Prove that exist bijection between inverse image of covering space

Question

Let $B$ be path-connected and $p:E\to B$ covering map (with $E$ as covering space). Prove that $\forall a,b\in B$ exist 1-1 injection correspondence between $p^{-1}(a)$ and $p^{-1}(b)$

I thought somehow taking the path between $a$ to $b$ and take on that the inverse means $p^{-1}(\gamma(t))$ but that's not well defined. How can I prove that?

Answer 1

I really like this proof because it uses a pretty neat trick:

Declare an equivalence relation on $B$ by saying $a \sim b$ if there exists a bijection between $p^{-1}(a)$ and $p^{-1}(b)$. Try to prove that the equivalence class $[x]$ of $x \in B$ is both open and closed. Since $B$ is path-connected (and hence connected), this implies that $[x] = B$, which completes the proof. To prove that $[x]$ is open, use the definition of a covering map. To prove it's closed, use the fact that the equivalence relation creates a partition of the set.

The set $p^{-1}(x)$ is often called the fiber above $x$, and this result is often stated by saying that all the fibers have the same cardinality.

Answer 2

I like the idea of using an algebraic model of a covering map, and one which has long been available is that of a covering morphism $q: H \to G$ of groupoids, dating back independently to C. Ehresmann and P.A. Smith, see also P.J. Higgins (1964) and his Categories and Groupoids, and the book by P. Gabriel and M. Zisman.

The condition for this is that for each $y \in Ob(H)$ and each $g\in G$ starting at $q(y)$ there is a unique $h \in H$ starting at $y$ such that $q(h)=g$. One proves by the usual methods that if $p: Y \to X$ is a covering map of spaces then the induced morphism of fundamental groupoids $q=\pi_1(p): \pi_1 Y \to \pi_1 X$ is a covering morphism. It was then of interest to this algebraic topologist to see how much you can do working with covering morphisms, and this is carried out in the book Topology and Groupoids, as it was in the 1968, 1988 editions. (Peter May's "Concise..." book goes some way in this direction, and does list the 1988 edition, as "idiosyncratic".)

In particular you easily find that for the above covering morphism $q: H \to G$ the groupoid $G$ operates on the fibres $q^{-1}(x), x \in Ob(G)$, and hence on a connected component of $G$ they all have the same cardinality.

This models the idea of the question.