Any acyclic resolution computes sheaf cohomology

Let $X$ be any topological space and $\mathscr{F}$ a sheaf on $X$ valued in $\textbf{Ab}$, the category of abelian groups. For now, any sheaf we will discuss will be valued in $\textbf{Ab}.$

A resolution of $\mathscr{F}$ is an exact sequence of the form $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{R}^{0} \rightarrow \mathscr{R}^{1} \rightarrow \cdots.$$ Let's state facts we will use without proofs.

Fact 1. The category of sheaves on $X$ (valued in $\textbf{Ab}$) has enough injectives. For our purpose, this just means the following: given any sheaf $\mathscr{F}$on $X,$ we may construct an injective resolution $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{I}^{0} \rightarrow \mathscr{I}^{1} \rightarrow \cdots$$ of sheaves on $X,$ meaning that this is a resolution where $\mathscr{I}^{j}$ are injective objects in the category of sheaves.

Fact 2. Given any two injective resolutions $0 \rightarrow \mathscr{F} \rightarrow \mathscr{I}^{\bullet}$ and $0 \rightarrow \mathscr{F} \rightarrow \mathscr{J}^{\bullet},$ we have $$H^{i}(\Gamma(X, \mathscr{I}^{\bullet})) \simeq H^{i}(\Gamma(X, \mathscr{J}^{\bullet}))$$ for all $i \geq 0.$ Hence, we will just define $$H^{i}(X, \mathscr{F}) := H^{i}(\Gamma(X, \mathscr{I}^{\bullet})),$$ which only make sense up to isomorphisms.

Remark. It is immediate that $$H^{0}(X, \mathscr{F}) = \Gamma(X, \mathscr{F})$$ because the global section functor is left-exact.

Fact 3. Given any short exact sequence $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0$$ of sheaves on $X,$ we have a long exact sequence $$\cdots \rightarrow H^{i}(X, \mathscr{A}) \rightarrow H^{i}(X, \mathscr{B}) \rightarrow H^{i}(X, \mathscr{C}) \rightarrow H^{i+1}(X, \mathscr{A}) \rightarrow \cdots.$$

Remark. We will sketch the proof of Fact 3 at the end of this posting. The definition using the Godement resolution of a sheaf makes Fact 3 almost automatically true, but then we need to switch the order of the exposition.

Lemma. Let $\mathscr{I}$ be any injective sheaf on $X.$ Then $$H^{i}(\Gamma(X, \mathscr{I})) = 0$$ for $i \geq 1.$

Proof. We may use the following injective resolution: $$0 \rightarrow \mathscr{I} \rightarrow \mathscr{I} \rightarrow 0 \rightarrow 0 \rightarrow \cdots,$$ which computes what we want. $\Box$

We say that a sheaf $\mathscr{A}$ is acyclic if $$H^{i}(X, \mathscr{A}) = 0$$ for all $i \geq 1.$ Note that we have just shown that injective sheaves are acyclic.

Theorem. Let $\mathscr{F}$ be a sheaf on a topological space $X.$ For any resolution $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{A}^{\bullet},$$ we have a natural map $$H^{i}(\Gamma(X, \mathscr{A}^{\bullet})) \rightarrow H^{i}(X, \mathscr{F}),$$ meaning it is a map in $\textbf{Ab}$ functorial in sheaves along with choices of resolutions.

Moreover, if each $\mathscr{A}^{j}$ is acyclic, then this map is an isomorphism.

Proof. We follow Theorem 3.13 in Wells. Take the exact sequence $$0 \rightarrow \mathscr{K}^{j} \rightarrow \mathscr{A}^{i} \rightarrow \mathscr{A}^{i+1},$$ which gives the exact sequence $$0 \rightarrow \Gamma(X, \mathscr{K}^{i}) \rightarrow \Gamma(X, \mathscr{A}^{i}) \rightarrow \Gamma(X, \mathscr{A}^{i+1}).$$ (Note that $\mathscr{K}^{0} = \mathscr{F}.$) Consider the induced short exact sequences $$0 \rightarrow \mathscr{K}^{i-1} \rightarrow \mathscr{A}^{i-1} \rightarrow \mathscr{K}^{i} \rightarrow 0.$$ Each such sequence induces the long exact sequence $$0 \rightarrow \Gamma(X, \mathscr{K}^{i-1}) \rightarrow \Gamma(X, \mathscr{A}^{i-1}) \rightarrow \Gamma(X, \mathscr{K}^{i}) \rightarrow H^{1}(X, \mathscr{K}^{i-1}) \rightarrow H^{1}(X, \mathscr{A}^{i}) \cdots.$$ Thus, we have $$H^{i}(\Gamma(X, \mathscr{A}^{\bullet})) = \frac{\Gamma(X, \mathscr{K}^{i})}{\mathrm{im}(\Gamma(X, \mathscr{A}^{i-1}) \rightarrow \Gamma(X, \mathscr{K}^{i}))} \rightarrow H^{1}(X, \mathscr{K}^{i-1}),$$ which we call $\alpha_{i,1},$ which is injective due to the suitable part of the long exact sequence above. Note that $\alpha_{i,1}$ is also surjective (and hence an isomorphism) if $H^{1}(X, \mathscr{A}^{i}) = 0.$

Fix $2 \leq j \leq i$ and consider the induced short exact sequences $$0 \rightarrow \mathscr{K}^{i-j} \rightarrow \mathscr{A}^{i-j} \rightarrow \mathscr{K}^{i-j+1} \rightarrow 0.$$ We get a long exact sequence $$\cdots \rightarrow H^{j-1}(X, \mathscr{A}^{i-j}) \rightarrow H^{j-1}(X, \mathscr{K}^{i-j+1}) \overset{\alpha_{i,j}}{\longrightarrow} H^{j}(X, \mathscr{K}^{i-j}) \rightarrow H^{j}(X, \mathscr{A}^{i-j}) \cdots.$$ Note that if $$H^{j-1}(X, \mathscr{A}^{i-j}) = 0 = H^{j}(X, \mathscr{A}^{i-j}),$$ then $\alpha_{i,j}$ is an isomorphism. Thus, we have constructed the composition $$H^{i}(\Gamma(X, \mathscr{A}^{\bullet})) \xrightarrow{\alpha_{i,1}} H^{1}(X, \mathscr{K}^{i-1}) \xrightarrow{\alpha_{i,2}} \cdots \xrightarrow{\alpha_{i,i-1}} H^{i-1}(X, \mathscr{K}^{1}) \xrightarrow{\alpha_{i,i}} H^{i}(X, \mathscr{F})$$ of the maps, which of which is an isomorphism if all $\mathscr{A}^{0}, \mathscr{A}^{1}, \dots$ are acyclic. Naturality should follow from Snake Lemma. This finishes the proof. $\Box$

Sketch of proof of Fact 3. Let $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0$$ be a short exact sequences of sheaves on $X.$ Choose injective resolutions $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{A}^{0} \rightarrow \mathscr{A}^{1} \rightarrow \cdots$$ and $$0 \rightarrow \mathscr{C} \rightarrow \mathscr{C}^{0} \rightarrow \mathscr{C}^{1} \rightarrow \cdots$$ using Fact 1. Considering the exact sequence $0 \rightarrow \mathscr{A} \rightarrow \mathscr{B},$ the injectivity of $\mathscr{A}^{0}$ induces a map $\mathscr{B} \rightarrow \mathscr{A}^{0}$ such that the map $\mathscr{A} \rightarrow \mathscr{A}^{0}$ factors as $$\mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{A}^{0}.$$ Using $\mathscr{B} \rightarrow \mathscr{A}^{0}$ and $\mathscr{B} \rightarrow \mathscr{C} \rightarrow \mathscr{C}^{0},$ we may induce a unique map $$\mathscr{B} \rightarrow \mathscr{A}^{0} \times \mathscr{C}^{0}$$ using the universal property of the product. If we pretend as if we can chase elements, we can show that $$0 \rightarrow \mathscr{B} \rightarrow \mathscr{A}^{0} \times \mathscr{C}^{0}$$ is exact. I remember Chapter 8 of MacLane's book allows us to do this in any general abelian category setting. I think in our case, it is easier because exactness can be checked at the level of stalks (which are abelian groups). Once we show this, we can construct a short exact sequence of injective resolutions, where taking the global sections at each level (except for the given sheaves) gives exact sequences. Then by applying Snake Lemma, we are done. $\Box$

Remark. The category $\textbf{Ab}$ can probably be replaced by any abelian category. The argument used in the (sketch of) proof of Fact 3 should be called "injective Horseshoe lemma".