Proposition 2.5.6.21 (00RC)

Proposition 2.5.6.21. Let $A_{\bullet }$ be a simplicial abelian group. Then:

The quotient map $\mathrm{C}_{\ast }(A) \twoheadrightarrow \mathrm{N}_{\ast }(A)$ induces an isomorphism on homology.
The inclusion map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A)$ induces an isomorphism on homology.
The subcomplex $\mathrm{D}_{\ast }(A) \subseteq \mathrm{C}_{\ast }(A)$ of Notation 2.5.5.5 is acyclic: that is, its homology groups are trivial.

Proof. By virtue of Proposition 2.5.6.17, assertions $(a)$, $(b)$, and $(c)$ are equivalent. It will therefore suffice to prove $(b)$. Note that the map $\widetilde{ \mathrm{N} }_{\ast }(A) \hookrightarrow \mathrm{C}_{\ast }(A)$ is the inclusion of a direct summand (Proposition 2.5.6.17) and is therefore automatically injective on homology. To show that it also induces a surjective map, it will suffice to show that every $n$-cycle $x \in \mathrm{C}_{n}(A)$ is homologous to an element of the subgroup $\widetilde{\mathrm{N}}_{n}(A)$. Let $i$ denote the smallest nonnegative integer for which the faces $d^{n}_ j(x)$ vanish for $i < j \leq n$; our proof will proceed by induction on $i$. If $i = 0$, then $x$ belongs to $\widetilde{\mathrm{N}}_{n}(A)$, and there is nothing to prove. Otherwise, let $y \in \mathrm{C}_{n}(A)$ denote the boundary given by $\partial ( s^{n}_{i}(x) )$. We then compute

\begin{eqnarray*} y & = & \partial (s^{n}_ i(x) ) \\ & = & \sum _{j = 0}^{n+1} (-1)^{j} (d^{n+1}_ j \circ s^{n}_ i)(x) \\ & = & (\sum _{j = 0}^{i-1} (-1)^{j} (s^{n-1}_{i-1} \circ d^{n}_ j)(x) ) + (-1)^{i} x + (-1)^{i+1} x + (\sum _{j = i+2}^{n} (-1)^{j} (s^{n-1}_{i} \circ d^{n}_{j-1})(x) ) \\ & = & s^{n-1}_{i-1}( \sum _{j=0}^{i-1} (-1)^{j} d^{n}_ j(x) ) \\ & = & s^{n-1}_{i-1}( (\sum _{j=0}^{i-1} (-1)^{j} d^{n}_ j(x) ) + (\sum _{j=i+1}^{n} (-1)^{j} d^{n}_ j(x) ) ) \\ & = & s^{n-1}_{i-1}( \partial (x) - (-1)^{i} d^{n}_ i(x) ) \\ & = & (-1)^{i+1} (s^{n-1}_{i-1} \circ d^{n}_ i)(x). \end{eqnarray*}

Set $x' = x + (-1)^{i} y$. For $j \geq i$ we compute

\begin{eqnarray*} d^{n}_ j(x') & = & d^{n}_ j(x) + (-1)^{i} d^{n}_ j(y) \\ & = & d^{n}_ j(x) - (d^{n}_ j \circ s^{n-1}_{i-1} \circ d^{n}_ i)(x) \\ & = & \begin{cases} d^{n}_ j(x) - d^{n}_ i(x) & \text{ if } j = i \\ d^{n}_ j(x) - (s^{n-2}_{i-1} \circ d^{n-1}_ i \circ d^{n}_ j)(x) & \text{ if } j > i \end{cases}\\ & = & 0. \end{eqnarray*}

Our inductive hypothesis then guarantees that $x'$ is homologous to an element of the subgroup $\widetilde{\mathrm{N}}_{n}(A)$. Since $x$ is homologous to $x'$, it follows that $x$ is also homologous to an element of the subgroup $\widetilde{\mathrm{N}}_{n}(A)$. $\square$