^n_0}$ and hence $$ (d_{s(U \times \delta[1]), n} \circ \sigma_n)(x_\alpha) = \sum\nolimits_{i = 0}^n (-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} - \sum\nolimits_{i = 0}^n (-1)^i (d^n_i(x))_{\alpha^n_0 \circ \delta^n_i} $$ To compute $d_{s(W), n}(x_\alpha)$, we have to omit all terms where $\alpha \circ \delta^n_i = \alpha^{n - 1}_0, \alpha^{n - 1}_n$. Hence we get $$ \begin{matrix} (\sigma_{n - 1} \circ d_{s(W), n})(x_\alpha) = \\ \sum\nolimits_{i = 0, \ldots, n\text{ and } \alpha \circ \delta^n_i \not = \alpha^{n - 1}_0\text{ or }\alpha^{n - 1}_n} \Big((-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} - (-1)^i (d^n_i(x))_{\alpha^{n - 1}_0} \Big) \end{matrix} $$ Clearly the difference of the two terms is the sum $$ \sum\nolimits_{i = 0, \ldots, n\text{ and } \alpha \circ \delta^n_i = \alpha^{n - 1}_0\text{ or }\alpha^{n - 1}_n} \Big((-1)^i (d^n_i(x))_{\alpha \circ \delta^n_i} - (-1)^i (d^n_i(x))_{\alpha^{n - 1}_0} \Big) $$ Of course, if $\alpha \circ \delta^n_i = \alpha^{n - 1}_0$ then the term drops out. Recall that $\alpha = \alpha^n_j$ for some $j \in \{1, \ldots, n\}$. The only way $\alpha^n_j \circ \delta^n_i = \alpha^{n - 1}_n$ is if $j = n$ and $i = n$. Thus we actually get $0$ unless $j = n$ and in that case we get $(-1)^n(d^n_n(x))_{\alpha^{n - 1}_n} - (-1)^n(d^n_n(x))_{\alpha^{n - 1}_0}$. In other words, we conclude the morphism $$ \delta(\sigma)_n : W_n \to (s(U)[-1] \oplus s(U)[-1])_n = U_{n - 1} \oplus U_{n - 1} $$ is zero on all summands except $U_n \cdot \alpha^n_n$ and on that summand it is equal to $((-1)^nd^n_n, -(-1)^nd^n_n)$. (Namely, the first summand of the two corresponds to the factor with $\alpha^{n - 1}_n$ because that is the map $[n - 1] \to [1]$ which maps everybody to $0$, and hence corresponds to $e_0$.) \medskip\noindent We obtain a canonical diagram $$ \xymatrix{ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & \diamond{s(U)} \ar[r] \ar[d]^{\Phi}& s(U)[-1] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & s(U \times \Delta[1]) \ar[r] \ar[d]^\Psi & s(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] & \diamond{s(U)} \ar[r] & s(U)[-1] \ar[r] & 0 } $$ We claim that $\Phi \circ \Psi$ is the identity. To see this it is enough to prove that the composition of $\Phi$ and $\delta(\sigma)$ as a map $s(U)[-1] \to s(W) \to s(U)[-1] \oplus s(U)[-1]$ is the identity in the first factor and minus identity in the second. By the computations above it is $((-1)^nd^n_0, -(-1)^nd^n_0) \circ (-1)^n s^n_n = (1, -1)$ as desired.