Lemma 76.9.6. Let $S$ be a scheme. Let $X \subset X'$ be a thickening of algebraic spaces over $S$. The functor

defines an equivalence of categories $X'_{\acute{e}tale}\to X_{\acute{e}tale}$.

Lemma 76.9.6. Let $S$ be a scheme. Let $X \subset X'$ be a thickening of algebraic spaces over $S$. The functor

\[ V' \longmapsto V = X \times _{X'} V' \]

defines an equivalence of categories $X'_{\acute{e}tale}\to X_{\acute{e}tale}$.

**Proof.**
The functor $V' \mapsto V$ defines an equivalence of categories $X'_{spaces, {\acute{e}tale}} \to X_{spaces, {\acute{e}tale}}$, see Theorem 76.8.1. Thus it suffices to show that $V$ is a scheme if and only if $V'$ is a scheme. This is the content of Lemma 76.9.5.
$\square$

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