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My problem concerns invertible elements in the Picard group.

The definition of invertible sheaf that I have is the following: let $X$ be a topological space, and $\mathcal{F}$ be a sheaf of $\mathcal{O}_{X}$-modules. We say that $\mathcal{F}$ is invertible if $X$ can be covered by open subsets $U_{i}$ for which $\mathcal{F} |_{U_{i}}$ is a free $\mathcal{O}_{X} |_{U_{i}}$-module of rank $1$.

Isomorphism classes of invertible sheaves on $X$ form an abelian group, with respect to the tensor product. This group is called Picard group, and it is denoted by $\mathsf{Pic} (X)$.

I would like to show that given a class in $\mathsf{Pic}(X)$ represented by $\mathcal{L}$, the corresponding invertible class is represented by $\mathcal{L}^{\vee}=\mathsf{Hom}(\mathcal{L},\mathcal{O}_{X})$. Can a prove of this fact be given only using basic properties of the tensor product?

Thank you.

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1 Answer 1

up vote 3 down vote accepted

If $F,G$ are sheaves of $\mathcal{O}_X$-modules, there is a canonical homomorphism $F \otimes \underline{\mathrm{Hom}}(F,G) \to G$, just evaluation. If $F=\mathcal{O}_X$, it is an isomorphism. Hence also when $F \cong \mathcal{O}_X$, and hence also (by the local nature of the homomorphism) when $F$ is invertible. For $G=\mathcal{O}_X$, we see that $\mathcal{L} \otimes \underline{\mathrm{Hom}}(\mathcal{L},\mathcal{O}_X) \cong \mathcal{O}_X$ for every invertible $\mathcal{L}$.

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