1
$\begingroup$

Let $p:E\rightarrow B$ be a fibration (i.e. have the homotopy lifting property with respect to all spaces), and $f: B'\rightarrow B$ and $g:B\rightarrow B'$ be homotopy inverses. Denote by $\pi_0\Gamma(B,E)$ the set of homotopy classes of sections of $p$, and likewise for other fibrations. I am interested in the following

Conjecture: There is a bijection $\beta:\pi_0\Gamma(B,E) \rightarrow \pi_0\Gamma(B',f^*E)$.

This would be a generalization of the elementary result $[B,X] \underset{\approx}{\xrightarrow{f^*}} [B',X]$, which is the case of trivial fibrations.

Some rough ideas:

  1. A good candidate for $\beta$ appears to be the induced map $f^*:\pi_0\Gamma(B,E) \rightarrow \pi_0\Gamma(B',f^*E)$ sending $\left[s:B\rightarrow E\right]$ to $\left[({\rm id}_{B'},s\circ f): B' \rightarrow f^*E\right]$, recalling $f^*E=B'\times_{f,p}E$. To prove $f^*$ is bijective, it would suffice to prove that the compositions $g^* \circ f^*$ and $f^* \circ g^*$ below are bijective: \begin{equation} \pi_0\Gamma(B,E) \xrightarrow{f^*} \pi_0\Gamma(B',f^*E) \xrightarrow{g^*} \pi_0\Gamma(B,g^*f^*E) \xrightarrow{f^*} \pi_0\Gamma(B',f^*g^*f^*E). \end{equation} It is easy to see that $\pi_0\Gamma(B,E)\approx\pi_0\Gamma(B,g^*f^*E)$. This is because $E\rightarrow B$ and $g^*f^*E\rightarrow B$ are pull-backs along homotopic maps $\rm id_{B}$ and $f\circ g$, and so are necessarily fiber homotopy equivalent (i.e. there eixst fiber-preserving maps between the total spaces whose compositions are homotopic to the identities via fiber-preserving maps); see e.g. Proposition 4.62 of Hatcher's "Algebraic Topology." For the same reason, $\pi_0\Gamma(B',f^*E) \approx \pi_0\Gamma(B',f^*g^*f^*E)$. However, I do not see why the bijections here should coincide with $g^* \circ f^*$ and $f^* \circ g^*$ above.

  2. I do not know if this is more suggestive, but in the diagram below, every diagonal map $B\rightarrow E$ making the lower right triangle commute gives rise to a map $B'\rightarrow E$ making the whole diagram commute. Assuming $\beta = f^*$, my conjecture reads: this assignment is bijective when we pass to homotopy classes. (The surjectivity is like a relative lifting problem.)

~

B'     E
|f     |p
v      v
B--id->B

Can we carry these ideas further? If not, do we still have a bijection $\beta$? Perhaps you have a counterexample in mind in which case the question would be addressed. Thanks in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

Have a look at this paper

R. Brown and P.R. Heath, ``Coglueing homotopy equivalences'', Math. Z. 113 (1970) 313-362.

available here. A special case is that you get a homotopy equivalence $f^* E \to E$. This should be enough for you.

May 14, 2015 (I have been on vacation.) I write [BH] for the cited paper.

Note that Corollary 1.4 of [BH] says under your conditions and with your notation that the map $f': f^*E \to E$ is a homotopy equivalence. Theorem 3.4 of [BH] gives you good control over homotopy inverses $g,g'$ of $f,f'$ and the homotopies to identities of $gf,fg,g'f',f'g'$. This should enable you to relate the homotopy classes of sections!

$\endgroup$
3
  • 1
    $\begingroup$ Thank you so much for your answer! This is a useful result, but how does it imply a bijection on $\pi_0\Gamma(-,-)$? To me the difficulty seems to be that there isn't an obvious map on $\pi_0\Gamma(-,-)$ induced by a map of fibrations, unless the map of fibrations is a pull-back. The paper says $(\bar f,f)$ is a homotopy equivalence $\bar p\rightarrow p$ of maps, but what this means is the existence of $(g_1,g): p \rightarrow \bar p$ satisfying certain properties, which is unfortunately not a pull-back. Possibly I'm just being dumb and there's a way around? $\endgroup$
    – user46652
    May 4, 2015 at 8:31
  • $\begingroup$ Thanks for expanding the answer! Let me try to be clearer. What we have is a map ${\bf f}:=(B'\xrightarrow{f} B,f^*E\xrightarrow{f'} E)$ between fibrations. Corollary 1.4 of [BH] says $f'$ is a homotopy equivalence, and Theorem 3.4 says $\bf f$ is a homotopy equivalence of maps. This gives us ${\bf g}:=(B\xrightarrow{g} B',E\xrightarrow{g'} f^*E)$ such that there are homotopies ${\bf gf} \simeq 1$, ${\bf fg} \simeq 1$ (over some maps between the base spaces, as specified in [BH]). $\endgroup$
    – user46652
    May 29, 2015 at 9:48
  • $\begingroup$ (cont'd) Now the problem is this (at least as it appears to me): to a section of $E\rightarrow B$ we can easily associate a section of $f^*E\rightarrow B'$ using the pull-back, but I don't know how we can even associate a section of $E\rightarrow B$ to a given section of $f^*E\rightarrow B'$, as $\bf g$ is not a pull-back. In other words, I don't see a natural way to define a map from the space of sections of $f^*E\rightarrow B'$ to that of $E\rightarrow B$. $\endgroup$
    – user46652
    May 29, 2015 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .