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This was a question I first asked on stack exchange, here. In my head it seems like a fairly reasonable thing to ask for, but I'm not aware of any construction in the literature.

Let $p:E\rightarrow X$ be a Serre fibration over a pointed, connected CW complex $X$ with strict fibre a CW complex $F=p^{-1}(\ast)$. Given another space $F'$ and a homotopy equivalence $F\simeq F'$, is it possible to construct a Serre fibration $p':E'\rightarrow X$ with strict fibre $F'=p'^{-1}(\ast)$? If $E'$ exists, is it then possible to extend the homotopy equivalence $F\simeq F'$ to a fibre homotopy equivalence $E\simeq E'$ over $X$?

If you assume some extra structure on the fibration $p$ and the homotopy equivalence $F\simeq F'$ then things work out fairly easily, but I'm interested in the more general case.

Really I would like both questions to be taken together, since as pointed out in the comments, the projection $X\times F'\rightarrow X$ trivially satisfies the requirements of the first question alone, and such an answer is not exactly what I was looking for.

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  • $\begingroup$ can you please recall what "strict fiber" means? $\endgroup$
    – Gael Meigniez
    Commented Jul 19, 2019 at 9:22
  • $\begingroup$ The simplicial analogue of this is closely related to univalence, and is shown in Theorem 3.4.1 of The simplicial model of Univalent Foundations (after Voevodsky) (Kapulkin–Lumsdaine 2012). Specifically, in the triangular-prism diagram in the proof of 3.4.1, take $B$ to be your $X$, $A$ to be the point of $X$, $\bar{E_2}$ to be your $E$ (so $E_2$ is the fiber $F$), and $E_1$ to be your $F'$. The extension $\bar{E_1}$ constructed there is then your desired $E'$. I don’t think it’s quite trivial to adapt the proof to the topological setting, though. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jul 19, 2019 at 9:26
  • $\begingroup$ @GaelMeigniez, by 'strict fibre' I mean the inverse image $p^{-1}(\ast)\subseteq E$, with $\ast\in X$ denoting the supplied basepoint. $\endgroup$
    – Tyrone
    Commented Jul 19, 2019 at 9:33
  • $\begingroup$ For the first question you presumably want to avoid a trivial answer such as taking the projection $F'\times X \to X$. $\endgroup$
    – Allen Hatcher
    Commented Jul 19, 2019 at 19:21

1 Answer 1

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You can make the construction as follows. There are three steps.

First make a Serre fibration over the unit interval, pi:T->I=[0,1] such that the fibre over 0 (resp. 1) is F (resp. F'). You will find how to make this in the general case; for example, for F=1 point and F'=the interval, here is a proof that the triangle does it.

Consider in the plane the triangle T: 0<=y<=x<=1 and the projection pi(x,y)=x. Given a CW-complex A, a continuous map from A to T:a->(x_0(a),y_0(a)) and a homotopy (x_t(a)) (0<=t<=1), consider K={(a,t)\in AxI/x_t(a)=0} and K_0=(Ax0)\cap K. On the complement (Ax0)\K_0 you have the slope function s_0(a):=y_0(a)/x_0(a); extend it to a continuous function s from (AxI)\K to I; the wanted homotopy is (a,t)->(x_t(a),s(a,t)x_t(a)) for (a,t)\notin K and (a,t)->(0,0) for (a,t)\in K. Hence, pi is a Serre fibration.

Second, let $f:X\rightarrow X$ be homotopic to the identity and contract some neighborhood $N$ of $*$ onto $*$. Pulling back your fibration $p$ through $f$, you are reduced to the case where $p$ is a projection $F\times N\rightarrow N$ over $N$.

Third, consider a function $g: N\rightarrow I$ whose value is $0$ on $\partial N$ and $1$ on $*$; define $E'$ over $N$ as the amalgamated product of $T$ with $N$ over $pi$ and $g$; define $E'$ as $E$ over $(X-N)$. If you have chosen $N$ to retract by deformation on $*$, then this $E'$ will have the property you want.

Maybe there is a more elegant way to present the construction (i'm not an expert in homotopy theory nor cw-complexes)

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  • $\begingroup$ Thanks, Gael. If you work out that last step I'd be i'd be interested in seeing it. $\endgroup$
    – Tyrone
    Commented Jul 20, 2019 at 17:52
  • $\begingroup$ Gael, does the projection of a right triangle onto its base not satisfy that? $\endgroup$
    – Connor Malin
    Commented Jul 25, 2019 at 16:10
  • $\begingroup$ After Peter's comment, I have merged my partial answers. Thanks, Peter. $\endgroup$
    – Gael Meigniez
    Commented Jul 26, 2019 at 10:52
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    $\begingroup$ Could someone elaborate on how to produce a fibration over $[0,1]$ with both $F$ and $F'$ as fibers, starting just from a homotopy equivalence $f$ between $F$ and $F'$? The triangle example doesn't seem that informative to me; it suggests that sometimes the mapping cylinder of $f$ will do the trick, but that doesn't work in general (e.g. it fails when $f$ is a constant map from a disk to itself). $\endgroup$
    – Dan Ramras
    Commented Sep 1, 2023 at 18:06
  • $\begingroup$ The question raised in my previous comment is addressed here: mathoverflow.net/questions/453933/… $\endgroup$
    – Dan Ramras
    Commented Sep 4, 2023 at 20:08

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