Lemma 67.5.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is surjective (in the sense of Section 67.3) if and only if $|f| : |X| \to |Y|$ is surjective.
67.5 Surjective morphisms
We have already defined in Section 67.3 what it means for a representable morphism of algebraic spaces to be surjective.
Proof. Namely, if $f : X \to Y$ is representable, then it is surjective if and only if for every scheme $T$ and every morphism $T \to Y$ the base change $f_ T : T \times _ Y X \to T$ of $f$ is a surjective morphism of schemes, in other words, if and only if $|f_ T|$ is surjective. By Properties of Spaces, Lemma 66.4.3 the map $|T \times _ Y X| \to |T| \times _{|Y|} |X|$ is always surjective. Hence $|f_ T| : |T \times _ Y X| \to |T|$ is surjective if $|f| : |X| \to |Y|$ is surjective. Conversely, if $|f_ T|$ is surjective for every $T \to Y$ as above, then by taking $T$ to be the spectrum of a field we conclude that $|X| \to |Y|$ is surjective. $\square$
This clears the way for the following definition.
Definition 67.5.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say $f$ is surjective if the map $|f| : |X| \to |Y|$ of associated topological spaces is surjective.
Lemma 67.5.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
$f$ is surjective,
for every scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is surjective,
for every affine scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is surjective,
there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is a surjective morphism,
there exists a scheme $U$ and a surjective étale morphism $\varphi : U \to X$ such that the composition $f \circ \varphi $ is surjective,
there exists a commutative diagram
where $U$, $V$ are schemes and the vertical arrows are surjective étale such that the top horizontal arrow is surjective, and
there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is surjective.
\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]
Proof. Omitted. $\square$
Lemma 67.5.4. The composition of surjective morphisms is surjective.
Proof. This is immediate from the definition. $\square$
Lemma 67.5.5. The base change of a surjective morphism is surjective.
Proof. Follows immediately from Properties of Spaces, Lemma 66.4.3. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)