We show that the site $X_{spaces, {\acute{e}tale}}$ is a “topological invariant”. It then follows that $X_{\acute{e}tale}$, which consists of the representable objects in $X_{spaces, {\acute{e}tale}}$, is a topological invariant too, see Lemma 75.8.2.

**Proof.**
The morphism $f$ is representable and a universal homeomorphism, see Morphisms of Spaces, Section 66.53.

We first prove that the functor is faithful. Suppose that $V', V$ are objects of $Y_{spaces, {\acute{e}tale}}$ and that $a, b : V' \to V$ are distinct morphisms over $Y$. Since $V', V$ are étale over $Y$ the equalizer

\[ E = V' \times _{(a, b), V \times _ Y V, \Delta _{V/Y}} V \]

of $a, b$ is étale over $Y$ also. Hence $E \to V'$ is an étale monomorphism (i.e., an open immersion) which is an isomorphism if and only if it is surjective. Since $X \to Y$ is a universal homeomorphism we see that this is the case if and only if $E_ X = V'_ X$, i.e., if and only if $a_ X = b_ X$.

Next, we prove that the functor is fully faithful. Suppose that $V', V$ are objects of $Y_{spaces, {\acute{e}tale}}$ and that $c : V'_ X \to V_ X$ is a morphism over $X$. We want to construct a morphism $a : V' \to V$ over $Y$ such that $a_ X = c$. Let $a' : V'' \to V'$ be a surjective étale morphism such that $V''$ is a separated algebraic space. If we can construct a morphism $a'' : V'' \to V$ such that $a''_ X = c \circ a'_ X$, then the two compositions

\[ V'' \times _{V'} V'' \xrightarrow {\text{pr}_ i} V'' \xrightarrow {a''} V \]

will be equal by the faithfulness of the functor proved in the first paragraph. Hence $a''$ will factor through a unique morphism $a : V' \to V$ as $V'$ is (as a sheaf) the quotient of $V''$ by the equivalence relation $V'' \times _{V'} V''$. Hence we may assume that $V'$ is separated. In this case the graph

\[ \Gamma _ c \subset (V' \times _ Y V)_ X \]

is open and closed (details omitted). Since $X \to Y$ is a universal homeomorphism, there exists an open and closed subspace $\Gamma \subset V' \times _ Y V$ such that $\Gamma _ X = \Gamma _ c$. The projection $\Gamma \to V'$ is an étale morphism whose base change to $X$ is an isomorphism. Hence $\Gamma \to V'$ is étale, universally injective, and surjective, so an isomorphism by Morphisms of Spaces, Lemma 66.51.2. Thus $\Gamma $ is the graph of a morphism $a : V' \to V$ as desired.

Finally, we prove that the functor is essentially surjective. Suppose that $U$ is an object of $X_{spaces, {\acute{e}tale}}$. We have to find an object $V$ of $Y_{spaces, {\acute{e}tale}}$ such that $V_ X \cong U$. Let $U' \to U$ be a surjective étale morphism such that $U' \cong V'_ X$ and $U' \times _ U U' \cong V''_ X$ for some objects $V'', V'$ of $Y_{spaces, {\acute{e}tale}}$. Then by fully faithfulness of the functor we obtain morphisms $s, t : V'' \to V'$ with $t_ X = \text{pr}_0$ and $s_ X = \text{pr}_1$ as morphisms $U' \times _ U U' \to U'$. Using that $(\text{pr}_0, \text{pr}_1) : U' \times _ U U' \to U' \times _ S U'$ is an étale equivalence relation, and that $U' \to V'$ and $U' \times _ U U' \to V''$ are universally injective and surjective we deduce that $(t, s) : V'' \to V' \times _ S V'$ is an étale equivalence relation. Then the quotient $V = V'/V''$ (see Spaces, Theorem 64.10.5) is an algebraic space $V$ over $Y$. There is a morphism $V' \to V$ such that $V'' = V' \times _ V V'$. Thus we obtain a morphism $V \to Y$ (see Descent on Spaces, Lemma 73.7.2). On base change to $X$ we see that we have a morphism $U' \to V_ X$ and a compatible isomorphism $U' \times _{V_ X} U' = U' \times _ U U'$, which implies that $V_ X \cong U$ (by the lemma just cited once more).

Pick a scheme $W$ and a surjective étale morphism $W \to Y$. Pick a scheme $U'$ and a surjective étale morphism $U' \to U \times _ X W_ X$. Note that $U'$ and $U' \times _ U U'$ are schemes étale over $X$ whose structure morphism to $X$ factors through the scheme $W_ X$. Hence by Étale Cohomology, Theorem 59.45.2 there exist schemes $V', V''$ étale over $W$ whose base change to $W_ X$ is isomorphic to respectively $U'$ and $U' \times _ U U'$. This finishes the proof.
$\square$

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