We will describe a functor in the reverse direction inspired by the construction of Lemma \ref{lemma-extension} (except that we throw in a sign to get the boundaries right). Let $A_\bullet$ be a chain complex with boundary maps $d_{A, n} : A_n \to A_{n - 1}$. For each $n \geq 0$ denote $$ I_n = \Big\{ \alpha : [n] \to \{0, 1, 2, \ldots\} \mid \Im(\alpha) = [k]\text{ for some }k \Big\}. $$ For $\alpha \in I_n$ we denote $k(\alpha)$ the unique integer such that $\Im(\alpha) = [k]$. We define a simplicial object $S(A_\bullet)$ as follows: \begin{enumerate} \item $S(A_\bullet)_n = \bigoplus_{\alpha \in I_n} A_{k(\alpha)}$, which we will write as $\bigoplus_{\alpha \in I_n} A_{k(\alpha)} \cdot \alpha$ to suggest thinking of ``$\alpha$'' as a basis vector for the summand corresponding to it, \item given $\varphi : [m] \to [n]$ we define $S(A_\bullet)(\varphi)$ by its restriction to the direct summand $A_{k(\alpha)} \cdot \alpha$ of $S(A_\bullet)_n$ as follows \begin{enumerate} \item $\alpha \circ \varphi \not \in I_m$ then we set it equal to zero, \item $\alpha \circ \varphi \in I_m$ but $k(\alpha \circ \varphi)$ not equal to either $k(\alpha)$ or $k(\alpha) - 1$ then we set it equal to zero as well, \item if $\alpha \circ \varphi \in I_m$ and $k(\alpha \circ \varphi) = k(\alpha)$ then we use the identity map to the summand $A_{k(\alpha \circ \varphi)} \cdot (\alpha \circ \varphi)$ of $S(A_\bullet)_m$, and \item if $\alpha \circ \varphi \in I_m$ and $k(\alpha \circ \varphi) = k(\alpha) - 1$ then we use $(-1)^{k(\alpha)} d_{A, k(\alpha)}$ to the summand $A_{k(\alpha \circ \varphi)}\cdot (\alpha \circ \varphi)$ of $S(A_\bullet)_m$. \end{enumerate} \end{enumerate} Let us show that $S(A_\bullet)$ is a simplicial object of $\mathcal{A}$. To do this, assume we have maps $\varphi : [m] \to [n]$ and $\psi : [n] \to [p]$. We will show that $S(A_\bullet)(\varphi) \circ S(A_\bullet)(\psi) = S(A_\bullet)(\psi \circ \varphi)$. Choose $\beta \in I_p$ and set $\alpha = \beta \circ \psi$ and $\gamma = \alpha \circ \varphi$ viewed as maps $\alpha : [n] \to \{0, 1, 2, \ldots\}$ and $\gamma : [m] \to \{0, 1, 2, \ldots\}$. Picture $$ \xymatrix{ [m] \ar[r]_\varphi \ar[d]_\gamma & [n] \ar[r]_\psi \ar[d]_\alpha & [p] \ar[d]^\beta \\ \Im(\gamma) \ar[r] & \Im(\alpha) \ar[r] & [k(\beta)] } $$ We will show that the restriction of the maps $S(A_\bullet)(\varphi) \circ S(A_\bullet)(\psi)$ and $S(A_\bullet)(\psi \circ \varphi)$. to the summand $A_{k(\beta)} \cdot \beta$ agree. There are several cases to consider \begin{enumerate} \item Say $\alpha \not \in I_n$ so the restriction of $S(A_\bullet)(\psi)$ to $A_{k(\beta)} \cdot \beta$ is zero. Then either $\gamma \not \in I_m$ or we have $[k(\gamma)] = \Im(\gamma) \subset \Im(\alpha) \subset [k(\beta)]$ and the subset $\Im(\alpha)$ of $[k(\beta)]$ has a gap so $k(\gamma) < k(\beta) - 1$. In both cases we see that the restriction of $S(A_\bullet)(\psi \circ \varphi)$ to $A_{k(\beta)} \cdot \beta$ is zero as well. \item Say $\alpha \in I_n$ and $k(\alpha) < k(\beta) - 1$ so the restriction of $S(A_\bullet)(\psi)$ to $A_{k(\beta)} \cdot \beta$ is zero. Then either $\gamma \not \in I_m$ or we have $[k(\gamma)] \subset [k(\alpha)] \subset [k(\beta)]$ and it follows that $k(\gamma) < k(\beta) - 1$. In both cases we see that the restriction of $S(A_\bullet)(\psi \circ \varphi)$ to $A_{k(\beta)} \cdot \beta$ is zero as well. \item Say $\alpha \in I_n$ and $k(\alpha) = k(\beta)$ so the restriction of $S(A_\bullet)(\psi)$ to $A_{k(\beta)} \cdot \beta$ is the identity map from $A_{k(\beta)} \cdot \beta$ to $A_{k(\alpha)} \cdot \alpha$. In this case because $\Im(\alpha) =[k(\beta)]$ the rule describing the restriction of $S(A_\bullet)(\psi \circ \varphi)$ to the summand $A_{k(\beta)} \cdot \beta$ is exactly the same as the rule describing the restriction of $S(A_\bullet)(\varphi)$ to the summand $A_{k(\alpha)} \cdot \alpha$ and hence agreement holds. \item Say $\alpha \in I_n$ and $k(\alpha) = k(\beta) - 1$ so the restriction of $S(A_\bullet)(\psi)$ to $A_{k(\beta)} \cdot \beta$ is given by $(-1)^{k(\beta)}d_{A, k(\beta)}$ to $A_{k(\alpha)} \cdot \alpha$. Subcases \begin{enumerate} \item If $\gamma \not \in I_m$, then both the restriction of $S(A_\bullet)(\psi \circ \varphi)$ to the summand $A_{k(\beta)} \cdot \beta$ and the restriction of $S(A_\bullet)(\varphi)$ to the summand $A_{k(\alpha)} \cdot \alpha$ are zero and we get agreement. \item If $\gamma \in I_m$ but $k(\gamma) < k(\alpha) - 1$, then again both restrictions are zero and we get agreement. \item If $\gamma \in I_m$ and $k(\gamma) = k(\alpha)$ then $\Im(\gamma) = \Im(\alpha)$. In this case the restriction of $S(A_\bullet)(\psi \circ \varphi)$ to the summand $A_{k(\beta)} \cdot \beta$ is given by $(-1)^{k(\beta)}d_{A, k(\beta)}$ to $A_{k(\gamma)} \cdot \gamma$ and the restriction of $S(A_\bullet)(\varphi)$ to the summand $A_{k(\alpha)} \cdot \alpha$ is the identity map $A_{k(\alpha)} \cdot \alpha \to A_{k(\gamma)} \cdot \gamma$. Hence agreement holds. \item Finally, if $\gamma \in I_m$ and $k(\gamma) = k(\alpha) - 1$ then the restriction of $S(A_\bullet)(\varphi)$ to the summand $A_{k(\alpha)} \cdot \alpha$ is given by $(-1)^{k(\alpha)} d_{A, k(\alpha)}$ as a map $A_{k(\alpha)} \cdot \alpha \to A_{k(\beta)} \cdot \beta$. Since $A_\bullet$ is a complex we see that the composition $A_{k(\beta)} \cdot \beta \to A_{k(\alpha)} \cdot \alpha \to A_{k(\gamma)} \cdot \gamma$ is zero which matches what we get for the restriction of $S(A_\bullet)(\psi \circ \varphi)$ to the summand $A_{k(\beta)} \cdot \beta$ because $k(\gamma) = k(\beta) - 2 < k(\beta) - 1$. \end{enumerate} \end{enumerate} Thus $S(A_\bullet)$ is a simplicial object of $\mathcal{A}$. \medskip\noindent Let us construct an isomorphism $A_\bullet \to N(S(A_\bullet))$ functorial in $A_\bullet$. Recall that $$ S(A_\bullet) =  N(S(A_\bullet)) \oplus D(S(A_\bullet)) $$ as chain complexes by Lemma \ref{lemma-decompose-associated-complexes}. On the other hand it follows from Remark \ref{remark-degenerate-subcomplex} and the construction of $S(A_\bullet)$ that $$ D(S(A_\bullet))_n = \bigoplus\nolimits_{\alpha \in I_n,\ k(\alpha) < n} A_{k(\alpha)} \cdot \alpha \subset \bigoplus\nolimits_{\alpha \in I_n} A_{k(\alpha)} \cdot \alpha $$ However, if $\alpha \in I_n$ then we have $k(\alpha) \geq n \Leftrightarrow \alpha = \text{id}_{[n]} : [n] \to [n]$. Thus the summand $A_n \cdot \text{id}_{[n]}$ of $S(A_\bullet)_n$ is a complement to the summand $D(S(A_\bullet))_n$. All the maps $d^n_i : S(A_\bullet)_n \to S(A_\bullet)_n$ restrict to zero on the summand $A_n \cdot \text{id}_{[n]}$ except for $d^n_n$ which produces $(-1)^n d_{A, n}$ from $A_n \cdot \text{id}_{[n]}$ to $A_{n - 1} \cdot \text{id}_{[n - 1]}$. We conclude that $A_n \cdot \text{id}_{[n]}$ must be equal to the summand $N(S(A_\bullet))_n$ and moreover the restriction of the differential $d_n = \sum (-1)^id^n_i : S(A_\bullet)_n \to S(A_\bullet)_{n - 1}$ to the summand $A_n \cdot \text{id}_{[n]}$ gives what we want! \medskip\noindent Finally, we have to show that $S \circ N$ is isomorphic to the identity functor. Let $U$ be a simplicial object of $\mathcal{A}$. Then we can define an obvious map $$ S(N(U))_n = \bigoplus\nolimits_{\alpha \in I_n} N(U)_{k(\alpha)} \cdot \alpha \longrightarrow U_n $$ by using $U(\alpha) : N(U)_{k(\alpha)} \to U_n$ on the summand corresponding to $\alpha$. By Definition \ref{definition-split} this is an isomorphism. To finish the proof we have to show that this is compatible with the maps in the simplicial objects. Thus let $\varphi : [m] \to [n]$ and let $\alpha \in I_n$. Set $\beta = \alpha \circ \varphi$. Picture $$ \xymatrix{ [m] \ar[r]_\varphi \ar[d]_\beta & [n] \ar[d]_\alpha \\ \Im(\beta) \ar[r] & [k(\alpha)] } $$ There are several cases to consider \begin{enumerate} \item Say $\beta \not \in I_m$. Then there exists an index $0 \leq j < k(\alpha)$ with $j \not \in \Im(\alpha \circ \varphi)$ and hence we can choose a factorization $\alpha \circ \varphi = \delta^{k(\alpha)}_j \circ \psi$ for some $\psi : [m] \to [k(\alpha) - 1]$. It follows that $U(\varphi)$ is zero on the image of the summand $N(U)_{k(\alpha)} \cdot \alpha$ because $U(\varphi) \circ U(\alpha) = U(\alpha \circ \varphi) = U(\psi) \circ d^{k(\alpha)}_j$ is zero on $N(U)_{k(\alpha)}$ by construction of $N$. This matches our rule for $S(N(U))$ given above. \item Say $\beta \in I_m$ and $k(\beta) < k(\alpha) - 1$. Here we argue exactly as in case (1) with $j = k(\alpha) - 1$. \item Say $\beta \in I_m$ and $k(\beta) = k(\alpha)$. Here the summand $N(U)_{k(\alpha)} \cdot \alpha$ is mapped by the identity to the summand $N(U)_{k(\beta)} \cdot \beta$. This is the same as the effect of $U(\varphi)$ since in this case $U(\varphi) \circ U(\alpha) = U(\beta)$. \item Say $\beta \in I_m$ and $k(\beta) = k(\alpha) - 1$. Here we use the differential $(-1)^{k(\alpha)} d_{N(U), k(\alpha)}$ to map the summand $N(U)_{k(\alpha)} \cdot \alpha$ to the summand $N(U)_{k(\beta)} \cdot \beta$. On the other hand, since $\Im(\beta) = [k(\beta)]$ in this case we get $\alpha \circ \varphi = \delta^{k(\alpha)}_{k(\alpha)} \circ \beta$. Thus we see that $U(\varphi)$ composed with the restriction of $U(\alpha)$ to $N(U)_{k(\alpha)}$ is equal to $U(\beta)$ precomposed with $d^{k(\alpha)}_{k(\alpha)}$ restricted to $N(U)_{k(\alpha)}$. Since $d_{N(U), k(\alpha)} = \sum (-1)^i d^{k(\alpha)}_i$ and since $d^{k(\alpha)}_i$ restricts to zero on $N(U)_{k(\alpha)}$ for $i < k(\alpha)$ we see that equality holds. \end{enumerate} This finishes the proof of the theorem.