# Fiber bundle with null-homotopic fiber inclusion

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 $p_*(\pi_n(E))$.

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?

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

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