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 
\begin{equation}
Fp= \{(e,\beta) | \beta(1)=p(e)\} \subset E \times F(I,B)
\end{equation}
 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.
 A: This is a very general fact about model categories and homotopy pullbacks, as evidenced by Zhen Lin's comment. It's also proven as a special case of Proposition 4.65 in Hatcher's book. Let me nevertheless spell out the argument precisely for topological spaces.
Define $E_p = \{ (y, \gamma) \in E \times B^{[0,1]} \mid p(y) = \gamma(0) \}$. There's a map (in fact a fibration) $q : E_p \to B$, $(y,\gamma) \mapsto \gamma(1)$, and the homotopy fiber is the fiber $F_p = q^{-1}(b_0)$. The inclusion $$i : F = p^{-1}(b_0) \to F_p$$ is given by $i(y) = (y, \mathrm{cst}_{b_0})$.
Define a homotopy $g_t : E_p \to B$, $(y,\gamma) \mapsto \gamma(t)$. Then $g_0(y, \gamma) = \gamma(0) = p(y)$, so $g_0$ lifts through $p$ by $\bar{g}_0 : E_p \to E$, $\bar{g}_0(y,\gamma) = y$ (i.e. $p \circ \bar{g}_0 = g_0$). Because $E \to B$ is a fibration, by the homotopy lifting property, there is a full lift $\bar{g}_t : E_p \to E$ of $g_t$ through $p$. In other words, $\bar{g}_t$ satisfies the following equation:
$$p(\bar{g}_t(y,\gamma)) = \gamma(t).$$
Now restrict everything to the fibers: let $h_t : F_p \to F_p$ be given by $h_t(y,\gamma) = \bigl(\bar{g}_t(y,\gamma), \gamma_{\mid [t,1]} \bigr)$ (because of the previous equation, this is in $F_p$). Then $h_0$ is the identity, whereas $h_1(y,\gamma) = (\bar{g}_1(y,\gamma), \mathrm{cst}_{b_0})$ is in the image of $i : F \to F_p$. And now that $h_t$ is a homotopy between $ih_1$ and the identity, while the restriction of $h_t$ is a homotopy between $h_1i$ and the identity. Thus $F$ and $F_p$ are homotopy equivalent.

Final remark: it's much simpler to prove that $F$ and $F_p$ are weakly homotopy equivalent, because the map $i$ induces easily an isomorphism on all homotopy groups. The square
$$\require{AMScd}
\begin{CD}
E @>>> E_p \\
@VVV @VVV \\
B @>>> B
\end{CD}$$
induces a map between the long exact sequences of the respective fibrations:
$$\begin{CD}
\dots @>>> \pi_n(F) @>>> \pi_n(E) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots \\
@. @VVV @VVV @VVV @VVV @. \\
\dots @>>> \pi_n(F_p) @>>> \pi_n(E_p) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots
\end{CD}$$
Since $E$ and $E_p$ are homotopy weakly homotopy equivalent ($PB$ is contractible), and so by the five lemma and induction, the maps $\pi_n(F) \to \pi_n(F_p)$ are isomorphisms.
