**Question.** Let *p* : *E* → *B* be a fibration, let *b* be a point in *B*, and let *F* be the fibre of *p* over *b*. Suppose *E* is path-connected. When is *F* path-connected?

**Answer.** Choose any point *e* in *F* and consider the long exact sequence of homotopy groups induced by *p*. Exactness of

_{1}(

*E*,

*e*) → π

_{1}(

*B*,

*b*) → π

_{0}(

*F*,

*e*) → π

_{0}(

*E*,

*e*)

implies *F* is path-connected if and only if the homomorphism π_{1}(*E*, *e*) → π_{1}(*B*, *b*) is surjective, e.g. when *B* is simply connected.

As far as answers go, this one is quite neat. But it leaves something to be desired: after all, a geometric question deserves a geometric answer. So let’s unfold the proof to see what’s going on “under the hood” (so to speak).

Choose a point *e*′ in *F*. Since *E* is path-connected, there is a path *α* from *e*′ to *e*. Then *p* ∘ *α* is a loop in *B* based at *b*. Suppose π_{1}(*E*, *e*) → π_{1}(*B*, *b*) is surjective, i.e. each loop in *B* based at *b* is path-homotopic to the image of some loop in *E* based at *e*. By replacing *α* if necessary, we may assume *p* ∘ *α* is path-homotopic to the trivial loop at *b*. Since *p* is a fibration, the homotopy lifting property implies *α* is homotopic to a path in *F* from *e*′ to *e*. Since *e*′ is arbitrary, this shows that *F* is path-connected.

Conversely, suppose *F* is path-connected. Let *γ* be a loop in *B* based at *b*. By the homotopy lifting property, there is a path *α* in *E* from *e*′ to *e* such that *p* ∘ *α* is path-homotopic to *γ*. Since *F* is path-connected, there is also a path *β* in *F* from *e* to *e*′, so the composite path *αβ* is a loop in *E* based at *e*. By definition, *p* ∘ *β* is the trivial loop based at *b*, so *p* ∘ *αβ* is path-homotopic to *γ*. Since *γ* is arbitrary, this shows that π_{1}(*E*, *e*) → π_{1}(*B*, *b*) is surjective.

” e.g. when E is simply connected.”

You mean when B is simply connected?