If $h : Y \to X$ is a covering map and $Y$ is connected, then the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$.

Question

In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition

Let $h: Y \to X$ be a covering. For any path $\gamma : [0,1] \to X$ and any $y \in h^{-1}(x) (x = \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to Y$ of $\gamma$ with $\hat{\gamma}(0) = y$. Furthermore, for any homotopy $\gamma_t (t \in [0,1])$ of $\gamma$ with $\gamma_t = \gamma(0)$ and $\gamma_t(1) = \gamma(1)$, there exists a unique lift of $\hat{\gamma_t}$ such that $\hat{\gamma_t}$ is the homotopy of $\hat{\gamma}$ with $\hat{\gamma_t}(0) = \hat{\gamma}(0)$ and $\hat{\gamma_t}(1) = \hat{\gamma}(1)$.

The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$."

I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different $x_1, x_2$ and their fibers $h^{-1}(x_1), h^{-1}(x_2)$ such that $y_1 \in h^{-1}(x_1)$ and $y_2 \in h^{-1}(x_2)$. Take two paths $\gamma_1, \gamma_2$ such that $\gamma_1(0) = x_1$ and $\gamma_2(0) = x_2$. Then we get two lifts $\hat{\gamma_1}, \hat{\gamma_2}$ with $\hat{\gamma_1}(0) = y_1$ and $\hat{\gamma_2}(0) = y_2$. Now, since our covering space is connected we can continuously deform $\hat{\gamma_1}$ into $\hat{\gamma_2}$ and conclude that every elements in $h^{-1}(x_1)$ is also in $h^{-1}(x_2)$ and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated.

Answer 1

This is true more generally if the base space $X$ is connected. It suffices to notice that the relation $$x\sim x'\text{ iff the fibers over }x\text{ and }x'\text{ have the same cardinality}$$ is a equivalence relation with open equivalence classes. Since the equivalence classes of an equivalence relation partition the underlying set, and because a connected space is, by definition, one that doesn't admit any non trivial partition by open sets, the equivalence relation to be trivial, i.e. there can only be one equivalence class. This of course means that all fibers must have the same cardinality.

The fact that the equivalence classes are open is also obvious. Indeed, let $x\in X$. By definition of being a covering map, there exists an open neighborhood $V$ of $x$, a discrete nonempty set $I$ and a homeomorphism $\phi:p^{-1}(V)\to V\times I$ that commutes to the projections. Then the fibers over the $x'\in V$ all have cardinality (equal to that of $I$), which proves that for all $x'\in V$, $x\sim x'$, and the equivalence classes are open.

Answer 2

First, there is an underlying assumption that $X$ is connected.

Having said that, define a function on $X$ where $N(x)$ equals the cardinality of $h^{-1}(x)$. The range of this function can be thought of as a set of cardinal numbers, and you want to give that set the discrete topology. For example, if you are willing to add the assumption that $X$ has a countable basis, then the range of $N(x)$ is the set $\{1,2,3,…\} \cup \{\infty\}$. Give that set the discrete topology.

You wish to prove that $N(x)$ is a constant function. Since the domain is connected and the range is discrete, it suffices to prove that $N(x)$ is a continuous function.

To prove that $N(x)$ is continuous it suffices prove that $N(x)$ is constant on any neighborhood $U$ of $x$ such that $U$ is evenly covered by the covering map $h$, and that should be very straightforward.

Answer 3

In greater generality, it is possible to construct a functor $\lambda: \Pi (X) \to hTop$, where $\Pi(X)$ is the fundamental groupoid of $X$ and $hTop$ is the naive homotopy category associated to $Top$, which acts on points as $b \mapsto p^{-1}(b)$ whenever $p$ is a fibration. It follows that in a fibration with path connected base space any two fibers are homotopy equivalent; being the fibers of a covering discrete spaces, you get the thesis by the induced isomorphism $\pi_0 (f^{-1}(x)) \simeq \pi_0 (f^{-1}(x'))$.