Let $(X,O_X)$ be a scheme and let $F$ be some locally free $O_X$-module of finite rank.
Let $\bar{F}=\operatorname{Hom_{O_X}}(F,O_X)$. It is also an $O_X$-module. Show that $\bar{\bar{F}}\cong F$.
Since $F$ is locally free, $X$ can be covered by $(U_i)_{i \in I}$ open subsets of $X$ such that $F_{U_i} \cong O_{X_{U_i}}^{n_i}$
To somewhat lighten up the notation, I shall henceforth write $U_i = U$, $n_i = n$ for some fixed $i$, and $\operatorname{Hom}_{O_X} = \operatorname{Hom}$.
The idea would be to show $\bar{\bar{F_U}} \cong F_U$ and then conclude by gluing the isomorphisms.
So, $\bar{F_U} = \operatorname{Hom}(F_U,O_{X_U})\cong\operatorname{Hom}(O_{X_U}^n,O_{X_U})\cong O_{X_U}^n$ since I just need to pick an element of $O_{X_U}$ for each $e_i=(0,0,\cdots,0,i,0,\cdots,0,0)$ and there are $n$ of them with no strings attached since $O_{X_U}^n$ is free as an $O_{X_U}$-module (no relations).
Now I get to $\bar{\bar{F_U}}$. I should have $\bar{\bar{F_U}}\cong \operatorname{Hom}(O_{X_U}^n,O_{X_U})\cong O_{X_U}^{n} \cong F_u$
This seemed plausible to me at first sight (though I'm not sure the gluing works, I just hope it does, as always). However, the fact that not only $\bar{\bar{F_U}}\cong F_u$ but also $\bar{\bar{F_U}}\cong \bar{F_u} \cong F_u$ befuddles me to say the least, because it's stronger than what I was supposed to prove. I mean, this comes from an exercise, if it were true, wouldn't they simply ask "show that $\bar{F_U} \cong F_U"$? This would make more sense, so I can only conclude that this approach is probably flawed… could somebody point out why this won't work and how I am supposed to tackle this? Does this have to do with the patching up of isomorphisms (which I avoided getting into out of fear)? Thank you.
Best Answer
The standard way to write down a morphism $\mathcal{F} \to (\mathcal{F}^*)^*$ is by sending a section $s\in \mathcal{F}(U)$ to the morphism $\mathcal{F}^*(U)\to \mathcal{O}_X(U)$ given by evaluation at $s$. One can check that this morphism is an isomorphism on stalks if $\mathcal{F}$ is locally free of finite rank, and thus the map is actually an isomorphism of sheaves.
The key issue in your proposed proof is the gluing. It is certainly true that every locally free sheaf of finite rank on any scheme $X$ is locally isomorphic to $\mathcal{O}_X^n$: this is what being locally free means, after all. But globally these sheaves can be very different, and this is all contained in what you skipped over by "hoping the gluing works". For instance, $\mathcal{O}_{\Bbb P^n}(1)$ has dual $\mathcal{O}_{\Bbb P^n}(-1)$, but these sheaves are very different: the former has global sections, while the latter doesn't, and it's all down to how the trivializations get glued together.