When is a fiber of a fiber bundle a retract of the total space?

Question

When is a fiber of a fiber bundle a retract of the total space? It clearly is the case for trivial bundles. There is probably a homological condition ... any help is welcome.

Answer 1

I don't have a full answer, but I thought I'd play around with the condition and see where it got me. I'm probably missing something very obvious which makes what follows a bit redundant, but let's go with it.

As far as I can see, the existence of a retraction $r \colon E \to F$ is a very strong condition. In particular, it means that you have a short exact sequence $$0 \to \pi_n(F) \to \pi_n(E) \to \pi_n(B) \to 0$$ for all $n\geq 1$, which comes from the fact that $i^*\colon \pi_n(F) \to \pi_n(E)$ is injective for all $n\geq 0$, which forces the connecting morphism $\partial \colon \pi_{n+1}(B) \to \pi_n(F)$ in the long exact sequence of a fibration to have image $0$ for all $n\geq 0$.

Now, we have a diagram

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\la}[1]{\kern-1.5ex\xleftarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} F & \stackrel{\ra{i}}{\la{r}} & E & \ra{f} & B \\ \da{=} & & \da{p} & & \da{=} \\ F & \stackrel{\ra{}}{\la{}} & F \times B & \ra{} & B \\ \end{array} $$ coming from the universal property of the product and some diagram chasing to show that the left square commutes (remember $p=(r,f)$ and $f \circ i$ is the constant map at the base-point of $B$). This induces a map between short exact sequences.

$$\begin{array}{c} 0& \ra{} & \pi_nF & \ra{} & \pi_nE & \ra{} & \pi_nB & \ra{} & 0 \\ & & \da{=} & & \da{p^*} & & \da{=} & & \\ 0& \ra{} & \pi_nF & \ra{} & \pi_n(F \times B) & \ra{} & \pi_nB & \ra{} & 0 \\ \end{array} $$

By the 5-lemma, $p^*$ is an isomorphism for all $n\geq 2$. So, the map $p$ is only obstructed from being a weak homotopy equivalence by the non-abelian nature of $\pi_1$ which can cause the 5-lemma to fail (and some path component stuff coming from $\pi_0$).

In particular, I feel pretty confident in saying that we can at least agree that $E$ and $F\times B$ have the same universal cover up to homotopy. It's possible that they really are homotopy equivalent, but I think it's likely that examples exist where $\pi_1$ can induce a map which isn't an isomorphism.