The Stacks project

Lemma 14.23.2. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$ and let $k$ be an integer. Let $E$ be the object described in Lemma 14.22.4. Then the complex $s(E)$ is acyclic.

Proof. For a morphism $\alpha : [n] \to [k + 1]$ we define $\alpha ' : [n + 1] \to [k + 1]$ to be the map such that $\alpha '|_{[n]} = \alpha $ and $\alpha '(n + 1) = k + 1$. Note that if the image of $\alpha $ is $[k]$ or $[k + 1]$, then the image of $\alpha '$ is $[k + 1]$. Consider the family of maps $h_ n : E_ n \to E_{n + 1}$ which maps the summand corresponding to $\alpha $ to the summand corresponding to $\alpha '$ via the identity on $A$. Let us compute $d_{n + 1} \circ h_ n - h_{n - 1} \circ d_ n$. We will first do this in case the category $\mathcal{A}$ is the category of abelian groups. Let us use the notation $x_\alpha $ to indicate the element $x \in A$ in the summand of $E_ n$ corresponding to the map $\alpha $ occurring in the index set. Let us also adopt the convention that $x_\alpha $ designates the zero element of $E_ n$ whenever $\mathop{\mathrm{Im}}(\alpha )$ is not $[k]$ or $[k + 1]$. With these conventions we see that

\[ d_{n + 1}(h_ n(x_\alpha )) = \sum \nolimits _{i = 0}^{n + 1} (-1)^ i x_{\alpha ' \circ \delta ^{n + 1}_ i} \]


\[ h_{n - 1}(d_ n(x_\alpha )) = \sum \nolimits _{i = 0}^ n (-1)^ i x_{(\alpha \circ \delta _ i^ n)'} \]

It is easy to see that $\alpha ' \circ \delta ^{n + 1}_ i = (\alpha \circ \delta _ i^ n)'$ for $i = 0, \ldots , n$. It is also easy to see that $\alpha ' \circ \delta ^{n + 1}_{n + 1} = \alpha $. Thus we see that

\[ (d_{n + 1} \circ h_ n - h_{n - 1} \circ d_ n)(x_\alpha ) = (-1)^{n + 1} x_\alpha \]

These identities continue to hold if $\mathcal{A}$ is any abelian category because they hold in the simplicial abelian group $[n] \mapsto \mathop{\mathrm{Hom}}\nolimits (A, E_ n)$; details left to the reader. We conclude that the identity map on $E$ is homotopic to zero, with homotopy given by the system of maps $h'_ n = (-1)^{n + 1}h_ n : E_ n \to E_{n + 1}$. Hence we see that $E$ is acyclic, for example by Homology, Lemma 12.13.5. $\square$

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