TITLE: Pullback commutes with dual for locally free sheaf of finite rank
QUESTION [19 upvotes]: Let $ f:X\rightarrow Y$ be a morphism of ringed spaces. Let $ \mathscr{E} $ be an $\mathcal{O}_Y$ module that is locally free of finite rank. I want to show that $ (f^{*}\mathscr{E})^\vee\cong f^*(\mathscr{E}^\vee)$ where $\vee$ denotes the dual.
I can easily see how stalks are isomorphic since direct limit and tensor product commute with direct sum. What I'm unable to do is construct a morphism from one of these modules to the other. Once I have it, I think I can show that it induces isomorphisms on stalks. Could you please help me?
This is a step in proving the projection formula, so I cannot use this formula here.

REPLY [3 votes]: You can use the isomorphism
$$
\mathcal F\otimes_{\mathcal R}\mathcal E^*=\mathcal Hom_{\mathcal R}(\mathcal E,\mathcal F)
$$
which is valid for every locally free $\mathcal R$-module. (The isomorphism follows from the fact that the hom sheaf and the tensor product are cocontinuous.) Similarly, because $f^{-1}$ is cocontinuous, $f^{-1}$ commutes with taking duals of locally free modules. 
Using the isomorphism, you may compute as follows:
$$
\begin{aligned}
f^*(\mathcal E^*)&=\mathcal O_X\otimes_{f^{-1}\mathcal O_Y}f^{-1}(\mathcal E)^*=\mathcal O_X\otimes_{f^{-1}\mathcal O_Y}(f^{-1}\mathcal E)^*\\
&=\mathcal Hom_{f^{-1}\mathcal O_Y}(f^{-1}\mathcal E,\mathcal O_X)\\
&=\mathcal Hom_{\mathcal O_X}(f^*\mathcal E,\mathcal O_X)=(f^*\mathcal E)^*.
\end{aligned}
$$