# If a fiber bundle with a connected total space admits a section, is the fiber connected?

Question. If a fiber bundle with a connected total space admits a section, is the fiber connected?

(Since there's a section the bundle is surjective, whence the base is connected as a continuous surjective image of a connected set. Thus the isomorphism type of the fiber is constant over the base.)

For covering maps this reduces to the following.

Fact. For a covering map with connected total space, it admits a section iff it's a homeomorphism.

To prove this one can show the image of any section is clopen upstairs. This approach does not generalize to the case of non-discrete fibers, since sections are generally far from having open image (consider sections of product projections).

As long as we're dealing with a sane base (paracompact Hausdorff), a fiber bundle is a Serre (even a Hurewicz) fibration and therefore has an associated long exact sequence of homotopy groups. This sequence shows that when the total space is connected and the base is simply connected, the fibers must also be connected. Thus an example must involve a non-simply connected base.

• This is true for locally trivial fiber bundles. For general bundles you might need some assumptions such as local connectivity. – Moishe Kohan May 31 at 5:16
• @MoisheKohan, to me a fiber bundle is defined as locally trivial. Why is the result true/how to prove it? What do you mean by "general bundle"? – Arrow May 31 at 6:06

I think this follows from the long exact sequence of homotopy groups for a fibration (assuming our bundle is locally trivial).

Let $$F\stackrel{\iota}{\to} E\stackrel{\rho}{\to} X$$ be a fibre bundle such that $$E$$ is path-connected, and let $$s\colon X\to E$$ be a section, i.e. $$\rho\circ s = id_X$$. Then for each $$n$$ we have $$\pi_n(\rho)\circ \pi_n(s) = id_{\pi_nX}$$, so in particular $$\pi_n(\rho)$$ must be surjective for all $$n$$. Therefore the connecting maps $$\pi_n X \to \pi_{n-1} F$$ in the long exact sequence are all $$0$$ and so $$\pi_n(\iota)$$ is injective for all $$n$$, so in particular $$\pi_0F \hookrightarrow \pi_0 E$$ so $$F$$ must be connected.

• Dear William, thanks for your answer! I have two questions. First, it seems your proof indeed works for any Serre fibration (not necessarily a fiber bundle). Is this correct? Second, what argument do you have in mind to prove surjectivity of $\pi_n(\rho)$ and exactness implies $\pi_n(\iota)$ is injective? Just that the cokernel of $\pi_n(\rho)$ is trivial so the long sequence has zeros in the right places? – Arrow May 31 at 14:11
• Yes, this should work for any Serre fibration with a section. You're also correct with your second question, I added a tiny bit of clarification to my answer. – William May 31 at 14:14
• "Connected" should be "path connected" for this argument (or you need to argue with Chech homotopy groups). – Moishe Kohan May 31 at 15:37
• @MoisheKohan thanks for noticing! I have no idea about Čech homotopy groups. Is it the same kind of argument? – Arrow May 31 at 20:26

Not always, a principal bundle has a section if and only if it is trivial.

https://en.wikipedia.org/wiki/Principal_bundle#Trivializations_and_cross_sections

• I don't follow. Suppose a principal bundle has a section. Then it's trivial, so the total space is the product of the base and the fiber. Since the total space is connected, so is the fiber. (Otherwise if $F\cong F_1\amalg F_2$ then $E\cong B\times F\cong B\times F_1\;\amalg\;B\times F_2$ is disconnected.) Thus in the trivial case, a connected total space implies connected fiber. I don't understand how to get a counterexample from your answer. – Arrow May 31 at 0:03
• there exist principal bundles which are not trivial and there fibers is conected. – Tsemo Aristide May 31 at 0:15
• Yes but this doesn't answer my question. I am asking whether a fiber bundle with a connected total space which admits a section must have connected fibers. Principal fiber bundles which are not trivial do not admit a section. – Arrow May 31 at 0:16