Homotopy equivalence of fibres of a Hurewicz fibration

Question

Let $p:E \rightarrow B$ be a Hurewicz fibration. It is known that if $B$ is path connected, then the fibres over any two points in $B$ are homotopy equivalent.

Question: Is there a simple proof of the above?

I understand that "simple" is subjective, but any proof at all (or reference) is welcome - the "simpler" the better.

Anyway, I am struggling to prove this myself. I get the idea that this shouldn't be difficult, but I can't really get anywhere at all. Given $b, b' \in B$, I am struggling to even write down a sensible map from $p^{-1}(b)$ to $p^{-1}(b')$.

My attempt:

Let $b,b' \in B$. I wish to construct a map $f: p^{-1}(b) \rightarrow p^{-1}(b')$ (which I hope to eventually prove to be a homotopy equivalence).

Trying to write a map $f$ down explicitly doesn't seem to help: Let $e \in p^{-1}(b)$. Now what? What should $f(e)$ be?

So instead we try to exploit universal properties, and infer the existence of $f$ or even a unique $f$.

$p^{-1}(b)$ is a pull-back of the diagram

$$E \rightarrow B \hookleftarrow *$$

where $*$ is included at $b$.

On the other hand, $p^{-1}(b')$ is a pull-back of $$E \rightarrow B \hookleftarrow *$$ where $*$ is included at $b'$.

The problem now is that my inclusions of $*$ aren't the same map, so it doesn't seem that I can exploit the universal property of the pullback in the way that I want.

I cannot think of a useful way to exploit the fact that $p$ is a Hurewicz fibration without first having maps between the fibres.

Answer 1

@Kevin Carlson gave the right hint, but I'll flesh out the details. Since $B$ is path-connected, we have that there is a path $\gamma$ in $B$ which starts at $b$ and ends at $b'$. Now consider the following diagram

$\require{AMScd}$ \begin{CD} p^{-1}(b) @>{g}>> E\\ @V{i_0}VV @VV{p}V\\ p^{-1}(b)\times I @>>{h}> B \end{CD}

Let $g$ denote the inclusion of the fiber above our point $b$ into the total space. We can see then that the composite $p\circ g$ sends everything to the point $b$. This means that $h(x,0)=b$ for any $x\in p^{-1}(b)$.

Let's define $h(x,t) = \gamma(t)$ to be our path between $b$ and $b'$. We now invoke the fact that $p$ is a Hurewicz fibration to see that it satisfies the homotopy lifting property. That is, there is a unique map $\widetilde{h}: p^{-1}(b) \times I \to E$ which commutes with the above diagram (I would draw it but AMScd is a horrible package).

Finally we have that $\widetilde{h}$ defines a homotopy between $p^{-1}(b)$ and $p^{-1}(b')$ in $E$.