There's something in the following passage in "Basic Algebraic Geometry" by Shafarevich that I don't understand.
We prove in addition that the inverse image $f^{-1}(Z)$ under any regular map $f:X\rightarrow Y$ ($X$ and $Y$ are quasiprojective varieties) of any closed subset $Z\subset Y$ is closed in $X$.
By definition of a regular map $f:X\rightarrow Y$, for any point $x\in X$ there are neighbourhoods $U$ of $x$ in $X$ and $V$ of $f(x)$ in $Y$ such that $f(U)\subset V \subset \mathbb{A}^m$ and the map $f:U\rightarrow V$ is regular. By Lemma 2 we can assume that $U$ is an affine variety. By Lemma 1 it is enough to check that $f^{-1}(Z)\cap U=f^{-1}(Z\cap V)$ is closed in U...
I don't understand why $f^{-1}(Z)\cap U=f^{-1}(Z\cap V)$. In fact I think $f^{-1}(Z\cap V)$ is bigger than $f^{-1}(Z)\cap U$.
Thanks in advance.
Let $x\in f^{-1}(Z)\cap U = f^{-1}(Z)\cap f^{-1}(V)$. We then have that $x\in f^{-1}(Z)$ and $x\in f^{-1}(V)$. Hence $f(x)\in Z$ and $f(x)\in V$. Thus we have that $f(x)\in Z\cap V$, meaning that $x\in f^{-1}(Z\cap V)$.
Conversely, suppose that $x\in f^{-1}(Z\cap V)$. We then have that $f(x)\in Z\cap V$. Hence $f(x)\in Z$ and $f(x)\in V$, so $x\in f^{-1}(Z)$ and $x\in f^{-1}(V)$, giving $x\in f^{-1}(Z)\cap f^{-1}(V)$.
Thus $f^{-1}(Z)\cap f^{-1}(V) = f^{-1}(Z\cap V)$.
Notice that Shafarevich restricts the map $f$ to $U$, which is what gives us that $f^{-1}(V) = U$ in the first sentence. Beyond that, it's just set theory, I think.