Hatcher problem 1.3.8

Question

Suppose that $X$ and $Y$ are path connected and locally path connected spaces and that $\tilde{X}$ and $\tilde{Y}$ are simply connected coverings of $X$ and $Y$, repsectively. Prove that if $X$ and $Y$ are homotopy equivalent then $\tilde{X}$ and $\tilde{Y}$ are also homotopy equivalent.

Here is my proof, wanted to see if there are some major flaws in it or places that need more detail. Thanks

We know that $X \simeq Y$, so there are functions $f$ and $g$ such that $f: X \to Y$ and $g : Y \to X$, so that $$ f\circ g \simeq Id_X \text{ and } g\circ f\simeq Id_Y. $$ Now, choose a point $x \in X$ and let $f(x)\in Y$. Let $p_1: \tilde{X}\to X$ and $p_2 \tilde{Y}\to Y$ be covering maps. Then we know that $|p_1^{-1}(x)| = |\pi_1(X)|$ and $|p_2^{-1}(f(x))| = |\pi_1(Y)|$. Now, since $X\simeq Y$, we know that $|\pi_1(X)| = |\pi_1(Y)|$, and thus the cardinality of the fibers of $x$ and $f(x)$ must be equal. Thus, we define $$ p_1^{-1}(x) = \{a_n\}_{n\in A} \text{ and } p_2^{-1}(f(x)) = \{b_n\}_{n\in A} $$ for distinct $a_n \in \tilde{X}$ and $b_n \in \tilde{Y},$ where $A$ is the same size of $\pi_1(X)$, $\pi_1(Y)$. Likewise, $y\in Y$ and $g(y) \in X$, we get $$ p_1^{-1}(g(y)) = \{a_n\}_{n\in A} \text{ and } p_2^{-1}(y) = \{b_n\}_{n\in A} $$ for distinct points in $y\in \tilde{Y}$ and $g(y) \in \tilde{X}.$

Now, let $U_x$ be an evenly covered neighborhood of $x$ contained in $X$ and $V_{f(x)}$ be an evenly covered neighborhood of $f(x)$ contained in $Y$. Denote the lift 
$$
\tilde{f}: \tilde{X} \to \tilde{Y}
$$
be given by $\tilde{f}(a_n) = ((p_2^{-1})_n \circ f \circ p_1)(a_n)$, where $(p_2)_n$ is $p_2$ restricted to the component of $p_2^{-1}(V_{f(x)})$ containing $b_n$. In a similar way, $U_{g(y)}$ be an evenly covered neighborhood of $g(y)$ contained in $X$ and let $V_y$ be an evenly covered neighborhood of $y$ contained in $Y$. Let 
$$\tilde{g}:\tilde{Y}\to \tilde{X}$$
be given by $\tilde{g}(b_n) = ((p_1^{-1})_n \circ g \circ p_2)(b_n)$, where $(p_1)_n$ is $p_1$ restricted to the component of $p_1^{-1}(U_{g(y)})$ containing $a_n$. Since $\tilde{f}$ and $\tilde{g}$ are compositions of continuous functions, thus they are continuous, and we have
\begin{align*}
    \tilde{f}\circ \tilde{g} &= ((p_2^{-1})_n \circ f \circ p_1) \circ ((p_1^{-1})_n \circ g \circ p_2) \\
    &= (p_2^{-1})_n \circ g \circ f \circ p_2 \\
    &\simeq (p_2)_n^{-1} \circ Id_Y \circ p_2 \\
    &= Id_Y
    \end{align*}
likewise
\begin{align*}
    \tilde{g}\circ \tilde{f} &= ((p_1^{-1})_n \circ g \circ p_2) \circ ((p_2^{-1})_n \circ \phi \circ p_1) \\
    &= (p_1^{-1})_n \circ g \circ f \circ p_1 \\
    &\simeq (p_1)_n^{-1} \circ Id_Y \circ p_1 \\
    &= Id_Y
    \end{align*}
This tells us that $\tilde{f}$ and $\tilde{g}$ are homotopy inverses of each other, and that $\tilde{f}$ is a homotopy equivalence from $\tilde{X}$ to $\tilde{Y}$. Thus, $\tilde{X}\simeq \tilde{Y}.$

Answer 1

At the moment I don't seem to understand why the map $\tilde{f}$ is correctly defined. You have $n$ possible preimages of $f\circ p_1(a_n)$ and you have to choose exactly one, and show that this choice results in a continuous map. This is where local path connectedness (which you haven't) comes in.

I think that this exercise may be solved in a much easier way using the theorem about lifting maps to the covering space (if I remember correctly, it was before this exercise in Hatcher). It states that a map $f: Z\to X$ from a path connected, locally path connected space lifts uniquely to the covering space $\tilde{X}$, and the lifting exists if the image of $\pi_1(Z)$ under $f$ is inside the image of $\pi_1(\tilde{X})$ under the covering map. And the final part of your current proof would go along the lines of the proof of this theorem anyway.