# Is the base of a disc bundle necessarily a strong deformation retract of the total space?

I am reading Algebraic Topology by E.H.Spanier and in the proof of the Thom-Gysin map for disc bundles (on page 260) he says that $p : E \to B$ is a deformation retraction. I do not understand how this is the case. How do we view $B$ as a subspace of $E$ in the first place ? And then how does $p$ become a deformation retraction ?

Also please advise some reference where I could learn basic properties of disc/sphere bundles. Thanks.

Edit : Here is the statement of the assumption part of Theorem 5.7.11 (in whose proof the statement appears) : Let $(\xi,U_\xi)$ be an oriented q-sphere bundle with base B and projection $\dot{p}=p|_\dot{E}:\dot{E} \to B$. Here $(E,\dot{E})$ is a fiberbundle pair with fiber $(D^{n+1},S^n)$ and $p: E \to B$ is the projection map.

• Isn't $S^n \to \{\ast\}$ a counterexample? – Michael Albanese Dec 4 '15 at 17:08
• Can you include the relevant portions of the text for completeness? I'm not going to go find a copy of Spanier. (This should include both the relevant assertion and the definitions of $E$ and $B$.) – user98602 Dec 4 '15 at 17:09
• @MikeMiller Sorry. I am going to edit the question in the light of your comment. Unfortunately Spanier does not give the definition of a sphere bundle, so I assume he is referring to a fiber bundle whose fiber is $S^n$. One can assume that the bundle is orientable. – user90041 Dec 4 '15 at 17:12
• @MikeMiller Edited the question now. I think the definition is that $(E, \dot{E})$ is a q-sphere if the fiber pair is $(D^{n+1},S^n)$ . – user90041 Dec 4 '15 at 17:20
• It might be that he demands that the transition functions of your disc bundle are linear, in which case you have a well-defined zero section. If not, to show that the projection is a deformation retraction (well, to make sense of that), you would need to start by showing it has a section, and presumably you'd want one whose image is in the interior of each fiber. – user98602 Dec 4 '15 at 17:32

First, I suspect Spanier wants all of his sphere and disc bundles to have linear transition maps, hence there is a canonical zero section $B \hookrightarrow E$. If not, then to make sense of his claim, you need a section; obstruction theory + the fact that $B$ is contractible guarantees that one exists, and indeed you can force it to be in the interior of each fiber.
Once you have a section $s$, note that both $p$ and $s$ are homotopy equivalences by Whitehead + the long exact sequence of homotopy groups of a fibration. Now recall one version of the Whitehead theorem: if $X \hookrightarrow Y$ is a cofibration and also a homotopy equivalence, there is a deformation retraction onto $X$. I would guess that any section of a disc bundle is a cofibration, but if not, pick your section above to be a good one. In any case, once you have this, Whitehead gives you a deformation retraction onto the image of the section, as desired.
• @Mike Miller I am still learning the area, but the proofs of characteristic classes and other obstructions I have seen all assume the existence of a section to start with (either by assuming linear transition functions or just implicitly assume without proof). How does one get around this bit of $S^1$ logic? – wanderingmathematician May 2 '18 at 15:52
• Actually, because $\pi_n D^k = 0$ for all $n > 0$, the obstruction cocycle is even zero - usually in obstruction theory you need to take a step back to take a step forward, and redefine your map over $X_n$ (without changing it over $X_{n-1}$) - but this isn't true here. You can always just extend your map to the next skeleton. You probably want to know the theorem that a fiber bundle over a contractible compact space is trivializable. – user98602 May 2 '18 at 18:23