# Existence of section on fiber bundles with contractible fibers.

I would like a proof (or a sketch of a proof or a reference) for the following fact:

If $$\pi \colon P \to M$$ is a fiber bundle of manifolds, and its fibers are contractible, then there exist a continuous sections.

If necessary you may suppose the fibers are discs or that the base space is compact.

The reason why I'm asking this question is that this fact is used in paper Representations of surface groups in complex hyperbolic space - D. Toledo to show the existence of equivariant maps.

Is there a geometrical reason for this fact to be true? I would like to have an intuition of this.

• Are you willing to assume that $M$ is homeomorphic to a (generalized) CW-complex ? (this is the case if, e.g. $M$ is compact and differentiable) If it's the case then it's cofibrant in the Quillen model structure on $\mathbf{Top}$, and $\pi$ is a trivial fibration (it's a fibration because it's a fiber bundle, and the homotopy long exact sequence of the fibration together with the contractibility hypothesis show that it's a weak equivalence), so you can lift $id_M :M\to M$ along $\pi$, so that gives you a section. There is probably a more elementary argument Feb 16, 2020 at 14:21
• The comment of @Max can also be summarized by saying: construct the section by induction up the skeleta of the CW structure on $M$. This kind of induction is summarized in the phrase obstruction theory, and it goes back to Eilenberg. Feb 16, 2020 at 16:00
• @LeeMosher : you're right that this is probably a better way to phrase this, especially if OP wants a geometric explanation ! Feb 16, 2020 at 16:03
• This can be done without choosing CW structures: If there exists a numerable covering of $M$ for which each restriction $P|_U\rightarrow U$ is a homotopy equivalence over $U$, then $P\rightarrow M$ is a homotopy equivalence over $M$. See Theorem 7.57 in James's book General Topology and Homotopy Theory. If your manifold is paracompact and the bundle locally trivial, all assumptions are met. The upside is that I think the section can be constructed to be smooth in this case. Feb 16, 2020 at 17:25
• @Tyrone, I'm ok with just the continuous section. A smooth section can be built from the continuous one by the technique on this link.
– Hugo
Feb 17, 2020 at 23:13

If the base space $$M$$ is a CW complex then the problem of Existence Of A Section is solved using Obstruction Theory. (Steenrod's book "The Topology of Fibre Bundles" gives a great exposition in the classical "cell-by-cell" case, and you can find a more modern approach using Postnikov towers in Hatcher.)

(Edit: See Tyrone's comment for an alternate approach which doesn't use CW structures.)

The very rough idea is that if $$\sigma\colon M^{(n)} \to P$$ is a section on the $$n$$-skeleton and $$D$$ is an $$(n+1)$$-cell then $$\sigma$$ can be extended over $$D$$ iff $$\sigma|_{\partial D}$$ is null-homotopic, and since $$D$$ is contractible we can push the problem into a fibre over a point $$x\in D$$ and get a homotopy class $$[\sigma_D] \in \pi_n(F_x)$$ such that $$\sigma$$ can be extended over $$D$$ iff $$[\sigma_D] = 0$$. Since $$F$$ is contractible this homotopy class vanishes, and we can always extend the section.

To finish the obstruction theory picture, as $$D$$ varies over the $$(n+1)$$-cells all of these homotopy classes incredibly form an "obstruction class"

$$\mathfrak{o}(\sigma)\in H^{n+1}(M;\{\pi_n(F_x)\})$$

where $$\{\pi_n(F_x)\}$$ is a "bundle of coefficients" (if $$M$$ is simply-connected then the coefficients are just $$\pi_n(F)$$). Then $$\sigma$$ can be extended to the $$(n+1)$$-skeleton if and only if $$\mathfrak{o}(\sigma)$$ vanishes. If $$F$$ is contractible then all of these groups vanish, so a global section can be inductively constructed on the skeleta of $$M$$.

• Very nice answer! It's more intuitive and geometric than I expected. Thank you.
– Hugo
Feb 17, 2020 at 23:08