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Let $p:E \rightarrow X$ be a covering space.

It is well known that if $X$ is connected, then all the fibers have the same cardinality. This can be seen as a simple consequence of the fact that the cardinality of $p^{-1}(x)$ is a locally constant map.

My question is, assuming $E$ is path-connected, how can I prove that all the fibers have the same cardinality? Can somebody help me?

Thanks.

-- EDIT --

Idea : If $E$ is path connected, then $X$ is connected and we're done. To see that $X$ is connected, let $x,y\in X$ and let $x_{0}\in p^{-1}(x)$ and $y_{0}\in p^{-1}(y)$. Since $E$ is path connected, let $\sigma$ be a path connecting $x_{0}$ to $y_{0}$. Then, $p\circ\sigma$ is a path connecting $x$ to $y$. Thus, $X$ is path connected and hence, connected.

valls
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1 Answers1

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another way you can say this by using uniqueness of path-lifting

given two points $x$ & $y \in X$ you can consider a path from $x$ to $y$ (this will exists since $E$ is path-connected and $p$ is continuous)...then given any point $x_0 \in p^{-1} (x)$ you can get a lift of that path starting from $x_0$ ...and by the uniqueness there will be a unique $y_0$ where the lift of the initial path will end..this gives a injective map from the set $p^{-1}(x) \to P^{-1} (y) $ ...and by using same argument we'll get a injective map $p^{-1}(y) \to P^{-1} (x) $ ...this will prove that all the fibers have same cardinality.

(but honestly speaking some how in this proof also I am using the fact that $p$ is locally homeomorphism...so this proof is not very far away from the proof you've mentioned)

Anubhav Mukherjee
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  • If I prove that $X$ is connected, it's done. So, let $x,y\in X$. Consider $x_{0}\in p^{-1}(x)$ and $y_{0}\in p^{-1}(y)$. Since $E$ is by hypothesis path connected, let $\sigma$ be a path connecting $x_{0}$ and $y_{0}$. Then, $p\circ\sigma$ is a path connecting $x$ and $y$. Thus, $X$ is connected. Is there something missing ? – valls Feb 28 '15 at 18:37
  • it is all right..but more over observe $p$ is continuous...and continuous image of a path-connected space is path-connected. – Anubhav Mukherjee Feb 28 '15 at 18:39
  • thanks for a "different" approach, though :) – valls Feb 28 '15 at 18:45