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I've been working through a proof of Chevalley's theorem for the special case of morphisms of finite type between Noetherian schemes as outlined in Görtz-Wedhorn Theorems 10.19 and 10.20 on page 248.

The claim of Theorem 10.19 is:

If $f:X\rightarrow Y$ is a dominant morphism of finite type between Noetherian schemes, then $f(X)$ contains a dense open subset of $Y$.

The first portion of the proof deals with the case where $Y$ is assumed to be irreducible, and everything seems clear to me there. However, it then continues into the case $Y$ is reducible by decomposing $Y = Y_1\cup\ldots\cup Y_n$ as a union of irreducible components (using that $Y$ is Noetherian) and then concludes by applying the previous case to the restricted morphisms $f^{-1}(Y_i)\rightarrow Y_i$. I'm concerned in that I seem to be unsure how these natural restricted maps should be defined!

It's clear to me how to define the restriction of a scheme morphism to an open set using the natural open subscheme structures for the open set and its inverse image. However, I don't see a good way to do this for closed subsets. If the schemes are affine, say $X = \text{Spec}(B)$, $Y = \text{Spec}(A)$, then if $Z\subseteq Y$ is closed, the scheme $\text{Spec}(A/a)$ would have the same underlying topological space as $Z$ for some ideal $a\subseteq A$. Then $f^{-1}(Z)$ would be the underlying topological space of the scheme $\text{Spec}(B/a^e)$, where $a^e$ is the extension of $a$ by the ring homomorphism corresponding to $f$. There is an induced ring homomorphism $A/a\rightarrow B/a^e$, which corresponds to a scheme morphism $\text{Spec}(B/a^e)\rightarrow \text{Spec}(A/a)$ compatible with $f$. However, I don't see why this morphism need be dominant.

Specifically, my questions are:

If $f:X\rightarrow Y$ is a scheme morphism, and $Z\subseteq Y$ a closed subset, is there a canonical way to define a restricted map $f^{-1}(Z)\rightarrow Z$? Can we do this using the reduced induced closed subscheme structures of $f^{-1}(Z), Z$? If so, what would the map of sheaves look like? Finally, if such a map is defined, is it dominant and of finite type so that the proof of the theorem may use it?

I apologize if these are silly questions; it feels that I'm missing something obvious. However, I haven't had any luck so far finding a reference describing this construction explicitly and as far as I can tell such restricted maps aren't mentioned earlier in G&W. This idea of restricting to a closed subset is also used in the proof of Theorem 10.20. Thanks very much!

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    $\begingroup$ $f^{-1}(Z)$ is defined as $X\times_YZ$ and the natural morphism is $pr_2:X\times_YZ\to Z$. $\endgroup$ – Armando j18eos Jun 9 '17 at 13:53
  • $\begingroup$ @Armandoj18eos Thanks! For some reason I wasn't thinking in terms of the fiber product. If I assume $f:X\rightarrow Y$ is of finite type between Noetherian schemes, I see how the same can be said for $pr_2$, and I think it is also straightforward to show that the dominance of $f$ implies that of $pr_2$ (using that these schemes are Noetherian). I would be happy to accept your comment as an answer! $\endgroup$ – catfish Jun 10 '17 at 17:18
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I extend my comment!


By definition $f^{-1}(Z)=X\times_YZ$ and $g=f_{|f^{-1}(Z)}$ is the natural projection $pr_2:X\times_YZ\to Z$; because for any $x\in X,\,f_x^{\sharp}:\mathcal{O}_{Y,f(x)}\to\mathcal{O}_{X,x}$ is a morphism of finite type of local Noetherian rings, by base change $\forall x\in f^{-1}(Z),\,g_z^{\sharp}=f_x^{\sharp}\otimes Id_{\displaystyle\mathcal{O}_{Y,f(x)/\mathcal{I}_{Z,f(x)}}}$ and it is a morphism of finite type: by definition $g$ is a morphism of finite type of (Noetherian) schemes.

By Noetherian hypothesys, one can assume that $Z$ is irreducible; by this lemma: $g$ is a dominant morphism if and only if the inverse image of generic point $\eta$ of $Z$ via $g$ is not empty.

If $g$ is not dominant then $g^{-1}(\eta)=\emptyset\iff\eta\notin g(g^{-1}(Z))\Rightarrow\eta\notin\overline{g(g^{-1}(Z))}$, that is there exists an open neighbourhood $U$ of $\eta$ such that $g^{-1}(U)=\emptyset$; by construction, there exists an open subset $V$ of $Y$ such that $U=V\cap Z$ and $\emptyset=g^{-1}(U)=f^{-1}(U)=f^{-1}(V\cap Z)=f^{-1}(V)\cap f^{-1}(Z)\Rightarrow f^{-1}(V)=\emptyset$: by dominance of $f$ this is a contradiction, then $g$ is dominat.

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    $\begingroup$ Thank you! One concern I have though: it doesn't seem to be the case that the projection need be surjective. Indeed if $Z$ isn't contained in $f(X)$ ($f$ need not be surjective), then certainly $g$ won't be surjective right? However if $Y$ is Noetherian with a minimal decomposition into irreducible components $Y = Y_1\cup\ldots\cup Y_n$, and if $Z= Y_1$ for example, then for any open subset $U\subseteq Z$ one can construct a nonempty open subset of $Y$ contained in $U$ using $Y\setminus (Y_2\cup\ldots\cup Y_n)$, and so the density of $f(X)$ in $Y$ guarantees that of $g(f^{-1}(Z))$ in $Z$. $\endgroup$ – catfish Jun 12 '17 at 3:03
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    $\begingroup$ Yes, you are right: $g$ needs not be surjective. The dominant morphisms are particular topological maps: you can use the Noetherian hypothesys and pass to irreducible case and apply the lemma. $\endgroup$ – Armando j18eos Jun 12 '17 at 9:12

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