Lemma 25.4.2. Let $\mathcal{C}$ be a site. Let $\mathcal{F} \to \mathcal{G}$ be a morphism of presheaves of sets. Denote $K$ the simplicial object of $\textit{PSh}(\mathcal{C})$ whose $n$th term is the $(n + 1)$st fibre product of $\mathcal{F}$ over $\mathcal{G}$, see Simplicial, Example 14.3.5. Then, if $\mathcal{F} \to \mathcal{G}$ is surjective after sheafification, we have

$H_ i(K) = \left\{ \begin{matrix} 0 & \text{if} & i > 0 \\ \mathbf{Z}_\mathcal {G}^\# & \text{if} & i = 0 \end{matrix} \right.$

The isomorphism in degree $0$ is given by the morphism $H_0(K) \to \mathbf{Z}_\mathcal {G}^\#$ coming from the map $(\mathbf{Z}_ K^\# )_0 = \mathbf{Z}_\mathcal {F}^\# \to \mathbf{Z}_\mathcal {G}^\#$.

Proof. Let $\mathcal{G}' \subset \mathcal{G}$ be the image of the morphism $\mathcal{F} \to \mathcal{G}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Set $A = \mathcal{F}(U)$ and $B = \mathcal{G}'(U)$. Then the simplicial set $K(U)$ is equal to the simplicial set with $n$-simplices given by

$A \times _ B A \times _ B \ldots \times _ B A\ (n + 1 \text{ factors)}.$

By Simplicial, Lemma 14.32.3 the morphism $K(U) \to B$ is a trivial Kan fibration. Thus it is a homotopy equivalence (Simplicial, Lemma 14.30.8). Hence applying the functor “free abelian group on” to this we deduce that

$\mathbf{Z}_ K(U) \longrightarrow \mathbf{Z}_ B$

is a homotopy equivalence. Note that $s(\mathbf{Z}_ B)$ is the complex

$\ldots \to \bigoplus \nolimits _{b \in B}\mathbf{Z} \xrightarrow {0} \bigoplus \nolimits _{b \in B}\mathbf{Z} \xrightarrow {1} \bigoplus \nolimits _{b \in B}\mathbf{Z} \xrightarrow {0} \bigoplus \nolimits _{b \in B}\mathbf{Z} \to 0$

see Simplicial, Lemma 14.23.3. Thus we see that $H_ i(s(\mathbf{Z}_ K(U))) = 0$ for $i > 0$, and $H_0(s(\mathbf{Z}_ K(U))) = \bigoplus _{b \in B}\mathbf{Z} = \bigoplus _{s \in \mathcal{G}'(U)} \mathbf{Z}$. These identifications are compatible with restriction maps.

We conclude that $H_ i(s(\mathbf{Z}_ K)) = 0$ for $i > 0$ and $H_0(s(\mathbf{Z}_ K)) = \mathbf{Z}_{\mathcal{G}'}$, where here we compute homology groups in $\textit{PAb}(\mathcal{C})$. Since sheafification is an exact functor we deduce the result of the lemma. Namely, the exactness implies that $H_0(s(\mathbf{Z}_ K))^\# = H_0(s(\mathbf{Z}_ K^\# ))$, and similarly for other indices. $\square$

Comment #206 by Rex on

Typo: "given by the morphsm"

Comment #6778 by Bogdan on

I think the reference in "Thus it is a homotopy equivalence (Simplicial, Lemma 14.32.3)" should refer to some other lemma.

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