Showing that for an $\mathcal{O}_X$-module $\mathcal{F}$ we have $\operatorname{Hom}(\mathcal{O}_X, \mathcal{F}) \cong \mathcal{F}(X)$.

Question

Let $I$ be an injective $\mathcal{O}_X$-module. For every open set $U \subset X$ and $\mathcal{O}_U$-module $\mathcal{F}$, define an $\mathcal{O}_X$-module $j_!(\mathcal{F})$ by

$$ j_!(\mathcal{F})(V) = \begin{cases} \mathcal{F}(V), & \text{if } V \subset U \\ 0, & \text{if } V \not\subset U. \end{cases} $$

I'm trying to show that $$\operatorname{Hom}_{\mathcal{O}_X}(j_!(\mathcal{O}_U), I) \cong I(U)$$

but I don't quite see how the map should be constructed.

For a ring $R$ and an $R$-module $M$ the isomorphism $\operatorname{Hom}(R,M) \cong M$ is given by $\varphi \mapsto \varphi(1)$, but here I have a morphism of sheaves $\varphi: j_!(\mathcal{O}_U) \to I$ i.e. a family $\varphi_V:j_!(\mathcal{O}_U)(V) \to I(V)$ so the same map doesn't really work. For each morphism I should get back a section of $I(U)$. Will

$$ \varphi\mapsto \varphi(U)(1) $$

work here? The $1$ is a bit ambiguous. If we have a sheaf of rings with unity then there exist such an element, but what if we consider only abelian groups, no such unity exists then?

Answer 1

In general, if $f : (X, \mathcal O_X) \to (Y, \mathcal O_Y)$ is an open immersion of ringed spaces with image $U$ and $\mathcal F$ is a sheaf of $\mathcal O_X$-modules, you can define the presheaf of $\mathcal O_Y$-modules by $$ f_{p!}\mathcal{F}(V) = \begin{cases} \mathcal{F}(f^{-1}V), & \text{if } V \subset U \\ 0, & \text{if } V \not\subset U. \end{cases} $$ The sheaf of $\mathcal O_Y$-modules associated is denoted by $f_!\mathcal F$. You can show that $f_!$ is a functor between the categories of sheaves of $\mathcal O_X$-modules and sheaves of $\mathcal O_Y$-modules (it is even an exact functor).

Exercise 1: The functor $f_!$ is a left-adjoint of the functor $f^*$, in other words for every sheaf of $\mathcal O_X$-modules $\mathcal F$ and for every sheaf of $\mathcal O_Y$-modules $\mathcal G$, we have a natural bijection : $$\operatorname{Hom}_{\mathcal O_X} (\mathcal F, f^*\mathcal G) \simeq \operatorname{Hom}_{\mathcal O_Y} (f_! \mathcal F, \mathcal G)$$

Exercise 2: There exists a natural bijection $$\operatorname{Hom}_{\mathcal O_X}(\mathcal O_X, \mathcal F) \simeq \mathcal F(X)$$

If we adopt the notation $j$ instead of $f$ and we denote the inclusion by $j : U \to X$, the two exercises should help you prove the isomorphism that you want.

Answer 2

Note that $I(U)\cong \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_U, I)$ (this is always true, despite the fact that $I$ is injective) with isomorphism given by the map $\varphi\mapsto \varphi(V)(1)$.
Thus it suffices to show that $\operatorname{Hom}_{\mathcal{O}_X}(j_!(\mathcal{O}_U), I) \cong \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_U, I)$ but this is precisely an expression of the fact that $I$ is an injective object (you have an obvious monomorphism $j_!(\mathcal{O}_U)\to \mathcal{O}_U$ so any morphism $j_!(\mathcal{O}_U)\to I$ can be extended to a unique morphism $\mathcal{O}_U \to I$).