Lemma 68.6.5. Let $S$ be a scheme. Let $W$ be an algebraic space over $S$. Let $G$ be a finite group acting freely on $W$. Let $U = W/G$, see Properties of Spaces, Lemma 65.34.1. Let $\chi : G \to \{ +1, -1\} $ be a character. Then there exists a rank 1 locally free sheaf of $\mathbf{Z}$-modules $\underline{\mathbf{Z}}(\chi )$ on $U_{\acute{e}tale}$ such that for every abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$ we have

**Proof.**
The quotient morphism $q : W \to U$ is a $G$-torsor, i.e., there exists a surjective étale morphism $U' \to U$ such that $W \times _ U U' = \coprod _{g \in G} U'$ as spaces with $G$-action over $U'$. (Namely, $U' = W$ works.) Hence $q_*\underline{\mathbf{Z}}$ is a finite locally free $\mathbf{Z}$-module with an action of $G$. For any geometric point $\overline{u}$ of $U$, then we get $G$-equivariant isomorphisms

where the second $=$ uses a geometric point $\overline{w}_0$ lying over $\overline{u}$ and maps the summand corresponding to $g \in G$ to the summand corresponding to $g(\overline{w}_0)$. We have

because $q_*\mathcal{F}|_ W = \mathcal{F} \otimes _\mathbf {Z} q_*\underline{\mathbf{Z}}$ as one can check by restricting to $U'$. Let

be the subsheaf of sections that transform according to $\chi $. For any geometric point $\overline{u}$ of $U$ we have

It follows that $\underline{\mathbf{Z}}(\chi )$ is locally free of rank 1 (more precisely, this should be checked after restricting to $U'$). Note that for any $\mathbf{Z}$-module $M$ the $\chi $-semi-invariants of $M[G]$ are the elements of the form $m \cdot \sum \nolimits _ g \chi (g) g$. Thus we see that for any abelian sheaf $\mathcal{F}$ on $U$ we have

because we have equality at all stalks. The result of the lemma follows by taking global sections. $\square$

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