Good evening! I am reading through J. May's book 'A coincise course in Algebraic Topology' and I am trying to prove the following statement (some Lemma at page 62):
If $p:E\to B$ is a fibration, then the inclusion $$\phi:p^{-1}(\ast)\to Fp$$ specified by $\phi(e)=(e,c_\ast)$ is a based homotopy equivalence (here $\ast\in B$ is the base point).
I am struggling first of all because I am relatively new to such massive use of algebraic homotopy inside topology and I am still really far from having a solid grasp on such topics.
For what concerns my thoughts, I recall that in general any continuous map $f:X\to Y$ factors through a homotopy equivalence $\nu:X\to Nf$ and a fibration $\rho: Nf\to Y$ by mean of the mapping path space $Nf$. In this special case where $p$ is a fibration, I wouldn't know how to use this additional information. Also, the fact that the game is with pointed maps somewhat bothers my untrained brain.
It can be useful to remind that the homotopy fiber $Fp$ is the pull-back coming from the following diagram: $\require{AMScd}$
\begin{CD}
Fp @>{}>> PB\\
@V{\pi}VV @V{p_1}VV\\
E @>{p}>> B
\end{CD}
where $PB$ is the path space on $B$ and $\pi$ and $p\_1$ are the obvious projections (both of them fibrations).
Finally, one can notice that $Fp\simeq \rho^{-1}(\ast).$
One idea that seems to be reasonable is to prove that $\phi$ and $\pi$ are the pair of homotopy equivalences I need to work with since at least the composition $\pi\circ\phi$ is trivial, but I don't know how to deal with the other one.
Is anybody willing to help me and give me at least a strategy to follow?