The Stacks project

Lemma 29.41.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. If $X$ is universally closed over $S$ and $f$ is surjective then $Y$ is universally closed over $S$. In particular, if also $Y$ is separated and locally of finite type over $S$, then $Y$ is proper over $S$.

Proof. Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to S$, $q : Y \to S$ the structure morphisms. Let $S' \to S$ be a morphism of schemes. The base change $f' : X_{S'} \to Y_{S'}$ is surjective (Lemma 29.9.4), and the base change $p' : X_{S'} \to S'$ is closed. If $T \subset Y_{S'}$ is closed, then $(f')^{-1}(T) \subset X_{S'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed. This proves the first statement. Thus $Y \to S$ is quasi-compact by Lemma 29.41.8 and hence $Y \to S$ is proper by definition if in addition $Y \to S$ is locally of finite type and separated. $\square$

Comments (4)

Comment #771 by Kestutis Cesnavicius on

Is the analogue of this for finite morphisms ("image of a finite morphism is finite") stated anywhere in the SP?

Comment #792 by on

Hi! Not yet I think. I did add a more precise statement of this lemma in this commit. Moreover, in that commit you can also find the statement for schemes whose analogue for algebraic spaces mentioned in my blog post from which what you say easily follows. But, yes, we can/should add a lemma of this type (and also for integral morphisms I guess).

Comment #5540 by Zhenhua Wu on

Locally of finite type will suffice as universally closed morphism is quasi-compact, it's best to change this description in order to be consistent with tag 0AH6.

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