Given a map $f: A \to B$ and a homotopy equivalence $g: C \to A$, I wish to show that $E_f \to B$ and $E_{fg} \to B$ are fiber homotopy equivalent, where, given a function $f: A \to B$, $E_f$ is the set of ordered pairs $(a,\gamma)$ such that $a \in A$ and $\gamma$ is a path in $B$ with $\gamma(0) = f(a)$. The fibration $E_f \to B$ then takes $(a, \gamma)$ to $\gamma(1)$.
$g:C \to A$ is a homotopy equivalence iff $C$ is a deformation retract of the the mapping cylinder $M_g$. This means that both $C$ and $A$ are deformation retracts of $M_g$, so this reduces to the case that $A$ is a deformation retract of $C$.
It is here that I get stuck. If it is assumed that $A$ is a subset of $C$ to which $C$ deformation retracts, then there is an inclusion $i: E_f \to E_{fg}$, and one can define $h:E_{fg} \to E_f$ by $h(c,\gamma) = (g(c),\gamma)$. Though $hi:E_f \to E_f$ is fiber homotopy equivalent to the identity function, $ih: E_{fg} \to E_{fg}$ is not; if one defines a homotopy on $E_{fg}$ from the deformation retraction of $C$ to $A$, this will take to $c$ to $g(c)$ (which takes care of the first coordinate), but will alter $\gamma$, as the definition of $E_{fg}$ requires that $\gamma(0) = fg(c)$.
More specifically, if $g_t:C \to A$ is the homotopy that retracts $C$ to $A$ (let $g_1$ be the same function as $g$), the induced $G_t:E_{fg} \to E_{fg}$ will be $G_t(c,\gamma) = (g_t(c),\lambda_{c,t} * \gamma)$, where $\lambda_{c,t} * \gamma$ is the inverse of the path traced out by $fg_t$ restricted to $c \in C$ from $0$ to $t$, followed by $\gamma$. The only way this works as a homotopy (as far as I can tell) is if $fg_t = fg$ for all $t$ (which makes $\lambda_{c,1}$ a constant path), and I see no reason to assume that such is the case.
I would greatly appreciate any help.
This won't be a complete answer but I hope points you in the right direction, so I can quickly give a response.
First I don't think it is best to mix the fibrations and cofibrations, i.e. to introduce the mapping cylinder, but just to work with cofibrations or fibrations.
This result is a special case of a ``coglueing theorem'' which says roughly that pullbacks by homotopy equivalences are homotopy equivalences, given a suitable fibration condition. More precisely, consider the diagram:
Suppose the front and back squares are pullbacks, $p,q$ are fibrations, $\phi, \phi_1, \phi_2$ are homotopy equivalences, and $\Phi$ is determined by $\phi, \phi_1, \phi_2$, i.e. the who;e diagram is commutative. THEN $\Phi$ is a homotopy equivalence. The details are available in this paper.
I'll leave you to work out the way to get from this your answer, or to see if it is in the paper.