Quasicompact morphism of schemes implies the closure of the set-theoretic image is the underlying set of the scheme-theoretic image

Question

This is Corollary 9.4.5 in Vakil's notes. Suppose $\pi:X\to Y$ is a quasicompact morphism of schemes, the closure of the set-theoretic image of $\pi$ is the underlying set of the scheme-theoretic image.

I followed his proof until the following

If $U$ is the complement of the closure of the set-theoretic image, $\pi^{-1}(U)=\emptyset$. Under the hypotheses, the scheme theoretic image can be computed locally, so the scheme-theoretic image is the empty set on $U$.

How does the locally computed scheme-theoretic image implies the image on $U$ is empty? It seems I lack a good understanding of computing scheme-theoretic images locally, that is, the scheme-theoretic image is constructed from the quasicoherent ideal sheaf,

Answer 1

The phrasing of theorem 9.4.4 (of which this is a corollary) should clear this up:

9.4.4. Theorem. Suppose $\pi:X\to Y$ is a morphism of schemes. If $X$ is reduced or $\pi$ is quasicompact, then the scheme theoretic image of $\pi$ may be computed affine-locally: on $\operatorname{Spec} A\subset Y$, it is cut out by the functions (elements of $A$) that pull back to the function $0$ (on $\pi^{-1}(\operatorname{Spec} A)$).

Let $\operatorname{Spec} A\subset A$ be an affine open set disjoint from the closure of the set-theoretic image of $\pi$, and suppose the theorem holds. Then $\pi^{-1}(\operatorname{Spec} A)=\emptyset$, and as $\mathcal{O}_X(\emptyset)$ is the zero ring, we have that every element of $A$ pulls back to the zero function. Therefore the ideal cutting out the scheme theoretic image of $\pi$ inside $\operatorname{Spec} A$ is $A$, and so the intersection of the scheme-theoretic image and $\operatorname{Spec} A$ is the empty set. Repeating this for an affine open cover of $U$, we get the result in the corollary.