Lemma 29.41.11. Let $S$ be a scheme. Let $f : X \to Y$ be a surjective universally closed morphism of schemes over $S$.

1. If $X$ is quasi-separated, then $Y$ is quasi-separated.

2. If $X$ is separated, then $Y$ is separated.

3. If $X$ is quasi-separated over $S$, then $Y$ is quasi-separated over $S$.

4. If $X$ is separated over $S$, then $Y$ is separated over $S$.

Proof. Parts (1) and (2) are a consequence of (3) and (4) for $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Schemes, Definition 26.21.3). Consider the commutative diagram

$\xymatrix{ X \ar[d] \ar[rr]_{\Delta _{X/S}} & & X \times _ S X \ar[d] \\ Y \ar[rr]^{\Delta _{Y/S}} & & Y \times _ S Y }$

The left vertical arrow is surjective (i.e., universally surjective). The right vertical arrow is universally closed as a composition of the universally closed morphisms $X \times _ S X \to X \times _ S Y \to Y \times _ S Y$. Hence it is also quasi-compact, see Lemma 29.41.8.

Assume $X$ is quasi-separated over $S$, i.e., $\Delta _{X/S}$ is quasi-compact. If $V \subset Y \times _ S Y$ is a quasi-compact open, then $V \times _{Y \times _ S Y} X \to \Delta _{Y/S}^{-1}(V)$ is surjective and $V \times _{Y \times _ S Y} X$ is quasi-compact by our remarks above. We conclude that $\Delta _{Y/S}$ is quasi-compact, i.e., $Y$ is quasi-separated over $S$.

Assume $X$ is separated over $S$, i.e., $\Delta _{X/S}$ is a closed immersion. Then $X \to Y \times _ S Y$ is closed as a composition of closed morphisms. Since $X \to Y$ is surjective, it follows that $\Delta _{Y/S}(Y)$ is closed in $Y \times _ S Y$. Hence $Y$ is separated over $S$ by the discussion following Schemes, Definition 26.21.3. $\square$

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