## 21.5 First cohomology and extensions

Lemma 21.5.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules on $\mathcal{C}$. There is a canonical bijection

$\mathop{\mathrm{Ext}}\nolimits ^1_{\textit{Mod}(\mathcal{O})}(\mathcal{O}, \mathcal{F}) \longrightarrow H^1(\mathcal{C}, \mathcal{F})$

which associates to the extension

$0 \to \mathcal{F} \to \mathcal{E} \to \mathcal{O} \to 0$

the image of $1 \in \Gamma (\mathcal{C}, \mathcal{O})$ in $H^1(\mathcal{C}, \mathcal{F})$.

Proof. Let us construct the inverse of the map given in the lemma. Let $\xi \in H^1(\mathcal{C}, \mathcal{F})$. Choose an injection $\mathcal{F} \subset \mathcal{I}$ with $\mathcal{I}$ injective in $\textit{Mod}(\mathcal{O})$. Set $\mathcal{Q} = \mathcal{I}/\mathcal{F}$. By the long exact sequence of cohomology, we see that $\xi$ is the image of a section $\tilde\xi \in \Gamma (\mathcal{C}, \mathcal{Q}) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{O}, \mathcal{Q})$. Now, we just form the pullback

$\xymatrix{ 0 \ar[r] & \mathcal{F} \ar[r] \ar@{=}[d] & \mathcal{E} \ar[r] \ar[d] & \mathcal{O} \ar[r] \ar[d]^{\tilde\xi } & 0 \\ 0 \ar[r] & \mathcal{F} \ar[r] & \mathcal{I} \ar[r] & \mathcal{Q} \ar[r] & 0 }$

see Homology, Section 12.6. $\square$

The following lemma will be superseded by the more general Lemma 21.12.4.

Lemma 21.5.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules on $\mathcal{C}$. Let $\mathcal{F}_{ab}$ denote the underlying sheaf of abelian groups. Then there is a functorial isomorphism

$H^1(\mathcal{C}, \mathcal{F}_{ab}) = H^1(\mathcal{C}, \mathcal{F})$

where the left hand side is cohomology computed in $\textit{Ab}(\mathcal{C})$ and the right hand side is cohomology computed in $\textit{Mod}(\mathcal{O})$.

Proof. Let $\underline{\mathbf{Z}}$ denote the constant sheaf $\mathbf{Z}$. As $\textit{Ab}(\mathcal{C}) = \textit{Mod}(\underline{\mathbf{Z}})$ we may apply Lemma 21.5.1 twice, and it follows that we have to show

$\mathop{\mathrm{Ext}}\nolimits ^1_{\textit{Mod}(\mathcal{O})}(\mathcal{O}, \mathcal{F}) = \mathop{\mathrm{Ext}}\nolimits ^1_{\textit{Mod}(\underline{\mathbf{Z}})}( \underline{\mathbf{Z}}, \mathcal{F}_{ab}).$

Suppose that $0 \to \mathcal{F} \to \mathcal{E} \to \mathcal{O} \to 0$ is an extension in $\textit{Mod}(\mathcal{O})$. Then we can use the obvious map of abelian sheaves $1 : \underline{\mathbf{Z}} \to \mathcal{O}$ and pullback to obtain an extension $\mathcal{E}_{ab}$, like so:

$\xymatrix{ 0 \ar[r] & \mathcal{F}_{ab} \ar[r] \ar@{=}[d] & \mathcal{E}_{ab} \ar[r] \ar[d] & \underline{\mathbf{Z}} \ar[r] \ar[d]^{1} & 0 \\ 0 \ar[r] & \mathcal{F} \ar[r] & \mathcal{E} \ar[r] & \mathcal{O} \ar[r] & 0 }$

The converse is a little more fun. Suppose that $0 \to \mathcal{F}_{ab} \to \mathcal{E}_{ab} \to \underline{\mathbf{Z}} \to 0$ is an extension in $\textit{Mod}(\underline{\mathbf{Z}})$. Since $\underline{\mathbf{Z}}$ is a flat $\underline{\mathbf{Z}}$-module we see that the sequence

$0 \to \mathcal{F}_{ab} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} \to \mathcal{E}_{ab} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} \to \underline{\mathbf{Z}} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} \to 0$

is exact, see Modules on Sites, Lemma 18.28.9. Of course $\underline{\mathbf{Z}} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} = \mathcal{O}$. Hence we can form the pushout via the ($\mathcal{O}$-linear) multiplication map $\mu : \mathcal{F} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} \to \mathcal{F}$ to get an extension of $\mathcal{O}$ by $\mathcal{F}$, like this

$\xymatrix{ 0 \ar[r] & \mathcal{F}_{ab} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} \ar[r] \ar[d]^\mu & \mathcal{E}_{ab} \otimes _{\underline{\mathbf{Z}}} \mathcal{O} \ar[r] \ar[d] & \mathcal{O} \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & \mathcal{F} \ar[r] & \mathcal{E} \ar[r] & \mathcal{O} \ar[r] & 0 }$

which is the desired extension. We omit the verification that these constructions are mutually inverse. $\square$

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