Let $f : X \to Y$ and $g : Z \to Y$ be continuous maps between finite CW complexes. If $f$ is a simple homotopy equivalence, are there conditions on $g$ which guarantee that its pullback $f'$ is a simple homotopy equivalence?

Since homotopy equivalences themselves are not even preserved under pullback, a natural condition is for $f$ or $g$ to be a Serre fibration. The assumption that $f$ is a Serre fibration is better for my purposes, so let's assume that.

A possible reduction of the problem goes like this: Assume $f$ is a Serre fibration, then the pullback is a homotopy pullback, and we can replace $f$ with a formal deformation (using the terminology in Cohen's book), that is $f = f_0 \circ f_1 \circ \ldots f_n$ where each $f_i$ is an elementary expansion or contraction. We can then reduce the question to when the pullback of an elementary expansion/contraction is itself a simple homotopy equivalence. We can write down a ladder of pullbacks of the $f_i$s factoring $f'$, and if each $f_i'$ is a simple homotopy equivalence then the composition of them is too. I'm not sure how much this reduction really helps though.

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    $\begingroup$ You didn't define $f'$. I would suggest you attempt to reduce your question to one about simple maps, as in Definition 1.1.5 of folk.uio.no/rognes/papers/plmf.pdf, and then apply Proposition 2.1.3(c) of the same reference. $\endgroup$ – skupers Sep 8 '19 at 18:23
  • $\begingroup$ I'm using the unfortunate but common notation that $f'$ is the pullback of $f$ along $g$. Translating the problem into one about simple maps sounds like a fruitful approach, thank you! $\endgroup$ – Joe Berner Sep 8 '19 at 18:37
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    $\begingroup$ Before you can ask if $f'$ is a simple equivalence, you have to know that the homotopy pullback $X \times_Y Z$ is equivalent to a finite CW-complex in a preferred way. Given that under the given hypotheses it need not be equivalent to a finite CW-complex at all, it seems unlikely that there are reasonable conditions that guarantee this. $\endgroup$ – Oscar Randal-Williams Sep 9 '19 at 6:48
  • $\begingroup$ Though not your question, if it suffices to work in a (reasonably nice) simplicial or PL category, you have that maps with contractible point inverses are simple and obviously the pullback along any map will give a map with contractible point inverses. $\endgroup$ – Connor Malin Aug 1 at 2:08

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