Let $(X,O_X)$ be a ringed space, $E$ - finite locally free $O_X$-module. Let $E^*=Hom_{O_X}(E, O_X)$. How to show, that $E^{**} = E$? It's clear, that locally $E|_U = O_X^n|_U$, and then $E^*|_U = O_X^n|_U$, $E^{**}|_U = O_X^n|_U$. But the second isomorphism is canonical, then we can glue isomorphisms via intersections. It's obvious with canonicity, but how to show it more formally.

• I guess you want to assume that $E$ is finite locally free as an $\mathscr{O}_X$-module? – Keenan Kidwell Nov 3 '12 at 17:14
• Yes. I think, it's wrong in general case. – user46336 Nov 3 '12 at 17:18

If $\mathscr{E}$ is a finite locally free $\mathscr{O}_X$-module, then in particular it is of finite presentation. This implies that for any $\mathscr{O}_X$-module $\mathscr{F}$, the natural map $\mathcal{H}om_{\mathscr{O}_X}(\mathscr{E},\mathscr{F})_x\rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{E}_x,\mathscr{F}_x)$ is an isomorphism.
With this in mind, the fact that the natural map $\mathscr{E}\rightarrow(\mathscr{E}^\vee)^\vee$ is an isomorphism should boil down to the fact that for a ring $R$ and a finite free $R$-module $M$, the canonical map from $M$ to its double $R$-linear dual is an isomorphism. Indeed, granted this fact, the stalks of the natural map from $\mathscr{E}$ to its double dual can be identified with the natural map from $\mathscr{E}_x$ to its double $\mathscr{O}_{X,x}$-linear dual. I'm using here that $\mathscr{E}^\vee$ is also finite locally free, hence finitely presented.
• I don't know this the fact about the isomorphism between stalk of inner $Hom$ and $Hom$ between stalks. What to read before Hartshorne to understand sheafs of modules better? – user46336 Nov 3 '12 at 17:24