Number of sheets of covering space I have a question related to a number of sheets of covering space. So assume $p:\tilde{X}\to X$ be a covering map where $\tilde{X}$ is path-connected. Then in exercise 53.3 inMunkres' topology book, if we have a finite fiber, say $k$ many, at some point in $X$, then actually any point has $k$ many fiber so that in fact, we can say that the number of sheets is unique. I wonder if this statement holds for the infinite case. So the statement would be like: $\textit{If $\tilde{X}$ is path-connected, then the cardinality of fiber at any point in $X$ is the same}$.
So that the number of sheets is unique and the same as the cardinality of the fiber of any point in $X$.
 A: $X$ is a connected space.
Let $|A|$ denote the cardinality of set $A$.
Let $U$ be an evenly covered open set (of $X$) and let open sets(of $\tilde{X}$ ) ${U_α}$, $α \in J$, an index set, be the partitions of $p^{-1}(U)$. Also $p$ restricted $U_α$ is a homeomorphism between $U_α$ and $U$.
Let $x \in U$. Note that $p^{-1}$ {$ x $} $\bigcap U_α$ has exactly one element. Hence $|p^{-1}$ {$ x $}$|$ = $|J|$. This means that for $a,b \in U$, $|p^{-1}$ {$a$}$|$ = $|p^{-1}$ {$b $}$|$.
Now let $a \in X$.
$A$={$x \in X$; $|$$p^{-1}$ {$x$}$|$ =$|$$p^{-1}$ {$a$}$|$  }. By above remarks both $A$ and $X\backslash A$ are open. Since $X$ is connected this means $X\backslash A$ is empty. Hence $|p^{-1}$ {$ x $}$|$  is same for every $x \in X$.
A: This is true, and there is a neat proof if you are aware of the existence and uniqueness of the lifts of paths:
If $x$ and $x'$ are points in a path connected $X$ and $\gamma$ is a path from $x$ to $x'$, then there is a unique lift $\widetilde\gamma$ of $\gamma$ to any $\widetilde{x} \in p^{-1}(x)$. By following this path to its endpoint you get a map on the fibers $p^{-1}(x) \rightarrow p^{-1}(x')$. By using uniqueness of lifts again we get that this map is injective, which gives that all fibers have the same cardinality.
