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
\xymatrix{ X_{S'} \ar[d]_{f_{S'}} \ar[r]_ p & X \ar[d]^{f} \\ Y_{S'} \ar[r]^{q} & Y. }
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
\xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[dl]^ q \\ & Z }
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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