Let $\mathscr{E}$ be a locally free sheaf (on a ringed space $X$) of rank $n$. Then I want to prove that $\mathscr{E} \cong \mathscr{E}^{\vee\vee}$. My argument for the proof is rather categorical and gets stuck halfway.
First, let $\{U_i\}$ be a trivialization of $\mathscr{E}$ (i.e. $\mathscr{E}|_{U_i} \cong \mathcal{O}_{U_i}^{\oplus n}$. Note that the collection
$$\mathcal{B} = \{U \in \text{Open}(X) : U \subseteq U_i \text{ for some } i\}$$
forms a basis of $X$. There is a natural map $\alpha : \mathscr{E} \to \mathscr{E}^{\vee\vee}$ on the basis $\mathcal{B}$. Let $V \in \mathcal{B}$. The $V$-component of $\alpha$ is a map
$$\alpha_V : \mathscr{E}(V) \to \mathcal{H}om(\mathcal{H}om(\mathscr{E}, \mathcal{O}_X)|_V, \mathcal{O}_V)$$
Given $s \in \mathscr{E}(V)$, define the natural transformation $\alpha_V(s) : \mathcal{H}om(\mathscr{E}, \mathcal{O}_X)|_V \to \mathcal{O}_V$ with $W$-component
$$\begin{array}{cccc}
\alpha_V(s)_W : & \hom(\mathscr{E}|_W, \mathcal{O}_W) & \to & \mathcal{O}_V(W) \\
& \beta : \mathscr{E}|_W \to \mathcal{O}_W & \mapsto & \beta_W(s|_W)
\end{array}$$
What makes me stuck is the inverse of this map. Here is what I tried. Now define a natural transformation $\rho: \mathscr{E}^{\vee\vee} \to \mathscr{E}$ on the basis $\mathcal{B}$ as follows. Given $V \in \mathcal{B}$, the $V$-component of $\rho$ is a map
$$\rho_V : \mathcal{H}om(\mathcal{H}om(\mathscr{E}, \mathcal{O}_X)|_V, \mathcal{O}_V) \to \mathcal{E}(V)$$
Now take a morphism $\theta : \mathcal{H}om(\mathscr{E}, \mathcal{O}_X)|_V \to \mathcal{O}_V$. Using the trivialization earlier, the $V$-component $\theta_V : \hom(\mathscr{E}|_V, \mathcal{O}_V) \to \mathcal{O}_V(V)$ sends the composition
$$\mathscr{E}|_V \xrightarrow{\cong} \mathcal{O}_V^{\oplus n} \xrightarrow{\mathrm{proj.}} \mathcal{O}_V$$
to an element in $\mathcal{O}_V(V)$. (I’m not sure which projection should be used, perhaps any of them works?) Now define $\rho_V(\theta)$ to be the image of this element under the diagonal map $\mathcal{O}_V(V) \to \mathcal{O}_V(V)^{\oplus n} \cong \mathscr{E}(V)$. (I hope the construction makes sense, as I’m so confused as well.) And I have no ideal on whether this is indeed the inverse map we want. (Again, the requirement of "choosing a projection" shocks my categorical mind so much that I think some mistake is made.)
Best Answer
I think it's better to reduce to checking that $\alpha$ is an isomorphism on stalks, which lets you fall back to a claim you should already be familiar with: for any finite free module $M$ over a ring $R$, $M\cong M^{**}$ canonically by the map sending $m$ to evaluation at $m$, which is exactly what $\alpha$ does on stalks. (See theorem 4.2 on page 7 of Keith Conrad's notes for a refresher, if you need it.) The meta-lesson with sheaves here is that it's often a good idea to define globally and check locally.