The Stacks project

Proof. Although not strictly necessary, we recommend reading Remark 31.12.9 before proceeding with the proof. Choose an open subscheme $j : U \to X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $j^*\mathcal{F}$ and $j^*\mathcal{G}$ are finite locally free, see Lemma 31.12.13. The map

is an isomorphism, because we may check it locally and it is clear when the modules are finite free. Observe that $j^*$ applied to the displayed arrow of the lemma gives the arrow we've just shown is an isomorphism (small detail omitted). Since $j^*$ defines an equivalence between coherent reflexive modules on $U$ and coherent reflexive modules on $X$ (by Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5), we conclude that the arrow of the lemma is an isomorphism too. If $\mathcal{F}$ has rank $1$, then $j^*\mathcal{F}$ is an invertible $\mathcal{O}_ U$-module and the reflexive module $\mathcal{F}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{O}_ X)$ restricts to its inverse. It follows in the same manner as before that $(\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{F}^\vee )^{**} = \mathcal{O}_ X$. In this way we see that we have inverses for the group law given in the statement of the lemma. $\square$