Proof.
We stress that this is not a triviality.
Assume (1). By More on Morphisms of Spaces, Lemma 76.9.6 the morphism $i$ induces an equivalence of small étale sites and in particular of topoi. Of course $i^\sharp $ is surjective with locally nilpotent kernel by definition of thickenings.
Assume (2). (This direction is less important and more of a curiosity.) For any étale morphism $Y' \to Z'$ we see that $Y = Z \times _{Z'} Y'$ has the same étale topos as $Y'$. In particular, $Y'$ is quasi-compact if and only if $Y$ is quasi-compact because being quasi-compact is a topos theoretic notion (Sites, Lemma 7.17.3). Having said this we see that $Y'$ is quasi-compact and quasi-separated if and only if $Y$ is quasi-compact and quasi-separated (because you can characterize $Y'$ being quasi-separated by saying that for all $Y'_1, Y'_2$ quasi-compact algebraic spaces étale over $Y'$ we have that $Y'_1 \times _{Y'} Y'_2$ is quasi-compact). Take $Y'$ affine. Then the algebraic space $Y$ is quasi-compact and quasi-separated. For any quasi-coherent $\mathcal{O}_ Y$-module $\mathcal{F}$ we have $H^ q(Y, \mathcal{F}) = H^ q(Y', (Y \to Y')_*\mathcal{F})$ because the étale topoi are the same. Then $H^ q(Y', (Y \to Y')_*\mathcal{F}) = 0$ because the pushforward is quasi-coherent (Morphisms of Spaces, Lemma 67.11.2) and $Y$ is affine. It follows that $Y'$ is affine by Cohomology of Spaces, Proposition 69.16.7 (there surely is a proof of this direction of the lemma avoiding this reference). Hence $i$ is an affine morphism. In the affine case it follows easily from the conditions in Section 91.9 that $i$ is a thickening of algebraic spaces.
$\square$
Comments (0)