# Showing that a homotopy fiber of a fibration is homotopy equivalent to the fiber of the base point.

Assume $(E, e_0)$ and $(B, b_0)$ are based spaces with the indicated base points.

Given a based fibration $p: E \rightarrow B$. We have the respective homotopy: fiber $$Fp= \{(e,\beta) | \beta(1)=p(e)\} \subset E \times F(I,B)$$ I wrote $F(I,B)$ as the path space for $B$ or space of based maps from the unit interval to $B$ that send $0$ to $b_0$.

Question: The fibre $F=p^{-1}(b_0)$ has an inclusion map $\phi: F \rightarrow Fp$ where $e \mapsto (e, c_{b_0})$ is an homotopy equivalence.

What I've thought of so far: The path mapping space $Np=\{(e, \beta)| \beta(1)=p(e)\}\subset E \times F(I_{+}, B)$. $I_{+}$ is a the union of the unit interval and some arbitrary disjoint base point.

We can replace $p: E \rightarrow B$ with the composition of $\nu:E \rightarrow Np$ with $\rho: Np \rightarrow B$. $\nu$ will send $e \mapsto (e, c_{p(e)})$ and $\rho$ will send $(e,\beta) \mapsto \beta(1)$.

The map $\nu$ is a homotopy equivalence since we can deform $Np$ to $\nu(E)$ using the homotopy which sends $(e, \beta(t), s)$ to $(e, \beta((1-s)t)$. We also have that $\rho^{-1}(b_0)$ is the homotopy fiber $Fp$.

It is difficult to pull off a similar deformation if there is one. This is where I am currently stuck and any hints on the matter will be helpful.

This is a problem from Peter May's A Concise Course in Algebraic Topology and this book makes no reference to Serre vibrations.

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Perhaps this answer is helpful. –  Zhen Lin Jun 21 '14 at 7:11