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
implies F is path-connected if and only if the homomorphism π1(E, e) → π1(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 π1(E, e) → π1(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 π1(E, e) → π1(B, b) is surjective.