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While reading Hatcher, he stated "If $p: \tilde{X} \rightarrow X$ is a covering space, then the cardinality of the set $p^{-1}(x)$ is locally constant. I have trouble seeing that this is the case.

I suppose we would want to take a neighborhood $U$ of $x$ then take another point $y \in U$. We would want to look at the path from $x$ to $y$ and this should lift to a path in $\tilde{X}$. But I fail to see how to proceed to show that the cardinalities of $p^{-1}(x)$ and $p^{-1}(y)$ should be the same. Moreover, how would we choose our $U$ 'sufficiently small' for this to be so? Is it the case that if $X$ is connected then all of the fibers have the same cardinality?

Kyle L
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In the definition of a covering space you have a covering of $X$ by open sets $U_i$ such that $p^{-1}(U_i)\cong U_i\times F$. Then the cardinality of $p^{-1}(x)$ is $|F|$ for any $x\in U_i$.

J126
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  • Ah, then the path lift puts $p^{-1}(y)$ in the same sheet so it works out just fine from there. We did not discuss the fiber like this when we discussed the material so it is much clear written this way. But why would $X$ being connected mean that this is globally constant? – Kyle L Mar 04 '15 at 03:09
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    @KyleL If $X$ is connected then, for each $U_i$, there is a $U_j$ such that $U_i\cap U_j\neq \emptyset$. For any point in the intersection the fiber $F$ must be the same. Hence it is the same fiber for all points in $U_i\cup U_j$. You can use that to argue that it is the same $F$ for all the $U_i$. – J126 Mar 04 '15 at 03:22
  • Ah, this was far more clever than the way I was trying to go about showing it by contradiction. I will sit down and flesh out the details. Thanks! – Kyle L Mar 04 '15 at 03:27