Lemma 85.10.1. In Situation 85.3.3 let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$. There is a complex
of $\mathcal{O}$-modules which forms a resolution of $\mathcal{O}$. Here $g_{n!}$ is as in Lemma 85.6.1.
Lemma 85.10.1. In Situation 85.3.3 let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$. There is a complex
of $\mathcal{O}$-modules which forms a resolution of $\mathcal{O}$. Here $g_{n!}$ is as in Lemma 85.6.1.
Proof. We will use the description of $g_{n!}$ given in Lemma 85.3.5. As maps of the complex we take $\sum (-1)^ i d^ n_ i$ where $d^ n_ i : g_{n!}\mathcal{O}_ n \to g_{n - 1!}\mathcal{O}_{n - 1}$ is the adjoint to the map $\mathcal{O}_ n \to \bigoplus _{[n - 1] \to [n]} \mathcal{O}_ n = g_ n^*g_{n - 1!}\mathcal{O}_{n - 1}$ corresponding to the factor labeled with $\delta ^ n_ i : [n - 1] \to [n]$. Then $g_ m^{-1}$ applied to the complex gives the complex
on $\mathcal{C}_ m$. In other words, this is the complex associated to the free $\mathcal{O}_ m$-module on the simplicial set $\Delta [m]$, see Simplicial, Example 14.11.2. Since $\Delta [m]$ is homotopy equivalent to $\Delta [0]$, see Simplicial, Example 14.26.7, and since “taking free abelian sheaf on” is a functor, we see that the complex above is homotopy equivalent to the free abelian sheaf on $\Delta [0]$ (Simplicial, Remark 14.26.4 and Lemma 14.27.2). This complex is acyclic in positive degrees and equal to $\mathcal{O}_ m$ in degree $0$. $\square$
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