let $p: E\rightarrow B$ continuous and surjective. Show that if U is connected,then the partition of $p^{-1}(U)$ into slices is unique.

Question

Let $p: E\rightarrow B$ coninuous and surjective and suppose that $U$ is an open set of $B$ that is evenly covered by p and U. Show that if U is connected,then the partition of $p^{-1}(U)$ into slices is unique.

So far I got this:

Let b be a fixed point in U and suppose that there are two slices: $p^{-1}(U)=\cup B_\alpha$ and $p^{-1}(U)=\cup B_\beta$.

now there is a point $b_\alpha \in \cup B_\alpha$ such that $p(b_\alpha)=b$ and there is a point $b_\beta \in \cup B_\beta$ such that $p(b_\beta)=b$ and the points lie in a connected space because of the homeomorfism between $U$ and $B_\beta$ and $B_\alpha$.

I need to prove that these partitions of slices are actually the same but i don't see how......

Any hints?

Kees

Answer 1

I am unable to add comment right below Kees Til, so I just attach my answer here.

This is an exercise from Munkres Topology see here maths.ed.ac.uk/~aar/papers/munkres2.pdf on page 178 of the pdf (or page 336 of the book, §53 Covering spaces, second exercise) and here is the solution math.cornell.edu/~erin/topology/munkres.pdf – derivative Apr 30 at 14:29

Be careful that the solution is incorrect in math.cornell.edu/~erin/topology/munkres.pdf

it does not help to show a bijecction between $\mathcal{A}$ and $\mathcal{B}$, the bijection only tells you that they have the same cardinality.

Suppose $\{V_{\alpha}\}$ is the partition of $f^{-1}(U)$, the correct way to prove the statement is, show that each $V_{\alpha}$ is a connected component of $f^{-1}(U)$, the uniqueness then follows inmediately.