In May's book, the translation of fibers specifies a functor. I am reading A Concise Course in Algebraic Topology written by J. P. May. On a page 53 of the book, he constructs the translation of fibers as follows:

Let $p:E\rightarrow B$ be a fibration with fiber $F_b$ over $b\in B$ and let $i_b :F_b\rightarrow E$ be the inclusion. For a path $\beta : I\rightarrow B$ from $b$ to $b'$
, the CHP gives a lift $\tilde{\beta}$ in the diagram
$\require{AMScd}$
\begin{CD}
F_b\times \{0\} @>a>> E\\
@V  V V\nearrow \tilde{\beta}   @VV p V\\
F_b\times I @>>{\beta\circ\pi_2}> B
\end{CD}
At time $t$, $\tilde{\beta}$ maps $F_b$ to the fiber $F_{\beta(t)}$. In particular, at $t=1$, this gives a map
$$\tau[\beta]\equiv[\tilde{\beta_1}]:F_b\rightarrow F_{b'},$$
which we call the translation of fibers along the path class $[\beta]$.

Here $\tilde{\beta_1}$ assigns each $e\in F_b$ to $\tilde{\beta}(e,1)$. Note that he says that our standing hypothesis that all spaces in sight are compactly generated allows the theory to be developed without further retractions on the given spaces.
In this situation, I want to prove the following statement:

Theorem. Lifting of equivalence classes of paths in $B$ to homotopy classes of maps of fibers specifies a functor $\tau :\prod(B)\rightarrow h\mathscr{U}$.

Here $\prod(B)$ is the fundamental groupoid for $B$, and $h\mathscr{U}$ is the homotopy category of the category of compactly generated spaces.
To do this, how do I show that it preserves a composition in the category?
 A: I do not believe the compact generation hypothesis is relevant in these considerations.
In a sense, proving functoriality here is "straightforward", but one has to understand the general technique. It has already been shown that to each path $\beta$ in $B$, you can find a lift $\overline{\beta}\colon F_b\times I\rightarrow E$ and that the homotopy class of the map $\overline{\beta_1}\colon F_b\rightarrow F_b^{\prime}$ is well-defined and only depends on the homotopy class of $\beta$. Assume a second path $\gamma$ in $B$ is given such that $\beta(1)=\gamma(0)$, i.e. these represent composable morphisms in $\Pi(B)$. It suffices to construct a lift $\overline{\gamma\beta}\colon F_b\times I\rightarrow E$, such that $\overline{(\gamma\beta)}_1\colon F_b\rightarrow F_{b^{\prime\prime}}$ is the composite $\overline{\gamma}_1\overline{\beta}_1$.
In cases one has a construction of this kind (I mean a construction where you make choices, which the result turns out to be independent of) and want to confirm some compatibility result (such as functoriality here), it's always the case that you will have to make sensible choices in the construction, which yield a sensible choice in the composite scenario.
In this case, this means we start with homotopy classes $[\beta],[\gamma]$ of paths, represented by $\beta,\gamma$, and choose the appropriate representative $\gamma\beta$ of $[\gamma][\beta]$. Then, we choose lifts $\overline{\beta},\overline{\gamma}$ and construct an appropriate lift $\overline{\gamma\beta}$. There's only one reasonable way to combine $\overline{\beta}$ and $\overline{\gamma}$. Namely, $\overline{\gamma}\circ(\beta_1\times\mathrm{id}_I)$ is a map $F_b\times I\rightarrow F_{b^{\prime}}\times I\rightarrow E$, which, restricted to $t=0$, equals $\beta_1$ (followed by inclusion), which is precisely $\overline{\beta}$, restricted to $t=1$. Thus, by gluing together the intervals at the respective endpoints and reparametrizing, this yields a map $\overline{\gamma\beta}\colon F_b\times I\rightarrow E$. It is the inclusion when restricted to $t=0$ by construction, and it lifts $\pi_2\circ(\gamma\beta)$, because $\overline{\beta}$ lifts $\pi_2\circ\beta$, $\overline{\gamma}$ lifts $\pi_2\circ\gamma$ and this gluing together intervals at the endpoints and reparametrizing is done in the same way as when we compose paths.
