The Stacks project

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24.4 Acyclicity

Let $\mathcal{C}$ be a site. For a presheaf of sets $\mathcal{F}$ we denote $\mathbf{Z}_\mathcal {F}$ the presheaf of abelian groups defined by the rule

\[ \mathbf{Z}_\mathcal {F}(U) = \text{free abelian group on }\mathcal{F}(U). \]

We will sometimes call this the free abelian presheaf on $\mathcal{F}$. Of course the construction $\mathcal{F} \mapsto \mathbf{Z}_\mathcal {F}$ is a functor and it is left adjoint to the forgetful functor $\textit{PAb}(\mathcal{C}) \to \textit{PSh}(\mathcal{C})$. Of course the sheafification $\mathbf{Z}_\mathcal {F}^\# $ is a sheaf of abelian groups, and the functor $\mathcal{F} \mapsto \mathbf{Z}_\mathcal {F}^\# $ is a left adjoint as well. We sometimes call $\mathbf{Z}_\mathcal {F}^\# $ the free abelian sheaf on $\mathcal{F}$.

For an object $X$ of the site $\mathcal{C}$ we denote $\mathbf{Z}_ X$ the free abelian presheaf on $h_ X$, and we denote $\mathbf{Z}_ X^\# $ its sheafification.

Definition 24.4.1. Let $\mathcal{C}$ be a site. Let $K$ be a simplicial object of $\textit{PSh}(\mathcal{C})$. By the above we get a simplicial object $\mathbf{Z}_ K^\# $ of $\textit{Ab}(\mathcal{C})$. We can take its associated complex of abelian presheaves $s(\mathbf{Z}_ K^\# )$, see Simplicial, Section 14.23. The homology of $K$ is the homology of the complex of abelian sheaves $s(\mathbf{Z}_ K^\# )$.

In other words, the $i$th homology $H_ i(K)$ of $K$ is the sheaf of abelian groups $H_ i(K) = H_ i(s(\mathbf{Z}_ K^\# ))$. In this section we worry about the homology in case $K$ is a hypercovering of an object $X$ of $\mathcal{C}$.

Lemma 24.4.2. 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 14.32.3). 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. $\square$

Lemma 24.4.3. Let $\mathcal{C}$ be a site. Let $f : L \to K$ be a morphism of simplicial objects of $\textit{PSh}(\mathcal{C})$. Let $n \geq 0$ be an integer. Assume that

  1. For $i < n$ the morphism $L_ i \to K_ i$ is an isomorphism.

  2. The morphism $L_ n \to K_ n$ is surjective after sheafification.

  3. The canonical map $L \to \text{cosk}_ n \text{sk}_ n L$ is an isomorphism.

  4. The canonical map $K \to \text{cosk}_ n \text{sk}_ n K$ is an isomorphism.

Then $H_ i(f) : H_ i(L) \to H_ i(K)$ is an isomorphism.

Proof. This proof is exactly the same as the proof of Lemma 24.4.2 above. Namely, we first let $K_ n' \subset K_ n$ be the sub presheaf which is the image of the map $L_ n \to K_ n$. Assumption (2) means that the sheafification of $K_ n'$ is equal to the sheafification of $K_ n$. Moreover, since $L_ i = K_ i$ for all $i < n$ we see that get an $n$-truncated simplicial presheaf $U$ by taking $U_0 = L_0 = K_0, \ldots , U_{n - 1} = L_{n - 1} = K_{n - 1}, U_ n = K'_ n$. Denote $K' = \text{cosk}_ n U$, a simplicial presheaf. Because we can construct $K'_ m$ as a finite limit, and since sheafification is exact, we see that $(K'_ m)^\# = K_ m$. In other words, $(K')^\# = K^\# $. We conclude, by exactness of sheafification once more, that $H_ i(K) = H_ i(K')$. Thus it suffices to prove the lemma for the morphism $L \to K'$, in other words, we may assume that $L_ n \to K_ n$ is a surjective morphism of presheaves!

In this case, for any object $U$ of $\mathcal{C}$ we see that the morphism of simplicial sets

\[ L(U) \longrightarrow K(U) \]

satisfies all the assumptions of Simplicial, Lemma 14.32.1. Hence it is a trivial Kan fibration. In particular it is a homotopy equivalence (Simplicial, Lemma 14.30.8). Thus

\[ \mathbf{Z}_ L(U) \longrightarrow \mathbf{Z}_ K(U) \]

is a homotopy equivalence too. This for all $U$. The result follows. $\square$

Lemma 24.4.4. Let $\mathcal{C}$ be a site. Let $K$ be a simplicial presheaf. Let $\mathcal{G}$ be a presheaf. Let $K \to \mathcal{G}$ be an augmentation of $K$ towards $\mathcal{G}$. Assume that

  1. The morphism of presheaves $K_0 \to \mathcal{G}$ becomes a surjection after sheafification.

  2. The morphism

    \[ (d^1_0, d^1_1) : K_1 \longrightarrow K_0 \times _\mathcal {G} K_0 \]

    becomes a surjection after sheafification.

  3. For every $n \geq 1$ the morphism

    \[ K_{n + 1} \longrightarrow (\text{cosk}_ n \text{sk}_ n K)_{n + 1} \]

    turns into a surjection after sheafification.

Then $H_ i(K) = 0$ for $i > 0$ and $H_0(K) = \mathbf{Z}_\mathcal {G}^\# $.

Proof. Denote $K^ n = \text{cosk}_ n \text{sk}_ n K$ for $n \geq 1$. Define $K^0$ as the simplicial object with terms $(K^0)_ n$ equal to the $(n + 1)$-fold fibred product $K_0 \times _\mathcal {G} \ldots \times _\mathcal {G} K_0$, see Simplicial, Example 14.3.5. We have morphisms

\[ K \longrightarrow \ldots \to K^ n \to K^{n - 1} \to \ldots \to K^1 \to K^0. \]

The morphisms $K \to K^ i$, $K^ j \to K^ i$ for $j \geq i \geq 1$ come from the universal properties of the $\text{cosk}_ n$ functors. The morphism $K^1 \to K^0$ is the canonical morphism from Simplicial, Remark 14.20.4. We also recall that $K^0 \to \text{cosk}_1 \text{sk}_1 K^0$ is an isomorphism, see Simplicial, Lemma 14.20.3.

By Lemma 24.4.2 we see that $H_ i(K^0) = 0$ for $i > 0$ and $H_0(K^0) = \mathbf{Z}_\mathcal {G}^\# $.

Pick $n \geq 1$. Consider the morphism $K^ n \to K^{n - 1}$. It is an isomorphism on terms of degree $< n$. Note that $K^ n \to \text{cosk}_ n \text{sk}_ n K^ n$ and $K^{n - 1} \to \text{cosk}_ n \text{sk}_ n K^{n - 1}$ are isomorphisms. Note that $(K^ n)_ n = K_ n$ and that $(K^{n - 1})_ n = (\text{cosk}_{n - 1} \text{sk}_{n - 1} K)_ n$. Hence by assumption, we have that $(K^ n)_ n \to (K^{n - 1})_ n$ is a morphism of presheaves which becomes surjective after sheafification. By Lemma 24.4.3 we conclude that $H_ i(K^ n) = H_ i(K^{n - 1})$. Combined with the above this proves the lemma. $\square$

Lemma 24.4.5. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K$ be a hypercovering of $X$. The homology of the simplicial presheaf $F(K)$ is $0$ in degrees $> 0$ and equal to $\mathbf{Z}_ X^\# $ in degree $0$.


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