If X -> Y -> X is the identity and Y is of finite type, can I say anything about X?

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.

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By the way, this condition is known as $X$ being a "retract" of $Y$ – David White Sep 14 '11 at 15:37

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.
thanks$\phantom{om nom}$ – Yosemite Sam Sep 14 '11 at 14:53