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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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# The Mapping Cylinder of a Pullback Square

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.