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The Theorem 1.9.5 in T. A. Springer's Linear Algebraic Groups (page 19) states that:

Let $\phi: X \rightarrow Y$ be a morphism of varieties. Then $\phi X$ contains a non-empty open subset of its closure $\overline{\phi X}$.

(Here, a variety is defined as a quasi-compact ringed space to have an open cover by affine subsets along with being separated.)

The first line of Springer's proof wrotes:

Using a covering of $Y$ by affine open sets, we reduce the proof to the case that $Y$ is affine.

I wonder how such reduction proceeds, particularly whether the preimage of a affine open subset of $Y$ is a subvariety of $X$. (Initially I wrongly claim that the preimage of a affine open subset of $Y$ is not necessarily quasi-compact.)

Hope for an answer, thanks in advance!

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Actually, the map $\phi:X\to Y$ is quasi-compact, so the preimage of any quasi-compact set (like an affine open set of $Y$) is again quasi-compact. The key result is the following:

Lemma (Stacks Project 03GI): Let $f:X\to Y$ and $g:Y\to Z$ be morphisms of schemes. If $g\circ f$ is quasi-compact and $g$ is quasi-separated, then $f$ is quasi-compact.

In our situation, we take $Z=\operatorname{Spec} k$, $g$ the structure morphism, $f=\phi$, and then we have that $g\circ\phi$ is equal to the structure morphism $X\to \operatorname{Spec} k$ by assumption that $\phi$ is a map of $k$-schemes. As a separated morphism is quasi-separated, we have that $g$ is quasi-separated, and because $X$ is quasi-compact and $\operatorname{Spec} k$ is a point, we have that $g\circ f$ is quasi-compact. Thus your situation satisfies the assumptions of the lemma and we may conclude that $\phi$ is quasi-compact.

(I do not have a copy of this book in front of me right now, but I remember that some of these algebraic groups texts are notorious for not doing things correctly from a modern algebraic-geometry perspective. If it's doing some wacky thing like only taking closed points, do not fear: these geometric objects embed as subspaces with the induced topology of the schemes in question, and the maps between the not-quite-schemey versions are exactly the restrictions of the scheme maps to these subspaces and everything is okay.)

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  • $\begingroup$ May I ask that utilizing the method above, whether we can prove the preimage of a subvariety is still a subvariety? $\endgroup$ – Wembley Inter Oct 23 '20 at 8:32
  • $\begingroup$ @WembleyInter What do you mean? What parts of that claim are tripping you up? This shows quasi-compactness, and any locally closed subscheme of a separated scheme is again separated, so that would seem to do it by your definitions. $\endgroup$ – KReiser Oct 23 '20 at 9:39
  • $\begingroup$ Ah... Now I get it. Thanks! $\endgroup$ – Wembley Inter Oct 23 '20 at 11:01
  • $\begingroup$ @WembleyInter if this resolves your issues, please consider accepting the answer. If not, is there something you feel is missing from the answer? $\endgroup$ – KReiser Oct 23 '20 at 22:12
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Actually, $\phi$ is a quasi-compact map.

Indeed, let $B$ be an affine base scheme, so that $X,Y$ are separated quasicompact over $B$. Then $\phi$ is the composition of $\phi_1: X= X \times_Y Y \rightarrow X \times_B Y$ and $\phi_2: X \times_B Y \rightarrow Y$.

$Y \rightarrow B$ is separated so $\phi_1$ is a closed immersion so is quasi-compact. But $\phi_2$ is a base change of $X \rightarrow B$ which is a quasi-compact map so is quasi-compact. So the composition $\phi$ of $\phi_1$ and $\phi_2$ is quasi-compact.

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