Simple fibre of fibration

Question

If $F \rightarrow E \rightarrow B $ is a fibration with path-connected fiber $F$ and $E$ is a simple space, is $F$ also necessarily simple? I can't think of any reason why should that be true on the other hand I am not capable of finding any counterexamples. Thank you for any answers.

Answer 1

I don't believe this is true. Let $0\rightarrow N\rightarrow G\rightarrow A\rightarrow 0$ be a central extension of an abelian group $A$ by an abelian group $N$ with $G$ nonabelian. Then there is a principal fibration sequence of Eilenberg Mac Lane spaces $K(N,1)\rightarrow K(G,1)\rightarrow K(A,1)$ which we can deloop to get a fibration sequence

$K(G,1)\rightarrow K(A,1)\rightarrow K(N,2)$

Now with $E=K(A,1)$, $F=K(G,1)$ we have $E$, $F$ connected and $E$ simple sine $\pi_1E$ abelian and $\pi_nE=0$, $n\geq 1$. However, $\pi_1F=G$ is nonabeliann so $F$ cannot be simple.