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Given two homotopy equivalent spaces $X$ and $Y$, does there always exist a Hurewicz fibration $p: E\rightarrow [0,1]$ with $p^{-1} (0) = X$ and $p^{-1} (1)=Y$?

This issue shows up in the accepted answer to this question: Replacing the Fibre of a Fibration.

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There is an old result due to Patricia Tulley which claims that this is possible.

P. Tulley, A strong homotopy equivalence and extensions for Hurewicz fibrations, Duke Math. J. 36(3): 609-619 (September 1969).

The main result of the paper is that, with some basic standing assumptions on the spaces involved, two Hurewicz fibrations over the same base are fiber-homotopy equivalent if and only if they are concordant (this being the relation of strong fiber-homotopy equivalence).

This result is then applied to the fibrations $X\rightarrow\ast\leftarrow Y$ when $X,Y$ are homotopy equivalent spaces. The result is a Hurewicz fibring $$p:Z\rightarrow I$$ with $p^{-1}(0)\cong X$ and $p^{-1}(1)\cong Y$.

Although the results for general fibrations require some compactness assumptions, the statement regarding $X,Y$ above is given with no additional assumptions. However, I haven't gone through the paper thoroughly to check all the details.

Edit: As pointed out by Cihan in the comments, Tulley would later go on to prove the result cited above by showing that any two Hurewicz fibrations over the same base space are fibre-homotopy equivalent if and only if they are strongly fibre-homotopy equivalent.

Theorem (Tulley): Suppose that $B$ is a space and $p_0:E_0\rightarrow B$ and $p_1:E_1\rightarrow B$ are Hurewicz fibrations. Then the following statements are equivalent.

  1. There is a map $f:E_0\rightarrow E_1$ which is a homotopy equivalence of spaces and satisfies $p_1\circ f=p_0$.
  2. There is a fibre-homotopy equivalence $E_0\simeq_BE_1$.
  3. There is a space $Z$ and a Hurewicz fibration $\pi:Z\rightarrow B\times I$ such that $E_0\cong Z|_{B\times 0}$ and $E_1\cong Z|_{B\times1}$. $\quad\blacksquare$

The equivalence of the first and third bullet points above is a well-known result not due to Tulley. The third bullet point is the definition of a strong fibre-homotopy equivalence. In any case, when $B$ is a point, it clearly leads to the desired outcome. For its proof consult:

P. Tulley McAuley, A note on paired fibrations, Proc. Amer. Math. Soc. 34 (1972), 534-540.

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    $\begingroup$ What warrants the skepticism? The paper is, after all, published in Duke. $\endgroup$
    – Pedro
    Commented Sep 4, 2023 at 10:06
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    $\begingroup$ @Pedro there's no doubt about the integrity of the paper nor the ability of its author - only that I may have misremembered something from the last time I read it (which was several years ago). $\endgroup$
    – Tyrone
    Commented Sep 4, 2023 at 11:06
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    $\begingroup$ Ok. I simply (mis)interpreted "claims" as a softer statement than "proves", hence my confusion. $\endgroup$
    – Pedro
    Commented Sep 4, 2023 at 11:21
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    $\begingroup$ In a 1972 paper called "A note on paired fibrations", the same author shows that for Hurewicz fibrations involving Hausdorff spaces, a fiber homotopy equivalence is always a strong fiber homotopy equivalence. $\endgroup$
    – Cihan
    Commented Sep 4, 2023 at 11:55
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    $\begingroup$ Hi Dan, There is a second paper by Tulley which I would like to thank @Cihan for bringing to my attention (see the edit). In her first paper, Tulley proves a restricted version of the statement and goes on to massage it in the special case that $B$ is a point. In the second paper the statement is an immediate corollary of a much stronger result. $\endgroup$
    – Tyrone
    Commented Sep 4, 2023 at 13:31

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