A map is a bundle and the isomorphism class depends on a homotopy class A bundle is a continuous map $p:Y\to X$ with a selected fiber $F$
so that for each $x\in X$, there is an open neighborhood $N_x$ of $x$ and a homeomorphism
$p^{-1}N_x\cong N_x\times F$ such that
$$(p^{-1}N_x\xrightarrow{\cong} N_x\times F\xrightarrow{\pi_1} N_x)=(p^{-1}N_x\xrightarrow{p}N_x).$$
It follows that $F\cong p^{-1}(x)$.
Now let $p$ be a bundle with fiber $F$, and $f:A\to X$ a continuous map, so we can form
the pullback of topological spaces
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
Z&\ra{}&Y\\
\da{p'}&&\da{p}\\
A&\ra{f}&X
\end{array}.
$$
How would I prove that $p'$ is again a bundle with fiber $F$?
I assume that we say let $a\in A$. Since $p$ is a bundle, there exists an open neighborhood $N_{f(a)}$ of $f(a)$ and a homeomorphism $p^{-1}N_{f(a)}\cong N_{f(a)}\times F$ such that
$$(p^{-1}N_{f(a)}\xrightarrow{\cong} N_{f(a)}\times F\xrightarrow{\pi_1} N_{f(a)})=(p^{-1}N_{f(a)}\xrightarrow{p}N_{f(a)}).$$ Then we look at $f^{-1}(N_{f(a)})$ which is open since $f$ is continuous and contains $a$. I have no idea how to proceed from here. I am assuming that the open neighborhood of $A$ we are looking for is in fact $f^{-1}(N_{f(a)})$.
 A: You have two completely separate questions going on here. I'll address the first. You should ask the other in a different question, though you should first look around MSE to see if something like this answers your question about homotopy invariance.
Pullback is a bundle
To prove that the pullback is again a fiber bundle with the same fiber, we make the following observations.
(1) The pullback of a trivial bundle is (canonically) trivial. If $B=Y\times F$, and $f:X\to Y$, then 
$$f^*B = X\times_Y (Y\times F)\simeq (X\times_Y Y)\times F \simeq X\times F.$$
(2) We can reinterpret the fiber bundle condition that every $x\in X$ has a neighborhood $U_x$ with $p^{-1}(U_x) \cong U_x\times F$ as saying that every $x\in X$ has a neighborhood $U_x$ such that the pullback $U_x\times_X B$ is trivializable.
(3) Then if $x\in X$, $y=f(x)\in Y$, $U_y\subseteq Y$ is an open neighborhood of $y$ on which $B$ is trivializable. Let $V_x = f^{-1}(U_y)$.
Then (omitting associativity isos)
$$
\begin{align}
V_x\times_X X\times_Y B
&\simeq 
V_x\times_Y B
\\
&\simeq 
V_x\times_{U_y} U_y \times_Y B
\\
&\cong 
V_x\times_{U_y} U_y \times F
\\
&\simeq 
V_x \times F
\\
\end{align}
$$
The first iso is the natural unit iso, the second iso is again the unit,
the third is by a choice of trivialization, and the fourth is the natural iso of observation (1). This completes the proof.
