Double dual of locally free $\mathcal{O}_X$-module of finite rank

Question

This question has been asked by many people and the answers are mostly suggesting verifying the isomorphism on stalks. But I really have no idea how to proceed: we define $\alpha:\mathcal{E} \to (\mathcal{E}^\vee)^\vee$ by $$\alpha_U:\mathcal{E}(U)\to(\mathcal{E}^\vee)^\vee(U)=Hom_{\mathcal{O}_X|_U}(\mathscr{H}om_{\mathcal{O}_X}(\mathcal{E},U)|_U,\mathcal{O}_X|_U)$$ which sends $s\in \mathcal{E}(U)$ to $f_s:\mathscr{H}om_{\mathcal{O}_X}(\mathcal{E},\mathcal{O}_X)|_U\to \mathcal{O}_X|_U$ where for $V\subseteq U$ any open subset, we have $$f_s(V):Hom_{\mathcal{O}_X|_V}(\mathcal{E}|_V,\mathcal{O}_X|_V)\to \mathcal{O}_X(V)$$ sending an $\mathcal{O}_V$-module morphism $\varphi:\mathcal{E}|_V\to \mathcal{O}_V$ to $\varphi(s|_V)$.
Then I get lost in verifying the isomorphism on stalks. Now we want to show that, for any $p\in X$, we have isomorphism on the stalks $$\alpha_p:\mathcal{E}_p\to ((\mathcal{E}^\vee)^\vee)_p$$ Assume we have $s_p,t_p\in \mathcal{E}_p$ such that $\alpha_p(s_p)=\alpha_p(t_p)$. Then since the trivialization $\{U_i\}_{i\in I}$ of $\mathcal{E}$ covers $X$, we may pick representatives of $s_p,t_p$ from some $\mathcal{E}(U)$ ($U=U_i$ for some $i$). And I am not sure what to do next. Maybe we need to use $\mathcal{E}_U\cong \oplus^n \mathcal{O}_X|_U$ and $\mathcal{E}_p\cong (\mathcal{E}|_U)_p\cong \oplus^n \mathcal{O}_{X,p}$?
Even thought I know that a finite module is isomorphic to its double dual in a natural way. I am not sure what to do next. This is a really stupid question. Thank you for any help!

Answer 1

I think it's totally normal to be confused by this; it's definitely no stupid question:P

You are supposed to use that the stalk $\mathcal{E}_p$ is isomorphic to $\mathcal{O}_{X,p}^{\oplus n}$ (when dealing with vector bundles this is the main item in your toolkit). Then by means of the commutative diagram $$ \require{AMScd} \begin{CD} \mathcal{E}_p @>{\alpha_p}>> (\mathcal{E^{\vee \vee}})_p\\ @V{\cong}VV @VV{\cong}V \\ \mathcal{O}_{X,p}^{\oplus n} @>{\alpha_p'}>> (\mathcal{O}_{X}^{\vee \vee})_p \end{CD} $$ the morphism $$ \alpha_p : \mathcal{E}_p \to (\mathcal{E}^{\vee \vee})_p $$ (where $\alpha'$ is defined just as $\alpha$ but with $\mathcal{O}_{X}^{\oplus n}$ instead of $\mathcal{E}$) identifies with (by TAG 01CP) the evaluation map $$ \mathcal{O}_{X,p}^{\oplus n} \to \text{Hom}_{\mathcal{O}_{X,p}}((\mathcal{O}_{X,p}^{\oplus n})^{\vee},\mathcal{O}_{X,p}) $$ which is an isomorphism, since as you mentioned for finite free modules over an arbitrary ring the evaluation map is an isomorphism.

Hope this helps:)