There is a bijection between irreducible components of the generic fiber and irreducible components passing through it.

Question

I have been working on the following problem from Ulrich-Görtz today and I can't seem to find a nice solution.

Let $f:X \rightarrow Y$ be a morphism of schemes and let $Y$ be irreducible and $\eta$ the generic point of $Y$. Show that there is a bijection between the irreducible components of $f^{-1}(\eta)$ and the irreducible components of $X$ passing through $f^{-1}(\eta)$ given by $$Z \rightarrow Z \cap f^{-1}(\eta).$$

What I have tried:

I basically think I would be able to prove this if I could show that the inverse map (I think it is an inverse at least) $W \rightarrow cl(W)$ where $W$ is an irreducible component of $f^{-1}(\eta)$ and $cl$ the closure, maps irreducible components to irreducible components. I have not been able to show this, or even that it maps irreducible subsets to irreducible subsets. The problem I encounter is of the following type:

Let $W \subset f^{-1}(\eta)$ be an irreducible component and suppose that $cl(W) = V_1 \cup V_2$ for $V_1$ and $V_2$ closed. Then $V_1 \cap f^{-1}{\eta} = V_2 \cap f^{-1}{\eta}$ (I think) , but why does this imply $V_1=V_2$?

I hope I am somewhat comprehensible, if not, please tell me and I will try to make my arguments more clear.

Answer 1

Apparently it is mentioned in EGA (what isn't?), Volume 1, Chapter 0.2, Irreducible spaces.

I will give the proof here for future reference and for my own sake.

Let $Y$ be an irreducible scheme with a generic point $\eta$ and $f:X \rightarrow Y$ be a morphism of schemes. For every irreducible component $Z$ of $X$ meeting $f^{-1}(\eta)$, $f(Z)$ is dense in $Y$ (since the image will contain the generic point). Now, if $Z$ is an irreducible component of $X$ and $f(Z)$ is dense in $Y$, letting $x$ be the generic point of $Z$, one must have that $\eta = f(x)$. This follows since $f(cl(x)) = f(Z) \subset cl(f(x)) = cl(f(Z)) = Y$. Further, $Z \cap f^{-1}(\eta)$ is the closure of $x$ in $f^{-1}(\eta)$ and thus irreducible, and any irreducible component in $f^{-1}(\eta)$ containing $x$ must be in $Z \cap f^{-1}(\eta)$ so that $Z \cap f^{-1}(\eta)$ is an irreducible component.

Since all irreducible components of $f^{-1}(\eta)$ are contained in an irreducible component of $X$, one sees that every irreducible component $Z$ of $X$ meeting $f^{-1}(\eta)$ admits a generic point which must be in $f^{-1}(\eta)$. Thus, we can find a bijective correspondence between them.

There is more to be filled in of course.