Bijection between pre-image in covering space and fundamental group

Question

Let $p:(\tilde{X},\tilde{x})\to (X,x)$ be a path-connected and locally path-connected covering space, where $X$ is path-connected, locally path-connected and semilocaly simply-connected. Consider the map $\partial :p^{-1}(x) \to \pi_1(X,x)$ be given by $y\mapsto [p\circ \tilde{\gamma}]$ where $\tilde{\gamma}$ is a path from $\tilde{x}$ to $y$. I think that $\partial $ is a bijection, where the inverse is given by $\partial^{-1}:\pi_1(X,x)\to p^{-1}(x),~[\gamma] \mapsto \tilde{\gamma}(1)$, where $\tilde{\gamma}$ is a lift of $\gamma$. Is this correct, or am I missing something?

Answer 1

Here's a counterexample: $\tilde X = X = S^1$, and $p(z)=z^2$, so $p$ is the map which wraps the circle around itself twice. This map is 2-to-1, so $p^{-1}(x)$ consists of exactly two points no matter what $x \in S^1$ you choose, but $\pi_1(S^1)=\mathbb{Z}$ is infinite.

For this counterexample, the map $\partial$ is not well-defined, because there are many paths $\tilde\gamma$ from $x$ to $y$, even in different path homotopy classes. Also, your map $\partial^{-1}$ is not one-to-one, because when you fix a base point $x \in S^1$, and lift the path that goes around $S^1$ one time, and the path that goes around three times, or any path that goes around an odd number of times, those lifted paths all start and end at the same point.

If, on the other hand, you had added the additional hypothesis that $\tilde X$ is simply connected, then the conclusion you want is true.