Definition 29.9.1. A morphism of schemes is said to be *surjective* if it is surjective on underlying topological spaces.

## 29.9 Surjective morphisms

Lemma 29.9.2. The composition of surjective morphisms is surjective.

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

Lemma 29.9.3. Let $X$ and $Y$ be schemes over a base scheme $S$. Given points $x \in X$ and $y \in Y$, there is a point of $X \times _ S Y$ mapping to $x$ and $y$ under the projections if and only if $x$ and $y$ lie above the same point of $S$.

**Proof.**
The condition is obviously necessary, and the converse follows from the proof of Schemes, Lemma 26.17.5.
$\square$

Lemma 29.9.4. The base change of a surjective morphism is surjective.

**Proof.**
Let $f: X \to Y$ be a morphism of schemes over a base scheme $S$. If $S' \to S$ is a morphism of schemes, let $p: X_{S'} \to X$ and $q: Y_{S'} \to Y$ be the canonical projections. The commutative square

identifies $X_{S'}$ as a fibre product of $X \to Y$ and $Y_{S'} \to Y$. Let $Z$ be a subset of the underlying topological space of $X$. Then $q^{-1}(f(Z)) = f_{S'}(p^{-1}(Z))$, because $y' \in q^{-1}(f(Z))$ if and only if $q(y') = f(x)$ for some $x \in Z$, if and only if, by Lemma 29.9.3, there exists $x' \in X_{S'}$ such that $f_{S'}(x') = y'$ and $p(x') = x$. In particular taking $Z = X$ we see that if $f$ is surjective so is the base change $f_{S'}: X_{S'} \to Y_{S'}$. $\square$

Example 29.9.5. Bijectivity is not stable under base change, and so neither is injectivity. For example consider the bijection $\mathop{\mathrm{Spec}}(\mathbf{C}) \to \mathop{\mathrm{Spec}}(\mathbf{R})$. The base change $\mathop{\mathrm{Spec}}(\mathbf{C} \otimes _{\mathbf{R}} \mathbf{C}) \to \mathop{\mathrm{Spec}}(\mathbf{C})$ is not injective, since there is an isomorphism $\mathbf{C} \otimes _{\mathbf{R}} \mathbf{C} \cong \mathbf{C} \times \mathbf{C}$ (the decomposition comes from the idempotent $\frac{1 \otimes 1 + i \otimes i}{2}$) and hence $\mathop{\mathrm{Spec}}(\mathbf{C} \otimes _{\mathbf{R}} \mathbf{C})$ has two points.

Lemma 29.9.6. Let

be a commutative diagram of morphisms of schemes. If $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact.

**Proof.**
Let $W \subset Z$ be a quasi-compact open. By assumption $p^{-1}(W)$ is quasi-compact. Hence by Topology, Lemma 5.12.7 the inverse image $q^{-1}(W) = f(p^{-1}(W))$ is quasi-compact too. This proves the lemma.
$\square$

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