A claim from Hatcher's Algebraic Topology P. 94 In Hatcher's Algebraic Topology P. 94, there is a claim:
For a map $f:A\rightarrow B$ between connected CW complexes, let $p:\tilde{M}_f \rightarrow M_f$ be universal cover of the mapping cylinder $M_f$. Then $\tilde{M}_f$ is itself the mapping cylinder of a map $\tilde{f}: p^{-1}(A)\rightarrow p^{-1}B$ since the line segments in the mapping cylinder structure on $M_f$ lift to line segments in $\tilde{M}_f$ defining a mapping cylinder structure.
For me, $\tilde{M}_f$ looks like a mapping cylinder. By to prove a space is a mapping cylinder is not easy. There are some problems to be solved. For example, how to define the $\tilde{f}$? How to give the homeomorphism between $\tilde{M}_f$ and $M_{\tilde{f}}$.
Can anyone give a rigorous proof for this claim?
I also hope to know the covering space $\bar{M}_f$  for $M_f$ if $\pi_1(\tilde{M}_f)$ is not trivial.
 A: Denote $I' = [0,1)$.
The complex $A$ is locally simply connected (as a CW complex). Each simply-connected open $U\subset A$ embeds in $M_f$, and $U\times I' \subset M_f$ lifts to $\widetilde{M_f}$ (its embedding in $M_f$ factors through $p$ since $\pi_1$ of the embedding is the trivial homomorphism with respect to any basepoint).
Fix $x\in U$ and $x' \in p^{-1}(x)$. Let $h_U$ be the embedding $U\times I' \hookrightarrow \widetilde{M_f}$ such that $h_U(x,0)=x'$.
If $U,V\subset A$ are two simply connected opens then the lifts of $U\times I',V\times I'$ to $\widetilde{M_f}$ have compatible product structures in the sense that for any $x\in U\cap V$ and any $t\in I'$ we have $h_U(x,t)=h_V(x,t)$ for $h_U,h_V$ lifting $x$ to the same point $x'$.
Identifying $A\times I'$ with the appropriate subset of $M_f$, this shows why $p^{-1}(A\times I') \cong p^{-1}(A)\times I'$: we essentially covered $p^{-1}(A)$ with opens (the lifts of simply connected opens of $A$) and showed that multiplying everything with a half-open interval gives an open cover of $p^{-1}(A\times I')$.
(In rather more fancy language, we showed $p^{-1}(A\times I')$ has the structure of a trivial fiber bundle over $p^{-1}(A)$ with fiber $I'$.)

It is clear how to obtain a map $F:p^{-1}(A)\times I \to \widetilde{M_f}$ from the above: we already constructed it on the subspace $p^{-1}(A)\times I'$ (on which it is an embedding,) and on $p^{-1}(A)\times\{1\}$ one maps to $p^{-1}(B)$. Now we need to check that
$$p^{-1}(A)\times I \bigsqcup p^{-1}(B) \overset{F\ \sqcup\ \mathrm{id}}{\to} \widetilde{M_f}$$
is a quotient map, or in other words that a subset of the range is open iff its preimage is open. One direction is clear, because the map above is continuous. You should be able to prove the other using the facts that $M_f$ is a quotient of $A\times I \sqcup B$ and that $p$ is a covering map.

To be explicit, the map $\tilde{f}$ for which $\widetilde{M_f}$ is the mapping cylinder can be defined as follows: for a point $x\in p^{-1}(A)$, use path lifting to lift the "vertical" segment in $M_f$ connecting $p(x)$ with $f(p(x))$ up to $\widetilde{M_f}$ such that $p(x)$ is lifted to $x$, and let $\tilde{f}(x)$ be this segment's other endpoint.
