If an open set is evenly covered by $p$, then its open subset is also evenly covered by $p$.

Question

If $U$ is an open set evenly covered by $p: E\to B$ and $W$ is an open set contained in $U$, then $W$ is also evenly covered by $p$.

I'm trying to prove this statement, but have a difficulty.

So by definition, we have a partition of $p^{-1}(U)$ into slices, say, $\{V_\alpha\}$. The obvious choice of a partition of $p^{-1}(W)$ would be $\{V_\alpha \cap p^{-1}(W)\}$, which is a collection of disjoint open sets in $E$. Clearly, the union of this collection is $p^{-1}(W)$, so we're left to show that for each $\alpha$, the restriction of $p$ to $V_\alpha \cap p^{-1}(W)$ is a homeomorphism of $V_\alpha \cap p^{-1}(W)$ onto $W$. This is what I'm struggling to show.

First, this restriction is clearly injective, since it is a restriction of an injective function, and so it has an inverse, which is again a restriction of the inverse of $p|V_\alpha$, so is continuous as well. But how can I show that the image is $W$? That is, $p(V_\alpha \cap p^{-1}(W))=W$? I would greatly appreciate any help on this.

Answer 1

Here are two general facts about a set theoretic map $f:X\to Y$ -

1. If $B\subseteq Y$ then $f(f^{-1}(B))=B$ when $f$ is surjective.

2. If $U,V\subseteq X$ then $f(U\cap V)=f(U)\cap f(V)$ when $f$ is injective.

Now in your case $p|_{V_\alpha}$ is a bijection. So, $$p(V_\alpha\cap p^{-1}(W))=p(V_\alpha)\cap p(p^{-1}(W))=U\cap W=W$$