When is a fiber of a fiber bundle a retract of the total space? 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.
 A: 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.
