TITLE: When does regularity of the base scheme imply regularity of the top scheme?
QUESTION [1 upvotes]: My problem is the following:
I have a finite surjective morphism $f: X\rightarrow Y$ of noetherian schemes and know that $Y$ is a regular scheme.
(Indeed, in my situation, the two schemes are topologically the same and the arrow is topologically the identity.)
I don't know if $f$ is étale or smooth. But I know that $f$ has a section.
I want to conclude somehow that $X$ is also a regular scheme. What further minimal conditions (on $f$ or $X$) would imply this?
I would be glad about just some hints of what one could do here.

REPLY [3 votes]: I may be misunderstanding the question, but it seems rather straightforward to me. 
a) As stated, without assuming, that $X$ is, say, reduced, it is certainly false:
Let $X=\mathrm{Spec}k[\varepsilon]=k[x]/(x^2)$, $Y=\mathrm{Spec} k$ and $f:X\to Y$ the structure map of $X$ as a $Y$-scheme. This is obviously a homeomorphism topologically and it has a section that maps $Y$ isomorphically to $X_{\mathrm{red}}$. 
b) It seems to me that pretty much this is the only thing that can go wrong: If $f$ has a section, then by definition of a section, $Y$ is isomorphic to the image of the section. 
As $Y$ is regular, it is necessarily reduced, so if $f$ is one-to-one, then $f$ and the section induce an isomorphism between $Y$ and $X_{\mathrm{red}}$. So, if $X$ is reduced, the statement is true, if $X$ is not reduced, then the question is whether there exists a morphism $X\to X_{\mathrm{red}}$ that's an isomorphism on $X_{\mathrm{red}}$. The above example shows that this can happen, so some assumption is required to rule out this possibility.
If $f$ is not one-to-one, something similar would still happen. The section would still induce an isomorphism between $Y$ and the image of the section which would have to be an irreducible component of $X_{\mathrm{red}}$. On the other hand, in this case you would probably want to assume that $X$ and $Y$ are irreducible, since otherwise you can just add an additional component that screws things up. However, then (assuming that $X$ is irreduciblke) this implies that $Y\simeq X_{\mathrm{red}}$ and you end up in the previous case again.
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