Theorem 24.10.1. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $i \geq 0$. The functors

are canonically isomorphic.

Theorem 24.10.1. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $i \geq 0$. The functors

\begin{eqnarray*} \textit{Ab}(\mathcal{C}) & \longrightarrow & \textit{Ab} \\ \mathcal{F} & \longmapsto & H^ i(X, \mathcal{F}) \\ \mathcal{F} & \longmapsto & \check{H}^ i_{\text{HC}}(X, \mathcal{F}) \end{eqnarray*}

are canonically isomorphic.

**Proof using spectral sequences..**
Suppose that $\xi \in H^ p(X, \mathcal{F})$ for some $p \geq 0$. Let us show that $\xi $ is in the image of the map $\check{H}^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F})$ of Lemma 24.5.3 for some hypercovering $K$ of $X$.

This is true if $p = 0$ by Lemma 24.5.1. If $p = 1$, choose a Čech hypercovering $K$ of $X$ as in Example 24.3.4 starting with a covering $K_0 = \{ U_ i \to X\} $ in the site $\mathcal{C}$ such that $\xi |_{U_ i} = 0$, see Cohomology on Sites, Lemma 21.8.3. It follows immediately from the spectral sequence in Lemma 24.5.3 that $\xi $ comes from an element of $\check{H}^1(K, \mathcal{F})$ in this case. In general, choose any hypercovering $K$ of $X$ such that $\xi $ maps to zero in $\underline{H}^ p(\mathcal{F})(K_0)$ (using Example 24.3.4 and Cohomology on Sites, Lemma 21.8.3 again). By the spectral sequence of Lemma 24.5.3 the obstruction for $\xi $ to come from an element of $\check{H}^ p(K, \mathcal{F})$ is a sequence of elements $\xi _1, \ldots , \xi _{p - 1}$ with $\xi _ q \in \check{H}^{p - q}(K, \underline{H}^ q(\mathcal{F}))$ (more precisely the images of the $\xi _ q$ in certain subquotients of these groups).

We can inductively replace the hypercovering $K$ by refinements such that the obstructions $\xi _1, \ldots , \xi _{p - 1}$ restrict to zero (and not just the images in the subquotients – so no subtlety here). Indeed, suppose we have already managed to reach the situation where $\xi _{q + 1}, \ldots , \xi _{p - 1}$ are zero. Note that $\xi _ q \in \check{H}^{p - q}(K, \underline{H}^ q(\mathcal{F}))$ is the class of some element

\[ \tilde\xi _ q \in \underline{H}^ q(\mathcal{F})(K_{p - q}) = \prod H^ q(U_ i, \mathcal{F}) \]

if $K_{p - q} = \{ U_ i \to X\} _{i \in I}$. Let $\xi _{q, i}$ be the component of $\tilde\xi _ q$ in $H^ q(U_ i, \mathcal{F})$. As $q \geq 1$ we can use Cohomology on Sites, Lemma 21.8.3 yet again to choose coverings $\{ U_{i, j} \to U_ i\} $ of the site such that each restriction $\xi _{q, i}|_{U_{i, j}} = 0$. Consider the object $Z = \{ U_{i, j} \to X\} $ of the category $\text{SR}(\mathcal{C}, X)$ and its obvious morphism $u : Z \to K_{p - q}$. It is clear that $u$ is a covering, see Definition 24.3.1. By Lemma 24.7.3 there exists a morphism $L \to K$ of hypercoverings of $X$ such that $L_{p - q} \to K_{p - q}$ factors through $u$. Then clearly the image of $\xi _ q$ in $\underline{H}^ q(\mathcal{F})(L_{p - q})$. is zero. Since the spectral sequence of Lemma 24.5.3 is functorial this means that after replacing $K$ by $L$ we reach the situation where $\xi _ q, \ldots , \xi _{p - 1}$ are all zero. Continuing like this we end up with a hypercovering where they are all zero and hence $\xi $ is in the image of the map $\check{H}^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F})$.

Suppose that $K$ is a hypercovering of $X$, that $\xi \in \check{H}^ p(K, \mathcal{F})$ and that the image of $\xi $ under the map $\check{H}^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F})$ of Lemma 24.5.3 is zero. To finish the proof of the theorem we have to show that there exists a morphism of hypercoverings $L \to K$ such that $\xi $ restricts to zero in $\check{H}^ p(L, \mathcal{F})$. By the spectral sequence of Lemma 24.5.3 the vanishing of the image of $\xi $ in $H^ p(X, \mathcal{F})$ means that there exist elements $\xi _1, \ldots , \xi _{p - 2}$ with $\xi _ q \in \check{H}^{p - 1 - q}(K, \underline{H}^ q(\mathcal{F}))$ (more precisely the images of these in certain subquotients) such that the images $d_{q + 1}^{p - 1 - q, q}\xi _ q$ (in the spectral sequence) add up to $\xi $. Hence by exactly the same mechanism as above we can find a morphism of hypercoverings $L \to K$ such that the restrictions of the elements $\xi _ q$, $q = 1, \ldots , p - 2$ in $\check{H}^{p - 1 - q}(L, \underline{H}^ q(\mathcal{F}))$ are zero. Then it follows that $\xi $ is zero since the morphism $L \to K$ induces a morphism of spectral sequences according to Lemma 24.5.3. $\square$

**Proof without using spectral sequences..**
We have seen the result for $i = 0$, see Lemma 24.5.1. We know that the functors $H^ i(X, -)$ form a universal $\delta $-functor, see Derived Categories, Lemma 13.20.4. In order to prove the theorem it suffices to show that the sequence of functors $\check{H}^ i_{HC}(X, -)$ forms a $\delta $-functor. Namely we know that Čech cohomology is zero on injective sheaves (Lemma 24.5.2) and then we can apply Homology, Lemma 12.11.4.

Let

\[ 0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0 \]

be a short exact sequence of abelian sheaves on $\mathcal{C}$. Let $\xi \in \check{H}^ p_{HC}(X, \mathcal{H})$. Choose a hypercovering $K$ of $X$ and an element $\sigma \in \mathcal{H}(K_ p)$ representing $\xi $ in cohomology. There is a corresponding exact sequence of complexes

\[ 0 \to s(\mathcal{F}(K)) \to s(\mathcal{G}(K)) \to s(\mathcal{H}(K)) \]

but we are not assured that there is a zero on the right also and this is the only thing that prevents us from defining $\delta (\xi )$ by a simple application of the snake lemma. Recall that

\[ \mathcal{H}(K_ p) = \prod \mathcal{H}(U_ i) \]

if $K_ p = \{ U_ i \to X\} $. Let $\sigma =\prod \sigma _ i$ with $\sigma _ i \in \mathcal{H}(U_ i)$. Since $\mathcal{G} \to \mathcal{H}$ is a surjection of sheaves we see that there exist coverings $\{ U_{i, j} \to U_ i\} $ such that $\sigma _ i|_{U_{i, j}}$ is the image of some element $\tau _{i, j} \in \mathcal{G}(U_{i, j})$. Consider the object $Z = \{ U_{i, j} \to X\} $ of the category $\text{SR}(\mathcal{C}, X)$ and its obvious morphism $u : Z \to K_ p$. It is clear that $u$ is a covering, see Definition 24.3.1. By Lemma 24.7.3 there exists a morphism $L \to K$ of hypercoverings of $X$ such that $L_ p \to K_ p$ factors through $u$. After replacing $K$ by $L$ we may therefore assume that $\sigma $ is the image of an element $\tau \in \mathcal{G}(K_ p)$. Note that $d(\sigma ) = 0$, but not necessarily $d(\tau ) = 0$. Thus $d(\tau ) \in \mathcal{F}(K_{p + 1})$ is a cocycle. In this situation we define $\delta (\xi )$ as the class of the cocycle $d(\tau )$ in $\check{H}^{p + 1}_{HC}(X, \mathcal{F})$.

At this point there are several things to verify: (a) $\delta (\xi )$ does not depend on the choice of $\tau $, (b) $\delta (\xi )$ does not depend on the choice of the hypercovering $L \to K$ such that $\sigma $ lifts, and (c) $\delta (\xi )$ does not depend on the initial hypercovering and $\sigma $ chosen to represent $\xi $. We omit the verification of (a), (b), and (c); the independence of the choices of the hypercoverings really comes down to Lemmas 24.7.2 and 24.9.2. We also omit the verification that $\delta $ is functorial with respect to morphisms of short exact sequences of abelian sheaves on $\mathcal{C}$.

Finally, we have to verify that with this definition of $\delta $ our short exact sequence of abelian sheaves above leads to a long exact sequence of Čech cohomology groups. First we show that if $\delta (\xi ) = 0$ (with $\xi $ as above) then $\xi $ is the image of some element $\xi ' \in \check{H}^ p_{HC}(X, \mathcal{G})$. Namely, if $\delta (\xi ) = 0$, then, with notation as above, we see that the class of $d(\tau )$ is zero in $\check{H}^{p + 1}_{HC}(X, \mathcal{F})$. Hence there exists a morphism of hypercoverings $L \to K$ such that the restriction of $d(\tau )$ to an element of $\mathcal{F}(L_{p + 1})$ is equal to $d(\upsilon )$ for some $\upsilon \in \mathcal{F}(L_ p)$. This implies that $\tau |_{L_ p} + \upsilon $ form a cocycle, and determine a class $\xi ' \in \check{H}^ p(L, \mathcal{G})$ which maps to $\xi $ as desired.

We omit the proof that if $\xi ' \in \check{H}^{p + 1}_{HC}(X, \mathcal{F})$ maps to zero in $\check{H}^{p + 1}_{HC}(X, \mathcal{G})$, then it is equal to $\delta (\xi )$ for some $\xi \in \check{H}^ p_{HC}(X, \mathcal{H})$. $\square$

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## Comments (2)

Comment #3532 by Kestutis Cesnavicius on

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