The Stacks project

Proof. Let $\mathcal{U}$ be an open covering of $X$. By Lemma 20.11.5 there is an exact sequence

Thus the map is injective. To show surjectivity it suffices to show that any element of $\check{H}^0(\mathcal{U}, \underline{H}^1(\mathcal{F}))$ maps to zero after replacing $\mathcal{U}$ by a refinement. This is immediate from the definitions and the fact that $\underline{H}^1(\mathcal{F})$ is a presheaf of abelian groups whose sheafification is zero by locality of cohomology, see Lemma 20.7.2. $\square$

Proof. We already know that $\check{H}^ n(X, -) \to H^ n(X, -)$ is an isomorphism of functors for $n = 0, 1$, see Lemma 20.16.1. The functors $H^ n(X, -)$ form a universal $\delta $-functor, see Derived Categories, Lemma 13.20.4. If we show that $\check{H}^ n(X, -)$ forms a universal $\delta $-functor and that $\check{H}^ n(X, -) \to H^ n(X, -)$ is compatible with boundary maps, then the map will automatically be an isomorphism by uniqueness of universal $\delta $-functors, see Homology, Lemma 12.12.5.

Let $0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$ be a short exact sequence of abelian sheaves on $X$. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ be an open covering. This gives a complex of complexes

which is in general not exact on the right. The sequence defines the maps

but isn't good enough to define a boundary operator $\delta : \check{H}^ n(\mathcal{U}, \mathcal{H}) \to \check{H}^{n + 1}(\mathcal{U}, \mathcal{F})$. Indeed such a thing will not exist in general. However, given an element $\overline{h} \in \check{H}^ n(\mathcal{U}, \mathcal{H})$ which is the cohomology class of a cocycle $h = (h_{i_0 \ldots i_ n})$ we can choose open coverings

such that $h_{i_0 \ldots i_ n}|_{W_{i_0 \ldots i_ n, k}}$ lifts to a section of $\mathcal{G}$ over $W_{i_0 \ldots i_ n, k}$. By Topology, Lemma 5.13.5 (this is where we use the assumption that $X$ is hausdorff and quasi-compact) we can choose an open covering $\mathcal{V} : X = \bigcup _{j \in J} V_ j$ and $\alpha : J \to I$ such that $V_ j \subset U_{\alpha (j)}$ (it is a refinement) and such that for all $j_0, \ldots , j_ n \in J$ there is a $k$ such that $V_{j_0 \ldots j_ n} \subset W_{\alpha (j_0) \ldots \alpha (j_ n), k}$. We obtain maps of complexes

In fact, the vertical arrows are the maps of complexes used to define the transition maps between the Čech cohomology groups. Our choice of refinement shows that we may choose

The cochain $g = (g_{j_0 \ldots j_ n})$ is not a cocycle in general but we know that its Čech boundary $\text{d}(g)$ maps to zero in $\check{\mathcal{C}}^{n + 1}(\mathcal{V}, \mathcal{H})$ (by the commutative diagram above and the fact that $h$ is a cocycle). Hence $\text{d}(g)$ is a cocycle in $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F})$. This allows us to define

Now, given an element $\xi \in \check{H}^ n(X, \mathcal{G})$ we choose an open covering $\mathcal{U}$ and an element $\overline{h} \in \check{H}^ n(\mathcal{U}, \mathcal{G})$ mapping to $\xi $ in the colimit defining Čech cohomology. Then we choose $\mathcal{V}$ and $g$ as above and set $\delta (\xi )$ equal to the image of $\delta (\overline{h})$ in $\check{H}^ n(X, \mathcal{F})$. At this point a lot of properties have to be checked, all of which are straightforward. For example, we need to check that our construction is independent of the choice of $\mathcal{U}, \overline{h}, \mathcal{V}, \alpha : J \to I, g$. The class of $\text{d}(g)$ is independent of the choice of the lifts $g_{i_0 \ldots i_ n}$ because the difference will be a coboundary. Independence of $\alpha $ holds¹ because a different choice of $\alpha $ determines homotopic vertical maps of complexes in the diagram above, see Section 20.15. For the other choices we use that given a finite collection of open coverings of $X$ we can always find an open covering refining all of them. We also need to check additivity which is shown in the same manner. Finally, we need to check that the maps $\check{H}^ n(X, -) \to H^ n(X, -)$ are compatible with boundary maps. To do this we choose injective resolutions

as in Derived Categories, Lemma 13.18.9. This will give a commutative diagram

Here $\mathcal{U}$ is an open covering as above and the vertical maps are those used to define the maps $\check{H}^ n(\mathcal{U}, -) \to H^ n(X, -)$, see Lemma 20.11.2. The bottom complex is exact as the sequence of complexes of injectives is termwise split exact. Hence the boundary map in cohomology is computed by the usual procedure for this lower exact sequence, see Homology, Lemma 12.13.12. The same will be true after passing to the refinement $\mathcal{V}$ where the boundary map for Čech cohomology was defined. Hence the boundary maps agree because they use the same construction (whenever the first one is defined on an element in Čech cohomology on a given open covering). This finishes our discussion of the construction of the structure of a $\delta $-functor on Čech cohomology and why this structure is compatible with the given $\delta $-functor structure on usual cohomology.

Finally, we may apply Lemma 20.11.1 to see that higher Čech cohomology is trivial on injective sheaves. Hence we see that Čech cohomology is a universal $\delta $-functor by Homology, Lemma 12.12.4. $\square$

Proof. We first prove this for $p = 0$. Injectivity follows from the definition of $\mathcal{F}|_ Z$ and holds in general (for any subset of any topological space $X$). Next, suppose that $s \in H^0(Z, \mathcal{F}|_ Z)$. Then we can find opens $U_ i \subset X$ such that $Z \subset \bigcup U_ i$ and such that $s|_{Z \cap U_ i}$ comes from $s_ i \in \mathcal{F}(U_ i)$. It follows that there exist opens $W_{ij} \subset U_ i \cap U_ j$ with $W_{ij} \cap Z = U_ i \cap U_ j \cap Z$ such that $s_ i|_{W_{ij}} = s_ j|_{W_{ij}}$. Applying Topology, Lemma 5.13.7 we find opens $V_ i$ of $X$ such that $V_ i \subset U_ i$ and such that $V_ i \cap V_ j \subset W_{ij}$. Hence we see that $s_ i|_{V_ i}$ glue to a section of $\mathcal{F}$ over the open neighbourhood $\bigcup V_ i$ of $Z$.

To finish the proof, it suffices to show that if $\mathcal{I}$ is an injective abelian sheaf on $X$, then $H^ p(Z, \mathcal{I}|_ Z) = 0$ for $p > 0$. This follows using short exact sequences and dimension shifting; details omitted. Thus, suppose $\overline{\xi }$ is an element of $H^ p(Z, \mathcal{I}|_ Z)$ for some $p > 0$. By Lemma 20.16.2 the element $\overline{\xi }$ comes from $\check{H}^ p(\mathcal{V}, \mathcal{I}|_ Z)$ for some open covering $\mathcal{V} : Z = \bigcup V_ i$ of $Z$. Say $\overline{\xi }$ is the image of the class of a cocycle $\xi = (\xi _{i_0 \ldots i_ p})$ in $\check{\mathcal{C}}^ p(\mathcal{V}, \mathcal{I}|_ Z)$.

Let $\mathcal{I}' \subset \mathcal{I}|_ Z$ be the subpresheaf defined by the rule

Then $\mathcal{I}|_ Z$ is the sheafification of $\mathcal{I}'$. Thus for every $(p + 1)$-tuple $i_0 \ldots i_ p$ we can find an open covering $V_{i_0 \ldots i_ p} = \bigcup W_{i_0 \ldots i_ p, k}$ such that $\xi _{i_0 \ldots i_ p}|_{W_{i_0 \ldots i_ p, k}}$ is a section of $\mathcal{I}'$. Applying Topology, Lemma 5.13.5 we may after refining $\mathcal{V}$ assume that each $\xi _{i_0 \ldots i_ p}$ is a section of the presheaf $\mathcal{I}'$.

Write $V_ i = Z \cap U_ i$ for some opens $U_ i \subset X$. Since $\mathcal{I}$ is flasque (Lemma 20.12.2) and since $\xi _{i_0 \ldots i_ p}$ is a section of $\mathcal{I}'$ for every $(p + 1)$-tuple $i_0 \ldots i_ p$ we can choose a section $s_{i_0 \ldots i_ p} \in \mathcal{I}(U_{i_0 \ldots i_ p})$ which restricts to $\xi _{i_0 \ldots i_ p}$ on $V_{i_0 \ldots i_ p} = Z \cap U_{i_0 \ldots i_ p}$. (This appeal to injectives being flasque can be avoided by an additional application of Topology, Lemma 5.13.7.) Let $s = (s_{i_0 \ldots i_ p})$ be the corresponding cochain for the open covering $U = \bigcup U_ i$. Since $\text{d}(\xi ) = 0$ we see that the sections $\text{d}(s)_{i_0 \ldots i_{p + 1}}$ restrict to zero on $Z \cap U_{i_0 \ldots i_{p + 1}}$. Hence, by the initial remarks of the proof, there exists open subsets $W_{i_0 \ldots i_{p + 1}} \subset U_{i_0 \ldots i_{p + 1}}$ with $Z \cap W_{i_0 \ldots i_{p + 1}} = Z \cap U_{i_0 \ldots i_{p + 1}}$ such that $\text{d}(s)_{i_0 \ldots i_{p + 1}}|_{W_{i_0 \ldots i_{p + 1}}} = 0$. By Topology, Lemma 5.13.7 we can find $U'_ i \subset U_ i$ such that $Z \subset \bigcup U'_ i$ and such that $U'_{i_0 \ldots i_{p + 1}} \subset W_{i_0 \ldots i_{p + 1}}$. Then $s' = (s'_{i_0 \ldots i_ p})$ with $s'_{i_0 \ldots i_ p} = s_{i_0 \ldots i_ p}|_{U'_{i_0 \ldots i_ p}}$ is a cocycle for $\mathcal{I}$ for the open covering $U' = \bigcup U'_ i$ of an open neighbourhood of $Z$. Since $\mathcal{I}$ has trivial higher Čech cohomology groups (Lemma 20.11.1) we conclude that $s'$ is a coboundary. It follows that the image of $\xi $ in the Čech complex for the open covering $Z = \bigcup Z \cap U'_ i$ is a coboundary and we are done. $\square$