7
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ 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 $\pi_{n-1}(F)\to\pi_n(B)$. $\endgroup$ Jul 16 '15 at 4:07
  • $\begingroup$ @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. $\endgroup$
    – 115465
    Jul 16 '15 at 17:20
  • 1
    $\begingroup$ 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 $D^n/\partial D^n\cong S^n$. $\endgroup$ Jul 19 '15 at 16:16
  • $\begingroup$ 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 $\pi_n(E, F) \simeq \pi_n(B)$ $\endgroup$
    – elidiot
    Jan 22 '20 at 16:31
  • $\begingroup$ I meant nullhomotopic relative to the inclusion of the base point* $\endgroup$
    – elidiot
    Jan 23 '20 at 11:16
6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Aug 4 '16 at 6:23
  • $\begingroup$ @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. $\endgroup$ Aug 4 '16 at 15:03
  • $\begingroup$ @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 $\partial \circ s = id_{\pi_{n-1}}(F)?$ $\endgroup$ Aug 9 at 8:53
1
$\begingroup$

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 $E$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.