Lemma 75.28.4. Let $S$ be a scheme. Let $f : X \to Y$ be a surjective finite locally free morphism of algebraic spaces over $S$. If $X$ has the resolution property, so does $Y$.
Proof. The condition means that $f$ is affine and that $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ Y$-module of positive rank. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module of finite type. By assumption there exists a surjection $\mathcal{E} \to f^*\mathcal{G}$ for some finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$. Since $f_*$ is exact (Cohomology of Spaces, Section 69.4) we get a surjection
\[ f_*\mathcal{E} \longrightarrow f_*f^*\mathcal{G} = \mathcal{G} \otimes _{\mathcal{O}_ Y} f_*\mathcal{O}_ X \]
Taking duals we get a surjection
\[ f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, \mathcal{O}_ Y) \longrightarrow \mathcal{G} \]
Since $f_*\mathcal{E}$ is finite locally free, we conclude. $\square$
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