Lemma 66.5.1. 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 66.3) if and only if $|f| : |X| \to |Y|$ is surjective.

## 66.5 Surjective morphisms

We have already defined in Section 66.3 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 65.4.3 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 66.5.2. 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 66.5.3. 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

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]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.

**Proof.**
Omitted.
$\square$

Lemma 66.5.4. The composition of surjective morphisms is surjective.

**Proof.**
This is immediate from the definition.
$\square$

Lemma 66.5.5. The base change of a surjective morphism is surjective.

**Proof.**
Follows immediately from Properties of Spaces, Lemma 65.4.3.
$\square$

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