Proof.
The morphism f is representable and a universal homeomorphism, see Morphisms of Spaces, Section 67.53.
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 67.51.2. 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 65.10.5) 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 74.7.2). 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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