How to show that $\mathcal{F} \otimes \mathcal{F}^\vee \cong \mathcal{O}_X$ Let $\mathcal{F}$ be a rank $1$ locally free sheaf. If we define $\mathcal{F}^\vee = Hom_{\mathcal{O}_X}(F, \mathcal{O}_X)$, then how would one go about showing that $\mathcal{F} \otimes \mathcal{F}^\vee \cong \mathcal{O}_X$? My thinking is that we can map $\mathcal{F} \otimes \mathcal{F}^\vee$ into $\mathcal{O}_X$ by through an element of $Hom_{\mathcal{O}_X}(F, \mathcal{O}_X)$, but I don't know what the formalization would look like.
 A: We have a canonical map $\phi \colon \mathcal{F} \otimes \mathcal{F}^{\vee} \to \mathcal{O}_X$ , which I'll call the evaluation map, obtained using the universal property of the tensor product of $\mathcal{O}_X$-modules : it is the map induced by the bilinear mapping 
$$\phi_{\text{bil}} \colon \mathcal{F}\times \mathcal{F}^{{\vee}} \to \mathcal{O}_X$$
such that 
$$(\phi_{\text{bil}})_U \colon \mathcal{F}(U) \times \mathrm{Hom}_{\mathcal{O}_U}(\mathcal{F}_{|U},\mathcal{O}_U) \to \mathcal{O}_U(U)$$
$$ (s,\xi)\mapsto\xi_{U}(s)$$
that is the expected evaluation pairing. Then it suffices to show that the map is an isomorphism on stalks. Since taking stalks commutes with $\mathcal{Hom}$ and with tensor product in the right way, we have to verify, using that $\mathcal{F}$ is locally free of rank $1$, the following statement of commutative algebra : let $A$ be a commutative ring and $M$ a free $A$-module of rank $1$. Then $M\otimes M^{\vee}\simeq A$, where the isomorphism is given by the evaluation map $x\otimes\xi \mapsto \xi(x) \in A$.
Actually, one can also work without stalks and show that it is an isomorphism when restricted to each open of an open cover of $X$, since $\mathcal{Hom}$ and tensor product commute with restriction to an open subset. It then suffices to show the result  for a free sheaf of rank $1$ :
So, let $\mathcal{F}$ be a free sheaf of rank $1$. Then the evaluation map is functorial in $\mathcal{F}$ so we get the following commutative diagram :
$$\require{AMScd}
\begin{CD}
\mathcal{F}\otimes\mathcal{F}^{\vee} @>>> \mathcal{O}_X \\
@V{\simeq}VV @VV{\mathrm{id}}V \\
\mathcal{O}_X\otimes\mathcal{O}_X^{\vee}  @>>> \mathcal{O}_X
\end{CD}$$
obtained by using the given isomorphism $\mathcal{F}\simeq\mathcal{O}_X$. Thus is suffices to show that the evaluation map for $\mathcal{O}_X$ is an isomorphism.
Note $\mathcal{O}_X\otimes_p\mathcal{O}_X^{\vee}$ the presheaf $U \mapsto \mathcal{O}_X(U)\otimes_{\mathcal{O}_X(U)}\mathcal{O}_X^{\vee}(U)$.Then $\phi$ is induced from a map $\phi_p \colon\mathcal{O}_X\otimes_p\mathcal{O}_X^{\vee}\to \mathcal{O}_X $ (with a similar definition) by UP of sheafication. We define a map $\psi_p \colon \mathcal{O}_X \to \mathcal{O}_X\otimes_p\mathcal{O}_X^{\vee}$ by
$$ (\psi_p)_U(x)=x\otimes \mathrm{id}_{\mathcal{O}_U}$$
We now have to check that both composites are the identity. 
First $$(\phi_p \circ \psi_p)_U (x)=(\phi_p)_U (x \otimes \mathrm{id}_{\mathcal{O}_U})=\mathrm{id}_{\mathcal{O}_U(U)}(x) = x $$ so $\phi_p\circ\psi_p=\mathrm{id}$.
Secondly, denote $\lambda$ the canonical bilinear map $\mathcal{O}_X\times\mathcal{O}_X^{\vee} \to \mathcal{O}_X\otimes_p\mathcal{O}_X^{\vee}$. It suffices, by UP of tensor product, to check $\psi_p \circ \phi_{\text{bil}}=\lambda$, i.e. that $(\psi_U)(\xi_{U}(s))= s\otimes \xi$. But
$$ (\psi_U)(\xi_{U}(s))=\xi_U(s)\otimes \mathrm{id}_{\mathcal{O}_U}=(\xi_U(1)\cdot s \otimes \mathrm{id}_{\mathcal{O}_U}) = \xi_U(1) \cdot (s \otimes \mathrm{id}_{\mathcal{O}_U}) = s \otimes (\xi_U(1)\cdot \mathrm{id}_{\mathcal{O}_U}) $$
We notice that $\xi_U(1)\cdot \mathrm{id}_{\mathcal{O}_U}$ is the map that sends $a \in \mathcal{O}_U(V)$ to 
$$a\cdot\xi_U(1)_{|V}=a\cdot\xi_V(1)=\xi_V(a)$$
 (by definition of the action of $\mathcal{O}_U(U)$ on $\mathcal{O}_X^{\vee}(U)$), so it is exactly $\xi$ and we are done. The isomorphism we obtain says that $\mathcal{O}_X\otimes_p\mathcal{O}_X^{\vee}$ is actually already a sheaf, so the sheafification morphism is an isomorphism and so is $\phi$.
