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Suppose I have a pullback square, which I think of as a map from the fibration $q:X\to A$ to the fibration $p:Y\to B$. Then there is an induced map $m: M \to N$ from the mapping cylinder $M$ of $X\to Y$ to the mapping cylinder $N$ of $A\to B$. Is $m$ a fibration?

Here's what I know: if we pull $p$ back by the canonical quotient $N \to B$ (which is a homotopy equivalence), we obtain a fibration $P \to N$, and the map $M\to N$ in question factors as $M \to P \to N$; the comparison map $M\to P$ is both a bijection and a homotopy equivalence; and I can show that if the image of $A\to B$ is a closed subset of $B$, then $M\to P$ is a closed map, and hence a homeomorphism.

Motivation: I'm reading the proof of Mather's First Cube Theorem. Mather proves that a certain map can be compressed (over the base) into such a mapping cylinder and concludes that the map is a weak fibration (satisfies the weak CHP), citing a Lemma to the effect that if a map deforms, over the base, into a fibration then it is a weak fibration. I can make this work by converting to a cofibration twice (!), but it feels artificial, and I thought maybe it was common knowledge to some people that these mapping cylinders are fibrations.

So, I would like honest Hurewicz fibrations.

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I thought about this, and couldn't get anywhere. But it seems to me that you ought to be able to get $M\to P$ to be a homeomorphism as long as you work in some category of compactly generated spaces (so that "pullback" always means pullback in compactly generated spaces). – Charles Rezk Sep 22 '10 at 17:28
I'd be satisfied with an answer for compactly generated spaces. – Jeff Strom Sep 22 '10 at 17:33
What do you want to do with this result? You must want more than just the long exact sequence in homotopy. Do you care about the difference between Serre fibrations and Hurewicz fibrations here? – Dan Ramras Sep 22 '10 at 21:48
Perhaps try to explicitly construct a lifting function for $m$ from those of $p$ and $q$, by subdividing the domain of a path in the mapping cylinder $N$? Then the image of $A$ in $B$ being closed would come into play. I'm sure there's a slicker way though... – Mark Grant Sep 23 '10 at 8:42
If $A$ and $B$ are connected can all the fibers of $m$ really be the same if the fibers of $p$ and $q$ differ? This does not seem intuitive at all to me but perhaps I am not thinking about it right. – Jeremy Brazas Oct 1 '10 at 19:10

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