7

It is an exercise from Hatcher (exercise 31, page 392):

For a fiber bundle $F \to E \xrightarrow{p} B$ such that the inclusion $F \hookrightarrow E$ is homotopic to a constant map, show that the long exact sequence of homotopy groups breaks up into split short exact sequences giving isomorphisms $\pi_n(B) \approx \pi_n(E) \oplus \pi_{n-1}(F)$.

Breaking up the long sequence is easy, it is a direct application of the hypothesis: since $i:F \to E$ is null-homotopic, $i_*$ is the null homomorphism and we have the following short exact sequences: $$0 \to \pi_n(E) \to \pi_n(B) \to \pi_{n-1}(F) \to 0 $$

But I couldn't split this short exact sequences. I know that it is suficient to construct a homomorphism $\gamma: \pi_n(B) \to \pi_n(E)$ such that $\gamma \circ p_*=Id_{\pi_n(E)}$. And this condition tells me how $\gamma$ should be on the range of $p_*$, but I don't know how to define it outside $p_*(\pi_n(E))$.

Edit: Reading Grumpy Parsnip's answer, two other questions came up:

  1. This is probably a dumb one. On the definition of $\partial$ on Grumpy's answer, he lifted a map $f:D^n \to B$ to $\bar{f}:D^n \to E$, but I'm not sure how this can be done. As far as I know the fiber bundle $p: E \to B$ has the homotopy lifting property with respect to disks $D^n$: given a homotopy $g_t:D^n \to B$ and a lift $\tilde{g}_0: D^n \to E$ of $g_0$, there is a homotopy $\tilde{g}_t: D^n \to E$ lifting $g_t$. This is exactly what we need to show that all these maps are well-defined, since they all use some sort of lifting. But I don't see how to use this property to define them.

  2. If such lifting always exist, wouldn't $\gamma: \pi_n(B) \to \pi_n(E)$ defined by $\gamma([f])=[\tilde{f}]$ (where $\tilde{f}$ is a lifting of $f$) be a splitting for the left side of the exact sequence?

Ugo Iaba
  • 1,207
  • 2
    I don't know if this will help, but note that there are two ways to prove a sequence splits. Maybe it's easier to construct a section $\pi_{n-1}(F)\to\pi_n(B)$. – Matt Samuel Jul 16 '15 at 04:07
  • @MattSamuel Sure, I'm aware of that. It is just that I'm more comfortable working with the left side. I dont completely understand the $\partial$ operator on the right side. Perhaps I should ask another question regarding the $\partial$ operator. – Ugo Iaba Jul 16 '15 at 17:20
  • 1
    For question $1$, a disk is an iterated product of intervals, so as long as you can lift a point, you can lift a disk. For question 2, the lift $\tilde{f}$ won't send the boundary to a point, so won't represent a map of a sphere $D^n/\partial D^n\cong S^n$. – Cheerful Parsnip Jul 19 '15 at 16:16
  • I think this is true whenever $E\to F$ is any null homotopic inclusion... Right? The statement is then $\pi_n(E, F) \simeq \pi_n(E) \times \pi_{n-1}(F)$. The case of a Serre fibration is a direct corollary of that, because $\pi_n(E, F) \simeq \pi_n(B)$ – elidiot Jan 22 '20 at 16:31
  • I meant nullhomotopic relative to the inclusion of the base point* – elidiot Jan 23 '20 at 11:16

2 Answers2

7

Here is how to construct a map from $\pi_{n-1}(F)\to \pi_n(B)$. Given a sphere $f\colon S^{n-1}\to F$, by hypothesis it bounds a disk $g\colon D^{n}\to E$. Consider the projection $\pi\circ g\colon D^n\to B$, where $\pi\colon E\to B$ is the projection map from the fiber bundle. Note that $\pi\circ g(\partial D^n)=\{*\}$ is a single point, so it represents a map of $S^{n}$ into $B$. I.e. it gives an element of $\pi_{n}(B)$ as desired. Now you have to show this map gives a well-defined homomorphism and is a splitting.

Edit: Responding to OP's comments, the right-hand splitting is the one you need. I doubt there is a natural left-hand splitting. The way you get the boundary operator in the long exact sequence is to take a map $f\colon S^n\to B$ representing your element of $\pi_n(B)$. You can think of it as a map $(D^n,\partial D^n)\to B$, where $\partial D^n$ maps to a point. Now by the homotopy lifting property for fiber bundles, you can lift $f$ to a map $\tilde{f}\colon D^n\to E$, but now $\tilde{f}(\partial D^n)$ will not be a point, but will instead lie in $\pi^{-1}(*)=F$. So $\tilde{f}$ give you an element of $\pi_{n-1}(F)$ as desired.

Eduardo Longa
  • 5,226
Cheerful Parsnip
  • 27,278
  • Where did we use that $F\to E\to B$ is a fiber bundle and not only a fibration, @GrumpyParsnip? Is that stronger assumption necessary? – iwriteonbananas Aug 04 '16 at 06:23
  • @iwriteonbananas: unless I'm missing something, this is true for fibrations. I assume Hatcher phrased it this way because he hadn't introduced fibrations yet in the text. – Cheerful Parsnip Aug 04 '16 at 15:03
  • @CheerfulParsnip First question: Who if $\tilde{f_0}$ in this diagram in your case? How do you prove the one you defined it's really a section? i.e $\partial \circ s = id_{\pi_{n-1}}(F)?$ – jacopoburelli Aug 09 '21 at 08:53
  • I've asked a separate question about the well-definedness and homomorphism properties here: https://math.stackexchange.com/questions/4318248/lifting-fibres-pi-n-to-bases-pi-n1-is-a-homomorphism – Descartes Before the Horse Nov 28 '21 at 15:16
1

To answer (2): I, too, feel more comfortable with the left hand side and so was reluctant to consider a map on the right hand side. However, I don't think it's so easy (obviously, the equivalent conditions of the splitting lemma (Hatcher p147) mean there must exist a suitable map if we do prove the sequence splits).

The map $p:E \to B$ is a covering map, mapping more than one element to an element $b \in B$. Your map $\gamma \colon B \to E$ inducing $\gamma_* \colon \pi_n(B) \to \pi_n(E)$ therefore cannot be well defined since there are multiple preimages in $E$.

Ali
  • 407