# Sheaf cohomology is independent of the flasque resolution

Let $\mathcal{F}$ be a sheaf on a topological space $X$. Through Godement resolution $\mathcal{F}$ admits an exact sequence of sheaves $0\rightarrow\mathcal{F}\rightarrow\mathcal{F}^{0}\rightarrow\mathcal{F}^{1}\rightarrow\mathcal{F}^{2}\rightarrow\cdots$, where $\mathcal{F}^{i},i=0,1,\ldots$ are flasque. Define the $i$-th cohomology group $H^{i}(X,\mathcal{F})$ of a sheaf $\mathcal{F}$ on $X$ as follows:

Suppose $0\rightarrow\mathcal{F}\rightarrow\mathcal{F}^{0}\rightarrow\mathcal{F}^{1}\rightarrow\mathcal{F}^{2}\rightarrow\cdots$ is a flasque resolution of $\mathcal{F}$. Then $H^{i}(X,\mathcal{F})$ is the $i$-th cohomology group of the complex $\mathcal{F}^{0}(X)\rightarrow\mathcal{F}^{1}(X)\rightarrow\mathcal{F}^{2}(X)\rightarrow\cdots$. Explicitly, $H^{i}(X,\mathcal{F}):=\frac{Ker(\mathcal{F}^{i}(X)\rightarrow\mathcal{F}^{i+1}(X))}{Im(\mathcal{F}^{i-1}(X)\rightarrow\mathcal{F}^{i}(X))}$, $i\geq 1$ and $H^{0}(X,\mathcal{F}):=Ker(\mathcal{F}^{0}(X)\rightarrow\mathcal{F}^{1}(X))$.

To check the well-definedness we need to show that for any other flasque resolution $0\rightarrow\mathcal{F}\rightarrow\tilde{\mathcal{F}^{0}}\rightarrow\tilde{\mathcal{F}^{1}}\rightarrow\tilde{\mathcal{F}^{2}}\rightarrow\cdots$ of $\mathcal{F}$, the $i$-th cohomology group of $\mathcal{F}$ on $X$ remains the same modulo isomorphism.

I have showed the well-definedness of the $0$-th cohomology group as follows. Since $0\rightarrow\mathcal{F}\rightarrow\mathcal{F}^{0}\rightarrow\mathcal{F}^{1}\rightarrow\mathcal{F}^{2}\rightarrow\cdots$ and $0\rightarrow\mathcal{F}\rightarrow\tilde{\mathcal{F}^{0}}\rightarrow\tilde{\mathcal{F}^{1}}\rightarrow\tilde{\mathcal{F}^{2}}\rightarrow\cdots$ are exact sequence of sheaves and $\mathcal{F}\rightarrow\mathcal{F}^{0},\mathcal{F}\rightarrow\tilde{\mathcal{F}^{0}}$ are injective sheaf morphisms, by exactness, $Ker(\mathcal{F}^{0}(X)\rightarrow\mathcal{F}^{1}(X))=Im(\mathcal{F}(X)\rightarrow\mathcal{F}^{0}(X))\simeq\mathcal{F}(X)\simeq Im(\mathcal{F}(X)\rightarrow\tilde{\mathcal{F}^{0}}(X))=Ker(\tilde{\mathcal{F}^{0}}(X)\rightarrow\tilde{\mathcal{F}^{1}}(X))$

But I don't know how to establish the well-definedness of the higher cohomology groups. Any help is appreciated.

• Using any flasque resolution, you get the same cohomology groups as using an injective resolution. The result is well known for injective resolutions, so you are done.
– MooS
Commented Nov 17, 2017 at 6:01
• Can you give a reference? I would like to establish this without going into acyclicity or derived functors.
– Sam
Commented Nov 17, 2017 at 7:06
• You can find this in Hartshorne for example. The fact that cohomology is independent of the choice of an injective resolution can be found in any book on homological algebra.
– MooS
Commented Nov 17, 2017 at 7:15
• Can you explain why any flasque resolution gives the same cohomology as an injective resolution?
– Sam
Commented Nov 17, 2017 at 21:13
• This is also in any book on honological algebra. Any acyclic resolution computes the cohomology.
– MooS
Commented Nov 17, 2017 at 21:30

Proving that any old acyclic resolution computes sheaf cohomology is not too bad. Note that flasque sheaves are acyclic, so we can take care of your question with this general approach. First, let $0 \rightarrow \mathscr{F} \rightarrow \mathscr{A}^{\bullet}$ be an acyclic resolution. Because $\Gamma(X,-)$ is left exact, we know that $$H^{0}(\Gamma(X,\mathscr{A}^{\bullet}))=\text{ker}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,\mathscr{A}^{1}) \right)=\Gamma(X,\mathscr{F})=H^{0}(X,\mathscr{F}).$$

Now consider the exact sequence

$$0 \longrightarrow \mathscr{F} \longrightarrow \mathscr{A}^{0} \longrightarrow K \longrightarrow 0.$$

From this, roll out the long exact sequence and observe we have isomorphisms $H^{i}(X,\mathscr{F}) \simeq H^{i-1}(X,K)$ for $i>1$, by the acyclicity of $\mathscr{A}^{0}$. However, $0 \rightarrow K \rightarrow \mathscr{A}^{1} \rightarrow \cdots$ is an injective resolution for $K$, so $H^{i-1}(X,K) \simeq H^{i}(\Gamma(X,\mathscr{A}^{\bullet}))$. In other words for all $i>1$ $$H^{i}(X,\mathscr{F}) \simeq H^{i}(\Gamma(X,\mathscr{A}^{\bullet})).$$

Now for the case where $i=1$. We know that

$$H^{1}(\Gamma(X,\mathscr{A}^{\bullet})) \simeq \Gamma(X,K)/\text{im}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,\mathscr{A}^{1}) \right).$$

But because $K$ is the image of $\mathscr{A}^{0} \rightarrow \mathscr{A}^{1}$, we know that $$\text{im}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,\mathscr{A}^{1}) \right) \simeq \text{im}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,K) \right),$$

and thus, $H^{1}(X,\mathscr{F}) \simeq H^{1}(\Gamma(X,\mathscr{A}^{\bullet})$.

• Let me point out your typo: $K$ is of course the image of $\mathscr A^0 \rightarrow \mathscr A^1$. $H^1(X,\mathscr F)$ is then the cokernel of $\Gamma(X,\mathscr A^0) \rightarrow \Gamma (X,K)$ Commented Jun 3, 2018 at 12:48
• Thanks for the catch! I've made the change. Commented Jun 3, 2018 at 16:29