# Given a scheme $X$ and invertible sheaf $L$ on $X$, show that $\mathcal{Hom}(L, \mathcal{O}_X)$ is invertible

The question is that given a scheme X and invertible sheaf L on X given by cocycles {$$\varphi_{ij}$$}, show that $$\mathcal Hom_{\mathcal O_X}(L, \mathcal O_X)$$ - the sheaf associated with $$U \mapsto \mathcal Hom_{\mathcal O_X}(L(U), \mathcal O_X(U))$$ - is invertible and given by the cocycles {$$\varphi^{-1}_{ij}$$}

To show invertibility, I know that for any two locally free sheafs F and G of rank 1, F $$\otimes$$ G is also invertible. So I tried showing that the following map is an isomorphism for $$\mathcal O_x$$-modules A and B

A$$\otimes \mathcal Hom_{\mathcal O_X}(A,B) \rightarrow B$$

where $$a \otimes f \mapsto f(a)$$

I have shown the surjectivity of this map, but I am having some trouble with injection and showing the commutativity with restriction maps (if this the best way to go about it). My hope was to then use this to say that if A = L and B = $$\mathcal O_X$$ then we get that $$\mathcal Hom_{\mathcal O_X}(L, \mathcal O_X) \cong \mathcal O_X$$

I found the following possible answer

$$\text{https://math.stackexchange.com/questions/2492225/hom-sheaf-invertible}$$

but I didn't understand why the isomorphism is clear. I also am not sure how to show the cocycles are exactly the ones given above.

• To be clear, I'm suggesting this approach in the case that $A = \mathcal{L}, B = \mathcal{O}_{X}$. (I don't think that your map is an isomorphism in general for $A$ and $B$.) Your sheaf is locally free of rank $1$, so you should try to think about what the stalk looks like. (In particular, you should try to reduce to showing that for any commutative ring $R$ and any $R$-module $L$ of constant rank $1$, the map $R \otimes_{R} Hom(L, R) \to R$ given by $x \otimes f \mapsto f(x)$ is an isomorphism.) – Alex Wertheim May 6 at 1:00