If $X$ is a variety, $Z\hookrightarrow X$ a closed subspace and $\jmath: U=X\setminus Z\hookrightarrow X$ the complementary open embedding, then apparently $$0 \ \longrightarrow \ \Gamma_Z(\mathcal{F}) \ \longrightarrow \ \mathcal{F} \ \longrightarrow\jmath_*\jmath^{-1}\mathcal{F} \ \longrightarrow \ 0$$ is exact for any injective sheaf $\mathcal{F}$. Here
- $\Gamma_Z(\mathcal{F})(V)$ are the sections of $\mathcal{F}(V)$ with support on $Z$, and
- $\jmath_*\jmath^{-1}\mathcal{F}(V)=\mathcal{F}(U\cap V)$.
I'm sure the reason this is true is because $R^1\Gamma_Z(\mathcal{F})=0$, but I don't know how to reduce to this. Perhaps it's also possible to check exactness on stalks (on $U$ it's obviously exact).
Could anyone help to resolve my confusion?
If $\mathcal{F}$ is injective, it is in particular flasque.
So let $V$ be any open subsets. We have $$\Gamma(V,\mathcal{F})\to\Gamma(V,j_*j^{-1}\mathcal{F})=\Gamma(V\cap U,\mathcal{F}_{|U})=\Gamma(V\cap U,\mathcal{F})$$
so since, $\mathcal{F}$ is flasque, this map is onto.
This means that the map $\mathcal{F}\to j_*j^{-1}\mathcal{F}$ is onto on every $V$ (or onto as a morphism of presheaves), but this immediately implies that this is onto as a map of sheaves.