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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.

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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.

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Thanks for your response! I still have a few questions, however. First, if one is to regard $E_f$ as a pullback ($P$ in the diagram), then $D$ in the diagram is the space of maps $[0,1] \to B$, $X$ is $B$, and $A$ is $B$. The pullback condition requires that $p(\gamma) = f(a)$; that is, $\gamma(1) = f(a)$, whereas the original condition was that $\gamma(0) = f(a)$. Second, if one has the homotopy equivalence between $E_{fg}$ and $E_f$ that results from this diagram, how does one know that it is fiber-preserving; that is, that $\gamma_t(1)$ is constant for all $t$? – Andrew Feb 25 '14 at 21:55
To keep with the diagram, which I copied from the paper, we rephrase your question as $g:Y \to B$ and $\varphi: B \to A$ is a homotopy equivalence. Also write $\Lambda B \to B$ for the path space so that $E_g$ is the pullback of this by $g$. Then $\varphi_1=1:Y \to Y$ and $f=\varphi g$. Since $\Phi$ is determined by the pullbacks, it is fibre preserving. Hope that helps. Note that the advantage of this proof is that it controls the homotopies. Reference 1. in the paper is now available as "Topology and Groupoids" (see my web page). – Ronnie Brown Feb 26 '14 at 12:34
Thanks again for your response, but unless I'm mistaken, the conditions of the problem were that $g:Y \to B$ is a homotopy equivalence and $\varphi:B \to A$ need not be, not the other way around. I also don't think that, in the context of this specific problem, $E_f$ denotes the pullback of the pathspace fibration, as the condition for ordered pairs in $E_f$ is that $\gamma(0) = f(a)$, but $\gamma(1)$ is mapped to B; if $E_f$ were the pullback, wouldn't it be required that $\gamma(1) = f(a)$, not $\gamma(0)$? – Andrew Feb 26 '14 at 17:27
With regard to the definition of $E_f$, the map $\Lambda B \to B$ may be chosen to be evaluation at $0$ or $1$. For the change to your actual question, you have I think to replace $(Q,E,Y,X,P,D,A,B)$ by $(E_{fg},\Lambda B, C,A.E_f,B,B)$. In the paper the key section is $3$. Special cases of the Coglueing Lemma appear in books (e.g. May "Concise...", p. 52) , which is also often proved in a Model Category, but without the detailed control of homotopies, which allows more general results, e.g. more than $2$ maps to $A$ and so more general limits than pullbacksd. – Ronnie Brown Feb 27 '14 at 11:26

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