If I have a covering map $p:E \rightarrow B$ and some connected set $U$, that is evenly covered, then $p^{-1}(U)$ as a partition into slices is unique.
Now, if I assume that $B$ is connected, then I want to show for a particular $b \in B$ with covered surrounding $U$ that $p^{-1}(U)$ has some fixed cardinality $k$ of sets which should then agree with $|p^{-1}(b)|=k$. (This one should then agree with the fibre cardinality of all the other elements in $U$.) But I am not so sure about this. I mean, since the partition into slices does not have to be unique in this case, there does not need to exist a unique partition into slices. Is this problem real or can it easily be resolved?
Best Answer
Let $p: E \rightarrow B$ be a covering map. For a fixed finite number $k$, define $A_k = \{ b \in B: |p^{-1}(b)|=k \}$. I claim this set is open and closed in $B$.
To see it is open, let $b \in A_k$ and let $U$ be an evenly covered neighbourhood of $b$. So $p^{-1}[U]= \cup_{i \in I} V_i$, where the $V_i$ are pairwise disjoint, and for every $i$, $p$ restricted to $V_i$ is a homeomorphism between $V_i$ and $U$. Then, as all these restrictions are bijections in particular, $b$ has one pre-image in each $V_i$, so $|I|=k$, and moreover, we know that $|p^{-1}(y)|=k$ as well for every $y \in U$, so that $p \in U \subset A_k$, making $b$ an interior point of $A_k$.
Suppose now that $b \notin A_k$, and we take an evenly covered neighbourhood $U$ of it, with the same notation as before. The same reasoning as the previous paragraph gives that all points in $U$ have the same number of preimages as $b$ has, so if $b \notin A_k$, we also have $U \subset B \setminus A_k$, which shows that all points not in in $A_k$ are also not in its closure, so $A_k$ is closed as well.
Now, as $B$ is connected, the only sets that can be open and closed are $B$ and $\emptyset$, so assuming there is some point with finite fibre, all other points have the same sized fibre as well.