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
and
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
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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