Lemma 2.5.6.16. Let $A_{\bullet }$ be a simplicial abelian group and let $n$ be a nonnegative integer. Then the map
\[ f: \bigoplus _{ \alpha : [n] \twoheadrightarrow [m] } \widetilde{\mathrm{N}}_{m}(A) \rightarrow A_{n} \quad \quad \{ x_{\alpha } \} \mapsto \sum \alpha ^{\ast }( x_{\alpha } ) \]
is an isomorphism of abelian groups. Here the direct sum is indexed by surjective nondecreasing maps $\alpha : [n] \rightarrow [m]$ for $0 \leq m \leq n$, and $\alpha ^{\ast }: A_{m} \rightarrow A_{n}$ denotes the associated group homomorphism.
Proof.
We first prove that $f$ is surjective. The proof proceeds by induction on $n$. By virtue of our inductive hypothesis, the image of $f$ contains the subgroups $\widetilde{\mathrm{N}}_{n}(A), \mathrm{D}_{n}(A) \subseteq \mathrm{C}_{n}(A) = A_ n$. It will therefore suffice to show that the composite map
\[ \widetilde{\mathrm{N}}_{n}(A) \hookrightarrow \mathrm{C}_{n}( A ) \twoheadrightarrow \mathrm{C}_{n}(A) / \mathrm{D}_{n}(A) \]
is surjective. Fix an element $\overline{x} \in \mathrm{C}_{n}(A) / \mathrm{D}_{n}(A)$. For each $x \in \mathrm{C}_{n}(A)$ representing $\overline{x}$, let $i_{x}$ be the smallest nonnegative integer such that $d^{n}_ j(x)$ vanishes for $i_{x} < j \leq n$. Without loss of generality, we may assume that $x$ is chosen so that $i = i_{x}$ is as small as possible. We wish to prove that $i = 0$ (so that $x$ belongs to $\mathrm{N}_{n}(A)$). Assume otherwise, and set $y = x - (s^{n-1}_{i-1} \circ d^{n}_ i)(x)$. Then $y$ is congruent to $x$ modulo $\mathrm{D}_{n}(A)$, and for $i \leq j \leq n$ we have
\begin{eqnarray*} d^{n}_ j(y) & = & d^{n}_ j(x) - (d^{n}_ j \circ s^{n-1}_{i-1} \circ d^{n}_ i)(x) \\ & = & d^{n}_ j( x) - \begin{cases} d^{n}_ i(x) & \text{ if } i = j \\ (s^{n-2}_{i-1} \circ d^{n-1}_{j-1} \circ d^{n}_ i )(x) & \text{ if } i < j. \end{cases}\\ & = & d^{n}_ j(x) - \begin{cases} d^{n}_ i(x) & \text{ if } i = j \\ (s^{n-2}_{i-1} \circ d^{n-1}_{i} \circ d^{n}_ j )(x) & \text{ if } i < j. \end{cases}\\ & = & 0. \end{eqnarray*}
It follows that $i_{y} < i = i_{x}$, contradicting our choice of $x$.
We now prove that $f$ is injective. Suppose otherwise, so that there exists a nonzero element
\[ \{ x_{\alpha } \} \in \bigoplus _{ \alpha : [n] \twoheadrightarrow [m] } \widetilde{\mathrm{N}}_{m}(A) \]
which is annihilated by $f$. Then there exists some surjective map $\beta : [n] \twoheadrightarrow [k]$ such that $x_{\beta }$ is nonzero. Assume that $k$ has been chosen as small as possible. Moreover, we may assume that $\beta $ is maximal among nondecreasing maps $[n] \twoheadrightarrow [k]$ such that $x_{\beta } \neq 0$: in other words, that for any other map $\alpha : [n] \twoheadrightarrow [k]$ satisfying $\beta (i) \leq \alpha (i)$ for $0 \leq i \leq n$, we either have $\beta = \alpha $ or $x_{\alpha } = 0$. Let $\gamma : [k] \rightarrow [n]$ be the map given by $\gamma (j) = \min \{ i \in [n]: \beta (i) = j \} $. Then $\gamma $ is a nondecreasing map satisfying $\beta \circ \gamma = \operatorname{id}_{[k]}$ and $\gamma (0) = 0$. We then have
\begin{eqnarray*} \gamma ^{\ast } f( \{ x_{\alpha } \} ) & = & \gamma ^{\ast }( \sum _{ \alpha : [n] \twoheadrightarrow [m] } \alpha ^{\ast }( x_{\alpha } )) \\ & = & \sum _{ \alpha : [n] \twoheadrightarrow [m] } ( \alpha \circ \gamma )^{\ast }( x_{\alpha } ). \end{eqnarray*}
We now inspect the summands appearing on the right hand side:
Let $\alpha : [n] \twoheadrightarrow [m]$ be a surjective nondecreasing function, and suppose that the composite map $[k] \xrightarrow {\gamma } [n] \xrightarrow {\alpha } [m]$ is not surjective. Then we can choose $0 \leq i \leq m$ such that $i$ does not belong the image of $\alpha \circ \gamma $. Then the homomorphism $(\alpha \circ \gamma )^{\ast }: A_{m} \rightarrow A_{k}$ factors through the face operator $d^{m}_ i: A_{m} \rightarrow A_{m-1}$. Note that we must have $i > 0$ (since $\gamma (0) = 0$ and $\alpha (0) = 0$), so that $x_{\alpha }$ is annihilated by $d^{m}_ i$ (by virtue of our assumption that $x_{\alpha }$ belongs to the subgroup $\mathrm{N}_{m}(A) \subseteq A_ m$) and therefore also by $(\alpha \circ \gamma )^{\ast }$.
Let $\alpha : [n] \twoheadrightarrow [m]$ be a surjective nondecreasing function, and suppose that the composite map $[k] \xrightarrow {\gamma } [n] \xrightarrow {\alpha } [m]$ is surjective but not injective. In this case, we must have $m < k$, so that $x_{\alpha }$ vanishes by virtue of the minimality assumption on $k$.
Let $\alpha : [n] \twoheadrightarrow [m]$ be a surjective map, and suppose that the composite map $[k] \xrightarrow {\gamma } [n] \xrightarrow {\alpha } [m]$ is bijective, so that $m = k$ and $\alpha \circ \gamma $ is the identity on $[k]$. For $0 \leq i \leq n$, we have $(\gamma \circ \beta )(i) \leq i$ (by the definition of $\gamma $), so that
\[ \beta (i) = ((\alpha \circ \gamma ) \circ \beta )(i) = (\alpha \circ (\gamma \circ \beta ))(i) \leq \alpha (i). \]
Invoking our maximality assumption on $\beta $, we conclude that either $\alpha = \beta $ or $x_{\alpha }$ vanishes.
Combining these observations, we obtain an equality
\[ x_{\beta } = \sum _{ \alpha : [n] \twoheadrightarrow [m] } ( \alpha \circ \gamma )^{\ast }( x_{\alpha } ) = \gamma ^{\ast } f( \{ x_{\alpha } \} ) = 0, \]
contradicting our choice of $\beta $.
$\square$