I try to understand the proof of the following lemma, which is taken from the book by William S. Massey, "Algebraic Topology: An Introduction" (page 153).
Lemma 3.4. If $(\tilde{X},p)$ is a covering space of $X$, then the sets $p^{-1}(x)$ for all $x\in X$ have the same cardinal number.
Here, I will just give you the proof.
Proof. Let $x_0,x_1\in X$. Let $f:I\to X$ be a path with $f(0)=x_0$ and $f(1)=x_1$. He now constructs a path $g$ from a point $y_0\in p^{-1}(x_0)$ to a point $y_1$ by lifting the path $f$ to a path $g$ in $\tilde{X}$. He then claims that $y_0\to y_1$ is the desired mapping and that using the inverse path $\hat{f}= f(1-t)$ gives a path $p^{-1}(x_1)\to p^{-1}(x_0)$. Finally, he claims that it is clear that the maps are inverses to one another. $\square$
Questions and Toughts.
(1) Since $X$ is path-connected, we can define a map $f:I\to X$, with $f(0)=x_0$, $f(1)=x_1$, is this correct?
(2) Since $(\tilde{X},p)$ is a covering space, there exists a point $y_0\in p^{-1}(x_0)$ (because of surjectivity). And by lifting property of paths, we can lift $f$ to a path $g:I\to \tilde{X}$, such that $pg=f$. He never says where the point $y_1$ lives, but I guess it is in $p^{-1}(x_1)$. Is all of this correct?
(3) I don't really understand the notation $y_0\to y_1$. I would like to write it as $y_0\mapsto y_1$, because I think we are only mapping the end-points of the paths to one-another. When someone uses this notation $y_0\to y_1$, I read it as a path from $y_0$ to $y_1$. And I don't think that is a mapping from $p^{-1}(x_0)\to p^{-1}(x_1)$, or maybe it is? Are we saying the same thing? Or maybe I am the one who gets this mapping wrong, how does it work?
(4) Is uniqueness of $g$ used somehow?
(5) For the inverse-map, I guess what we are doing is to consider the same path, and then we map $y_1$ to $y_0$ instead. This would give us that $p^{-1}(x_0)\to p^{-1}(x_1)\to p^{-1}(x_0)$ gives the following mapping $y_0\mapsto y_1\mapsto y_0$ and the other way $y_1\mapsto y_0\mapsto y_1$, hence, they are the identity maps. Is this correct?
Best wishes,
Mr. J
(1) Correct.
(2) Since, $f(1) = p \circ g(1), g(1) \in p^{-1}(x_1)$.
(3) The mapping from $p^{-1}(x_0)$ to $p^{-1}(x_1)$ is defined as $y_0 \mapsto y_1.$
(4) The map would not be well-defined if it wasn't for uniqueness of $g$.
(5) Correct.