# Why does the intersection change to a union in $r^{-1}(\bigcap r(V_i\cap W_i))=\bigcup V_i\cap W_i$?

Let $q: X\to Y$ and $r:Y\to Z$ be covering maps, $p=r\circ q$. If $r^{-1}(z)$ is finite for each $z$ in $Z$, $p$ is a covering map.

There is a proof on ask a topologist, but I can't follow why $r^{-1}(\bigcap r(V_i\cap W_i))=\bigcup V_i\cap W_i$?

Proof: Fix $z \in Z$. It has a neighbourhood $U_z$ such that $U_z$ is evenly covered by finitely many open sets $V_1,\dots,V_n$ and $r$, where $r^-1(z) = \{y_1,\dots,y_n\}$ and each $y_i$ is in $V_i$. Each $V_i$ is mapped homeomorphically onto $U_z$ by $r$. Now, each $y_i$ has a evenly covered neighbourhood as well. Say, $y_i$ has a neighbourhood $W_i$ such that $W_i$ is evenly covered by $q$. So $q^{-1}[W_i]$ is a disjoint union of open sets $O_{i,j}$ ($j$ running), and each maps $O_{i,j}$ is mapped homeomorphically by $q$ onto $V_i$. We shrink the $U_z$ to $U'_z = \bigcap_{i=1}^n r[V_i \cap W_i]$.

All the $V_i \cap W_i$ are mapped by $r$ into $U_z$ and as $r$ is open on $V_i$ we get that these sets are open, and all contain $z$. So $U'_z$ is still evenly covered by $r$, but the inverse image now is $\bigcup_{i=1}^n V_i \cap W_i$ and each of which is evenly covered by q...

This last line confuses me. How does $r^{-1}(\bigcap r(V_i\cap W_i))=\bigcup V_i\cap W_i$? Don't inverse images pass over intersections? Why does the $\bigcap$ change to a $\bigcup$?

That line is indeed confusing since it appears as though the $\bigcup$ arises from the $\bigcap$. But they are completely unrelated. The formula actually isn't quite correct. Let's see why:
For sake of simplicity, I assume that $W_i⊆V_i$ (if $W_i$ is evenly covered by $q$, then so is every subset of $W_i,$ so we can assume that it is a subset of $V_i$). Let $r_i$ denote the restriction of $r$ to $V_i$, so every $r_i$ is a homeomorphism onto its image $U$. Now $$U'=\bigcap_{i=1}^n r[W_i]$$ an open subset of $U$ which contains $z$. Then $$r^{-1}[U'] = \bigcup_{i=1}^n r_i^{-1}[U']$$ and each $r_i^{-1}[U']$ is an open subset of $W_i$, but it is in general not equal to $W_i$. But they assume that $r_i^{-1}[U']=W_i$, that is why they arive at $r^{-1}[U']=\bigcup_1^n W_i$