Should a closed immersion of schemes be closed map of topological spaces by definition?

Question

In Section 9.1 of his "Foundations of Algebraic Geometry" (version August 2022), Ravi Vakil defines a closed embedding (closed immersion) of schemes to be an affine morphism
$ \pi: X\rightarrow Y$ such that for every affine open subset $\operatorname{Spec} B\subset Y$, with $\pi^{-1}(\operatorname{Spec} B)\cong \operatorname{Spec} A$, the map of the corresponding rings $B\rightarrow A$ is surjective. Then in the following Exercise 9.1.A, he suggests to prove that these conditions entail that image of $\pi^{\#}$ is a closed subset in $Y$ as topological space. The property of being an affine morphism means that inverse images of affine open subsets of $Y$ are affine subschemes in $X$.

I don't see a way to prove this, as I don't understand the connection between the property of the pullback map of global sections $ \pi^{\#}:B\rightarrow A$, which is $$ \pi^{\#}:A=\Gamma(\operatorname{Spec} B, \mathscr{O}_{\operatorname{Spec} B})\rightarrow \Gamma(\operatorname{Spec} A, \mathscr{O}_{\operatorname{Spec} A})=B $$ being surjective (i.e., every regular function on X can be extended to $\pi(X)\subset Y$), and the topological properties of $\pi$ (being a closed map, or at least mapping $X$ to the closed subset of $Y$)

Now, in any other source I saw, the closed immersion maps $X$ homeomorphically to a closed subset of $Y$ by definition. There is even a discussion, where it is asked whether this condition is a necessary part of definition, and one of the answers argues that without it the surjectivity of $\pi^{\#}$ wouldn't be equivalent to surjectivity on the stalks (of $\pi^{\#}_x$ for $x \in X$). However, this is not really a proof that the topological condition is a necessary part of definition - nor does it offer how to solve Vakil`s question.

So my question is - should the topological condition ($\pi(X)$ is closed in $Y$) be part of the definition of the closed embedding, and if not - how does it follow from the definition?

Thank you in advance!

Answer 1

If $f:A\to B$ is a surjective ring map, then $\operatorname{Spec} f: \operatorname{Spec} B\to\operatorname{Spec} A$ is a homeomorphism on to it's image, which is exactly the closed subscheme $\operatorname{Spec} (A/\ker f)\subset\operatorname{Spec} f$. (See for instance here, or you can factor $A\to B$ as $A\to A/\ker f \cong B$ and note that since $\operatorname{Spec}$ is a functor, that means $\operatorname{Spec} f: \operatorname{Spec} B\to\operatorname{Spec} A$ factors as $\operatorname{Spec} B \cong \operatorname{Spec} (A/\ker f)\to\operatorname{Spec} A$.)

Now use the fact that if $X$ is a topological space, $S\subset X$ a subset, and $\{U_i\}_{i\in I}$ an open cover of $X$, the set $S$ is closed in $X$ iff $S\cap U_i$ is closed in $U_i$ for all $i\in I$. Applying this in our case to $\pi(X)\subset Y$ and the open cover consisting of all the affine open subschemes of $Y$, we see that $\pi(X)\subset Y$ must be closed.

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The question "should the topological condition be a part of the definition" depends on the other parts of the definition. Here, using the setup that Vakil provides, we don't need to add it, as we can prove it from the other portions of the definition. On the other hand, if one only takes the condition that the induced morphism of sheaves is surjective as in your linked question, then one does need to add it, as shown by the linked answer.