$Hom(\mathscr{O}_U,\mathscr{I})=\mathscr I(U)$

Question

I am studying algebraic geometry using Hartshorne's book, and in the lema 2.4 from chapter 3 we have that $(X,\mathscr O_X)$ is a ringed space, $\mathscr I$ is an injective $\mathscr O_X$-module, and that $\mathscr O_U$ is the sheaf $\mathscr O_X$ restrict to an open set $U\subset X$, and extended by zero outside $U$.

I understood the proof, but in the end of the proof he writes the equality $$Hom(\mathscr{O}_U,\mathscr{I})=\mathscr I(U),$$ and I can't see how to proof this.

Thanks in advance.

Answer 1

Giving an element in $\operatorname{Hom}(\mathcal{O}_U,\mathscr{I})$ is the same as giving maps $\mathcal{O}_U(V)\to \mathscr{I}(V)$ of $\mathcal{O}_U(V)$-modules, for every open $V\subset U$, which are compatible with restriction. But a morphism of $\mathcal{O}_U(V)$-modules $\mathcal{O}_U(V)\to \mathscr{I}(V)$ is uniquely determined by the image of $1_V\in \mathcal{O}_U(V)$ which is equal to $(1_U)|_V$. Hence, an element $\operatorname{Hom}(\mathcal{O}_U,\mathscr{I})$ is uniquely determined by the image of $1_U\in \mathcal{O}_U(U)$ under the map on global sections $\mathcal{O}_U(U)\to \mathscr{I}(U)$.

Answer 2

Like almost always in algebraic geometry, such things are a formal consequence of the ring theoretic statements, if you just know your functors. You should know $\operatorname{Hom}_R(R,M)=M$ for an $R$-module $M$. The version with sheaves is just a formal consequence:

It is a formal consequence that $\operatorname{Hom}_{\mathcal O_X}(\mathcal O_X,-)$ is the global sections functor, see this answer

Using this, you obtain the result:

Let $j:U \hookrightarrow X$ be the open immersion. Note that $\mathcal O_U = j_!j^{-1}\mathcal O_X$ by defintion. Note that $j_!$ is left adjoint to $j^{-1}$ and $j^{-1}$ is left adjoint to $j_*$. Thus we have

$$\operatorname{Hom}_{\mathcal O_X}(\mathcal O_U,-)=\operatorname{Hom}_{\mathcal O_X}(j_!j^{-1}\mathcal O_X,-)=\operatorname{Hom}_{\mathcal O_{X|U}}(j^{-1}\mathcal O_X,j^{-1}(-))\\=\operatorname{Hom}_{\mathcal O_X}(\mathcal O_X,j_*j^{-1}(-))=\Gamma(X,j_*j^{-1}(-))=\Gamma(U,j^{-1}(-))=\Gamma(U,-).$$

This is what we wanted to show.