The Stacks project

75.8 Topological invariance of the étale site

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.

Theorem 75.8.1. 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

\[ V \longmapsto V_ X = X \times _ Y V \]

defines an equivalence of categories $Y_{spaces, {\acute{e}tale}} \to X_{spaces, {\acute{e}tale}}$.

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. $\square$

Lemma 75.8.2. With assumption and notation as in Theorem 75.8.1 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 69.15.4 and Proposition 69.15.2. $\square$

reference

Remark 75.8.3. A universal homeomorphism of algebraic spaces need not be representable, see Morphisms of Spaces, Example 66.53.3. In fact Theorem 75.8.1 does not hold for universal homeomorphisms. To see this, let $k$ be an algebraically closed field of characteristic $0$ and let

\[ \mathbf{A}^1 \to X \to \mathbf{A}^1 \]

be as in Morphisms of Spaces, Example 66.53.3. 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

\[ Y_1 \xrightarrow {1} \mathbf{A}^1_ k \to X \quad \text{and}\quad Y_2 \xrightarrow {-1} \mathbf{A}^1_ k \to X. \]

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

\[ \xymatrix{ Y \ar[d] & Y_1 \amalg Y_2 \ar[l] \ar[d] \\ X & \mathbf{A}^1_ k \ar[l] } \]

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 64.14.4) 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.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05ZG. Beware of the difference between the letter 'O' and the digit '0'.