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Given a fiber bundle $p:\mathbb R^n\rightarrow B$ with a connected fiber $F$, why must the base $B$ be simply connected?

If $p$ was a covering map then it is a principal $\pi_1(B)$-bundle. Since the fiber is connected and $\pi_1(B)$ is discrete then $B$ is simply connected.

But how to prove that for general fiber bundles?

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If $F \to E \to B$ is a fibration, there is a long exact sequence is homotopy

$$\dots \to \pi_{n+1}(B) \to \pi_n(F) \to \pi_n(E) \to \pi_n(B) \to \pi_{n-1}(F) \to \dots$$

Every fibre bundle is a fibration, so applying this sequence to the fibre bundle $F \to \mathbb{R}^n \to B$, we get

$$\dots \to \pi_2(B) \to \pi_1(F) \to \pi_1(\mathbb{R}^n) \to \pi_1(B) \to \pi_0(F) \to \pi_0(\mathbb{R}^n) \to \pi_0(F) \to 0$$

As $\mathbb{R}^n$ is contractible, $\pi_1(\mathbb{R}^n) = 0$ and as $F$ is connected, $\pi_0(F) = 0$. By exactness, $\pi_1(B) = 0$. As $\mathbb{R}^n$ is connected, $\pi_0(\mathbb{R}^n) = 0$ so again by exactness, $\pi_0(B) = 0$. Therefore $B$ is simply connected.

By the contractibility of $\mathbb{R}^n$, it actually follows that $\pi_{n+1}(B) = \pi_n(F)$ for all $n \geq 0$.

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  • $\begingroup$ Awesome, thanks Michael $\endgroup$
    – Ronald
    May 13, 2017 at 23:59

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