# 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.

• Is $\cal F$ a line bundle? – Lord Shark the Unknown Apr 5 at 17:59
• If you can construct a map $\mathcal{F} \otimes \mathcal{F} \check{} \to \mathcal{O}_X$ (which is equivalent to giving an $\mathcal{O}_X$-bilinear map $\mathcal{F} \times \mathcal{F} \check{} \to \mathcal{O}_X$) and then show that it acts as an isomorphism on each stalk, then that map will be an isomorphism. – Daniel Schepler Apr 5 at 18:02
• @LordSharktheUnknown $\mathcal{F}$ is a rank 1 locally free sheaf. – Smash Apr 5 at 18:04

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 $$\phi$$.