Restriction of a scheme morphism to a closed subset 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!
 A: 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.
