Surjectivity of sending loop from fundamental group to the endpoint of its lift?

Question

This question refers to the following map defined on a fundamental group of some topological space $X$ covered by a covering map $p:\bar{X}\rightarrow X$:

$$r: \pi_{1}(X, x_{0}) \rightarrow p^{-1}(x_{0}),r[f] = \bar{f}(1)$$

where $\bar{f}: [0,1] \rightarrow \bar{X}$ is the unique lift of $f$, a loop in $X$.

I (and I assume most students learning the introduction to algebraic topology) have seen this map in the proof that $\pi_1(S^1)\cong\mathbb{Z}$, and have seen some of its properties proved for that specific case. However, the linked question claims $r$ has the following 2 properties in the general case as well:

1. Surjectivity if $\bar{X}$ is path connected
2. Injectivity if $\bar{X}$ is simply connected

I have managed to generalize the proof given for the specific case of $\mathbb{R}$ to show injectivity for simply connected spaces. However, the proof I saw for surjectivity is too specific for $\mathbb{R}$ and I wasn't able to think of a proof for the general case.

Is the general case truly correct (as mentioned in the linked question), and if so, can someone provide a proof or an idea for a proof?

Answer 1

Yes it is: first we need to fix $y_0\in p^{-1}(x_0)$ as the starting point of the lifts used by $r$. Then, for $x\in p^{-1}(x_0)$, since $\overline{X}$ is path-connected, there is a path $\gamma$ with starting point $y_0$ and end point $x$.

Then $p\circ \gamma$ is a path that starts at $p(y_0) = x_0$ and ends at $p(x)=x_0$, in other words a loop. Now $p\circ\gamma$ has a unique lift starting at $y_0$, so i must be $\gamma$, thus $r[p\circ\gamma] = x$, so $r$ is surjective

Answer 2

In order for the lift of $f$ to be unique, we need to fix a point $\widetilde{x_{0}} \in p^{-1}(x_{0})$, and then require that $\overline{f}(0) = \widetilde{x_{0}}$. Once this is done, given any other point $y \in p^{-1}(x_{0})$, path connectivity will give a path $\alpha$ from $\widetilde{x_{0}}$ to $y$. This path can then be dropped down to a loop in $X$ at $x_{0}$, namely $\alpha \circ p$. This loop $\alpha \circ p$ will satisfy (by uniqueness) the condition $r(\alpha \circ p) = y$, proving surjectivity.