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 76.8.2.
Theorem 76.8.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is integral, universally injective and surjective. The functor defines an equivalence of categories $Y_{spaces, {\acute{e}tale}} \to X_{spaces, {\acute{e}tale}}$.
\[ V \longmapsto V_ X = X \times _ Y V \]
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
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
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
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. $\square$
Lemma 76.8.2. With assumption and notation as in Theorem 76.8.1 the equivalence of categories $Y_{spaces, {\acute{e}tale}} \to X_{spaces, {\acute{e}tale}}$ restricts to equivalences of categories $Y_{\acute{e}tale}\to X_{\acute{e}tale}$ and $Y_{affine, {\acute{e}tale}} \to X_{affine, {\acute{e}tale}}$.
Proof. This is just the statement that given an object $V \in Y_{spaces, {\acute{e}tale}}$ we have $V$ is a(n affine) scheme if and only if $V \times _ Y X$ is a(n affine) scheme. Since $V \times _ Y X \to V$ is integral, universally injective, and surjective (as a base change of $X \to Y$) this follows from Limits of Spaces, Lemma 70.15.4 and Proposition 70.15.2. $\square$
Remark 76.8.3. A universal homeomorphism of algebraic spaces need not be representable, see Morphisms of Spaces, Example 67.53.3. In fact Theorem 76.8.1 does not hold for universal homeomorphisms. To see this, let $k$ be an algebraically closed field of characteristic $0$ and let be as in Morphisms of Spaces, Example 67.53.3. Recall that the first morphism is étale and identifies $t$ with $-t$ for $t \in \mathbf{A}^1_ k \setminus \{ 0\} $ and that the second morphism is our universal homeomorphism. Since $\mathbf{A}^1_ k$ has no nontrivial connected finite étale coverings (because $k$ is algebraically closed of characteristic zero; details omitted), it suffices to construct a nontrivial connected finite étale covering $Y \to X$. To do this, let $Y$ be the affine line with zero doubled (Schemes, Example 26.14.3). Then $Y = Y_1 \cup Y_2$ with $Y_ i = \mathbf{A}^1_ k$ glued along $\mathbf{A}^1_ k \setminus \{ 0\} $. To define the morphism $Y \to X$ we use the morphisms These glue over $Y_1 \cap Y_2$ by the construction of $X$ and hence define a morphism $Y \to X$. In fact, we claim that is a cartesian square. We omit the details; you can use for example Groupoids, Lemma 39.20.7. Since $\mathbf{A}^1_ k \to X$ is étale and surjective, this proves that $Y \to X$ is finite étale of degree $2$ which gives the desired example. More simply, you can argue as follows. The scheme $Y$ has a free action of the group $G = \{ +1, -1\} $ where $-1$ acts by swapping $Y_1$ and $Y_2$ and changing the sign of the coordinate. Then $X = Y/G$ (see Spaces, Definition 65.14.4) and hence $Y \to X$ is finite étale. You can also show directly that there exists a universal homeomorphism $X \to \mathbf{A}^1_ k$ by using $t \mapsto t^2$ on affine spaces. In fact, this $X$ is the same as the $X$ above.
\[ \mathbf{A}^1 \to X \to \mathbf{A}^1 \]
\[ Y_1 \xrightarrow {1} \mathbf{A}^1_ k \to X \quad \text{and}\quad Y_2 \xrightarrow {-1} \mathbf{A}^1_ k \to X. \]
\[ \xymatrix{ Y \ar[d] & Y_1 \amalg Y_2 \ar[l] \ar[d] \\ X & \mathbf{A}^1_ k \ar[l] } \]
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