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:
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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.
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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 $E\rightarrow B$.)
Best Answer
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).