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I previously asked this on Math.SE but didn't receive a satisfactory answer.

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 vague ideas:

  1. It was pointed out by an author of [R. Brown and P.R. Heath, "Coglueing homotopy equivalences'', Math. Z. 113 (1970) 313-362] that the canonical projection $f':f^*E \rightarrow E$ is a homotopy equivalence (Corollary 1.4). Furthermore, there exists a map $g':E \rightarrow f^*E$, making the obvious diagram involving $g$ commute, such that $g'\circ f'$ and $f'\circ g'$ are homotopic to the identities via maps that factor through the bases (Theorem 3.4). This is an interesting result, but the issue is, unlike $f'$, that $g'$ doesn't seem to induce a map between sections in a natural way, so I don't know how this may be applied to my conjecture.

  2. There's an 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 that $f^*E=B'\times_{f,p}E$. One may try to prove $f^*$ is bijective. To do so, 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} Since $E\rightarrow B$ and $g^*f^*E\rightarrow B$ are pull-backs along homotopic maps, they are fiber homotopy equivalent (i.e. there eixst fiber-preserving maps between the total spaces, the compositions of which are homotopic to the identities via fiber-preserving maps), by e.g. Proposition 4.62 of Hatcher's "Algebraic Topology." It follows that \begin{equation} \pi_0\Gamma(B,E)\approx\pi_0\Gamma(B,g^*f^*E). \end{equation} Similarly, \begin{equation} \pi_0\Gamma(B',f^*E) \approx \pi_0\Gamma(B',f^*g^*f^*E). \end{equation} However, it is not known whether these bijections are given by $g^*\circ f^*$ and $f^* \circ g^*$.

Thank you in advance!


EDIT 6/5/2015: Upon encouragement by Dan Ramras, I made a renewed effort to carry my second idea further. I think the conjecture holds at least in "favorable cases," but I'm not sure how to conveniently characterize such cases, or if a more general proof is possible.

Our task boils down to the following. Let $p:E\rightarrow B$ be a fibration, and $F_t: B\rightarrow B$ be a homotopy such that $F_0={\rm id}_B$. I shall use $p_t$ to denote the pull-back fibration $F_t^*E\rightarrow B$. On the one hand, to each section $s\in \Gamma(B,E)$ of $p$ we can associate the section $({\rm id}_B, s \circ F_1)\in \Gamma(B, F_1^*E)$ of $p_1$. On the other hand, by the aforementioned Proposition 4.62 there is a fiber homotopy equivalence $\Phi:E\rightarrow F_1^*E$, so to each $s\in \Gamma(B,E)$ we can also associate the section $\Phi\circ s\in \Gamma(B, F_1^*E)$. The second way of associating sections is guaranteed to induce a bijection $\pi_0\Gamma(B,E)\xrightarrow{\approx} \pi_0\Gamma(B, F_1^*E)$, since $\Phi$ is a fiber homotopy equivalence. The task now is to show that the first way of associating sections induces the same map $\pi_0\Gamma(B,E)\rightarrow \pi_0\Gamma(B, F_1^*E)$. It suffices to show, given each $s\in \Gamma(B,E)$, that $\Phi\circ s$ and $({\rm id}_B, s \circ F_1)$ are in the same homotopy class of sections.

Let us recall the construction of $\Phi$. Regarding $F$ as a map $B\times I \rightarrow B$, there is the pull-back $\pi:F^*E\rightarrow B\times I$ of $p$ along $F$. Let \begin{eqnarray} L: E\times I &\rightarrow& B\times I \\ (e,t) &\mapsto& (p(e), t), \end{eqnarray} which can be thought of as a homotopy of maps $E\rightarrow B\times I$. Now consider the homotopy lifting problem \begin{eqnarray} E\times \{0\} &\xrightarrow{\widetilde L_0}&~ F^*E \\ \downarrow~~~~~~& &~~~\downarrow\pi \\ E\times I ~~~& \xrightarrow{~L~} & B\times I \end{eqnarray} where $\widetilde L_0$ is the obvious injection $E\times\{0\} \xrightarrow{\approx} F_0^*E \hookrightarrow F^*E$. Let $\widetilde L: E\times I \rightarrow F^*B$ be the lift of $L$ extending $\widetilde L_0$. Then we define $\Phi$ as the restriction of $\widetilde L $ to $t=1$, i.e. \begin{eqnarray} \Phi: E &\rightarrow& F_1^*E \\ e &\mapsto& \widetilde L(e,1). \end{eqnarray} Remarkably, by the proof of Proposition 4.62, the homotopy class $[\Phi]\in\pi_0\Gamma(B,F_1^*E)$ of $\Phi$ is independent of the choice of the lift $\widetilde L$. Therefore, it suffices prove the following: given each $s\in \Gamma(B,E)$, there exists such a choice of $\widetilde L$ that $\widetilde L(s(-),1) = ({\rm id}_B, s\circ F_1) \in \Gamma(B, F_1^*B)$. The nice thing is this choice can depend on $s$.

Thus suppose $s$ is given, and we will construct an $\widetilde L$ in two steps. In the first step, define \begin{eqnarray} \psi: s(B) \times I &\rightarrow& F^*E \\ (s(b), t) &\mapsto& \left((b, (s\circ F_t)(b), t\right). \end{eqnarray} This is well-defined as one can verify $(b, (s\circ F_t)(b)$ is indeed in $F_t^*E$. Noting that $\psi(s(b),0) = ((b,s(b)), 0)$, we paste $\psi$ and $\widetilde L_0$ to obtain \begin{eqnarray} \widetilde L_0 \cup \psi: (E\times\{0\}) \cup (s(B)\times I) \rightarrow F^*E. \end{eqnarray} $\widetilde L_0 \cup \psi$ certainly extends $\widetilde L_0$, and it lifts $L$ because $\pi \left((b, (s\circ F_t)(b), t\right) = (b,t) = L(s(b),t)$. In the second step, we have to solve the following homotopy lifting problem for the pair $(E,s(B))$ (or "homotopy lifting extension problem"): \begin{eqnarray} (E\times\{0\}) \cup (s(B)\times I) &\xrightarrow{\widetilde L_0 \cup \psi}&~ F^*E \\ \downarrow~~& &~~~\downarrow\pi \\ E\times I & \xrightarrow{~~~L~~~} & B\times I \end{eqnarray} This is where I had to make some favorable assumptions. Let us assume that every element of $\pi_0\Gamma(B,E)$ has a representative $s$ such that $(E,s(B))$ can be given a CW pair structure. By using a different representative if necessary we can assume that the given $s$ has this property. Now, a fibration is a Serre fibration, and a Serre fibration has the homotopy lifting property with respect to all CW pairs. Therefore the desired $\widetilde L:E\times I \rightarrow F^*E$ exists.

(To complete the proof of the original conjecture, of course, the same favorable assumptions should be made about $f^*E\rightarrow B'$ as well as $E\rightarrow B$.)

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  • $\begingroup$ Have you tried carefully tracing through the proof of 4.62 to extract formulas for the bijections it gives you? This might take some effort, but I would expect it to lead to a proof of your conjecture. $\endgroup$ – Dan Ramras Jun 4 '15 at 18:51
  • $\begingroup$ @DanRamras Thanks for your comment! Following your advice I made another attempt to relate the construction in 4.62 to that mentioned in my second idea. However, I was only able to prove the conjecture under certain assumptions, which might have been unnecessary. I've just edited my post. Let me know if you have further ideas, or if you spot any gap in my partial proof. Thanks again! $\endgroup$ – user46652 Jun 6 '15 at 5:50
  • $\begingroup$ I only skimmed what you wrote, but it looks to me like you only really need s to be a closed cofibration at the end (using the Strom model structure on Top). $\endgroup$ – Dan Ramras Jun 6 '15 at 6:10
  • $\begingroup$ @DanRamras I guess this is what you were pointing at. $s$ is a closed cofibration iff $(E,s(B))$ is an NDR pair, which in turn implies $(E\times I, (E\times\{0\})\cup(s(B)\times I))$ is a DR pair, and a fibration has the lifting property wrt all DR pairs. But I'm confused as to whether all these hold for arbitrary topological spaces. The source where I read this only dealt with compactly generated Hausdorff spaces (Felix et al., "Rational Homotopy Theory," Proposition 2.1). I can certainly compromise if necessary, but it'd be better to be free of such restrictions. $\endgroup$ – user46652 Jun 7 '15 at 2:41
  • $\begingroup$ @DanRamras I realized I wouldn't need to check all those mentioned in my previous comment. It'd be enough to show that $(E\times\{0\})\cup(s(B)\times I)\rightarrow E\times I$ (which is already acyclic/a homotopy equivalence) is a (Hurewicz) cofibration. Then we could use the axioms for model structure. $\endgroup$ – user46652 Jun 7 '15 at 2:54
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Here is one way of proving the conjecture is true in general, using the modern method of weak factorization systems.

A weak factorization system has at its core two classes of maps the left class and the right class and they satisfy lifting properties with respect to each other. Here the right class is the class of Hurewicz fibrations. The left class is the class of trivial Hurewicz cofibration (i.e. DR pair). These classes define each other. The left class consists of all maps with the left lifting property with respect to all maps in the right class; the right class consists of all maps with the right lifting property with respect to the left class. So trivial Hurewicz cofibrations have the left-lifting property with respect to the Hurewicz fibrations. They are exactly the Hurewicz cofibrations which are also homotopy equivalences.

We won't really need the general notion, just a few special cases. You already know some trivial Hurewicz cofibrations. For example for any space $B$, the inclusion $$B \times \{0\} \to B \times [0,1]$$ has the left lifting property with respect to any Hurewicz fibration (by definition), hence this is the first example of a trivial Hurewicz cofibration.

Also the trivial Hurewicz cofibrations are closed under several operations: composition, taking retracts, and cobase change (aka pushouts). Using these we can form new examples of trivial Hurewicz cofibrations. Let $f: B' \to B$ be any map. Then the inclusion of $B$ in the mapping cylinder $$B \to B' \times [0,1] \cup^{B'\times \{1\}} B$$ is an example (a pushout of our previous example). We can also consider the inclusion on the other side $$B' \times \{0\} \to B' \times [0,1] \cup^{B'\times \{1\}} B$$ If the map $f$ is a homotopy equivalence then this too is a trivial Hurewicz cofibration, though this takes a bit more work to see. The hard part is proven in prop 7 of these notes, as well as many textbooks.

Now let us first consider the special case of your conjecture where $f: B' \to B$ is a trivial Hurewicz cofibration.

Lemma: If $f: B' \to B$ is a trivial Hurewicz cofibration and $E \to B$ is a Hurewicz fibration, then we get an induced bijection: $$ f^*: \pi_0 \Gamma(B, E) \to \pi_0 \Gamma(B', f^*E) $$

Proof: First let's show surjectivity. A section of $f^*E$ is the same as a map $s: B' \to E$ such that $ps = f$. This is a triangle which we can enlarge into a square where the left edge is $f$, the right edge is $p: E \to B$ and the bottom is the identity on $B$. This square is a "lifting problem". Now we use the property that trivial Hurewicz cofibrations have the left lifting property with respect to Hurewicz fibrations to solve the lifting problem. This solution is a map $B \to E$ which is exactly a section of $E$ which restricts on $B'$ to the original section.

Injectivity is proven by the same argument but using the fact that $$ B \times \{0,1\} \cup^{B' \times \{0,1\}} B' \times I \to B \times I$$ is also a trivial Hurewicz cofibration. [We leave this part as an exercise]. QED.

Now using this lemma we can prove the general conjecture as follows. We will construct a space Z with two trivial Hurewicz cofibrations $$ i:B \to Z $$ $$ j:B' \to Z $$ and a map $k:Z \to B$ such that the composite $$B \to Z \to B $$ is the identity map on $B$ and the composite $$B' \to Z \to B $$ is our map $f$. Once we have a space with these maps, the previous lemma shows that the maps between homotopy classes of sections $i^*$ and $j^*$ are bijections. Since $k^* i^* = 1^*$ is also a bijection, this means that $k^*$ is a bijection too, and hence $f^* = k^* j^*$ is a bijection, which is what we wanted to show.

Such a space $Z$ can be constructed as $$ B' \times [0,1] \cup^{B' \times 1} B \times [1,2] $$

that is we glue $B' \times I$ to $B \times I$ at one end via the map $f$. The maps $i,j$ are given by including $B'$ and $B$ at 0 and 2, respectively (the ends of the "cylinder"). The map $k$ is given by projecting $B$ (by using $f$ for points on the first half of the cylinder).

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  • $\begingroup$ Thanks for writing out this beautiful proof with such clarity! I have two minor questions. 1) In the proof of the lemma, did you mean to say, for injectivity, we need the fact that $(B\times\{0,1\})\cup(B'\times I)\rightarrow B\times I$ is a trivial Hurewicz cofibration? This would follow from Corollary 8 of the notes you cited. 2) Does the proof essentially depend on the compactly generated weakly Hausdorff assumption? I'm guessing the answer is yes, because that appears to be the working assumption of the notes, where Prop. 5 says every Hurewicz cofibration is closed. $\endgroup$ – user46652 Jul 11 '15 at 23:28
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    $\begingroup$ Yes on the first point. I will edit. For the second point, my point set topology here is a little rusty, but I don't think we need the compactly generated weak Hausdorff assumption here, though many references will make that assumption. The point is that there is a similar weak factorization system on the category of all topological spaces where instead of just Hurewicz cofibrations we use closed Hurewicz cofibrations. Proving this is part of constructing the "Strom model structure". Finally, all of the cofibrations we use above are variations on mapping cylinders and so should be closed. $\endgroup$ – Chris Schommer-Pries Jul 13 '15 at 7:05
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Rather than write out a solution I would like to make some suggestions that I hope would enable you to solve the problem. I am open for further discussion.

It seems to me that one of the difficulties you have is the lack of a notation for some of the concepts that are likely to crop in the solution. Thus you have a notation for classes of sections, but not for more general situations.

So here is my suggestion, including changing a little the notation in the cited paper.

Let $p:D \to A, u: Z \to A$ be maps. Write $[Z,D;u]$ for the set of homotopy classes of maps $f:Z \to D$ such that $pf=u$ and each homotopy projects down to the constant homotopy on $u$. Thus classes of sections are the case $Z=A, u=1_A$. The first result is that if $p$ is a fibration and $\theta: u \simeq v$ is a homotopy then we have a bijection $\theta_{\#}: [Z,D;u] \to [Z,D;v]$ giving an operation of the groupoid of such homotopies on the sets of homotopy classes.

Now look at the results in the paper and see if you can see how this operation changes if you replace $p$ by a homotopy equivalent $q: E \to B$.

The paper gives a reference 1. which is available in a new edition titled Topology and Groupoids. The dual problem to yours could also be of interest. That will involve retractions.

Later: What your question exposes is that the paper cited does not discuss at all is how $[Z,D;u]$ depends on maps $(D \to A) \to (E \to B)$, and homotopies of such. So this is something you could try to write up.

July 9, 2015: here is some more information.

Consider a diagram as follows:

$$ \begin{matrix} && D & \xrightarrow{e} & E & \xrightarrow{d} & D \\ & f \nearrow & \downarrow p & & \downarrow q && \downarrow p\\ Z & \xrightarrow{u} &A & \xrightarrow{b} & B & \xrightarrow{a} & A \end{matrix}$$

We suppose $p,q$ are fibrations and $e,b$ give a fibre homotopy equivalence with homotopy inverse $d,a$. We are considering lifts $f$ of $u$. Note that if $u=1: A \to A$ then $f$ will be a section of $p$. We then get an induced diagram

$$ \begin{matrix} [Z,D;u] & \xrightarrow{\alpha} & [Z,E;bu] && \\ & \cong \searrow & \downarrow \rho & \searrow \cong & \\ && [Z,D;abu] & \xrightarrow{\gamma} & [Z,E;babu] \end{matrix} $$ where $\alpha$ is induced by $e,b$ and $\gamma $ is induced by $d,a$.

The first slanting arrow comes from the homotopy $u \simeq abu$ and the second from the homotopy $1 \simeq ba$. The diagram is commutative because the homotopies are part of the fibre homotopy equivalence. So $\rho \alpha$ and $\gamma \rho$ are isomorphisms. It is now a standard elementary categorical result that $\alpha, \rho, \gamma$ are isomorphisms.

When one examines the last part of the argument, one finds that curious "conjugacies" come in, see the cited paper.

As you see, to get the result on sections, i.e. when $u=1: A \to A$ , one has to introduce a more general concept and notation.

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  • $\begingroup$ Thanks for continuing to help! This does give me some insights into what's going on. I'll have to see what I can do with it. $\endgroup$ – user46652 Jun 7 '15 at 8:46
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    $\begingroup$ You can contact me by email if you want. One seems to need the old trick: if abc is defined in a category, and ab, bc are isomorphisms, then a,b,c, are iso also. $\endgroup$ – Ronnie Brown Jun 7 '15 at 13:00

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