MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've encountered this problem, where I know everything in site (no pun intended) to be locally of finite type over my ground field, but I really need quasi-compactness.

Say I have two morphisms $a: X \to Y$ and $b: Y \to X$ such that $ba = id$ and say I know thaht $Y$ is of finite type, can I say anything about $X$?

This question comes from a moduli problem. I'm trying to prove that X is of finite type by knowing that Y is of finite type and that any family for X actually comes from a family for Y. So boundedness of families of Y should imply boundedness for families of X.

share|cite|improve this question
By the way, this condition is known as $X$ being a "retract" of $Y$ – David White Sep 14 '11 at 15:37
up vote 5 down vote accepted

Yes, it follows easily that if $Y$ is of finite type then $X$ is of finite type and one only needs the surjectivity of $b$:

Let $\{U_{\alpha}\}_{\alpha \in A}$ be an affine open cover of $X$, so $\lbrace b^{-1}(U_{\alpha})\rbrace_{\alpha \in A}$ is an open cover of $Y$. Since $Y$ is of finite type, hence quasi-compact, it follows that this has a finite subcover, say $\lbrace b^{-1}(U_{\beta})\rbrace_{\beta \in B}$, where $B$ is a finite subset of $A$. Since $b$ is surjective, it follows that $\lbrace U_{\beta}\rbrace_{\beta \in B}$ is an open cover of $X$. Thus, $X$ is covered by finitely many affine open sets, so is of finite type.

share|cite|improve this answer
thanks$\phantom{om nom}$ – Yosemite Sam Sep 14 '11 at 14:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.