Timeline for Is it true that all sphere bundles are boundaries of disk bundles?

12 events

when toggle format	what		by	license	comment
Sep 9, 2011 at 2:40	vote	accept	user16750
Sep 7, 2011 at 22:41	answer	added	Ryan Budney		timeline score: 6
Sep 7, 2011 at 22:04	comment	added	Igor Belegradek		There is a restriction maps $r: Diff(D^{k+1})\to Diff(S^k)$, which induces a map on classifying spaces $Br: BDiff(D^{k+1})\to BDiff(S^k)$. Smooth bundles over $M$ are homotopy classes of mapps into the suitable classifying space. What you are asking is whether any map from $M$ to $BDiff(S^k)$ is homotopic to a map that can be lifted is the image of $Br$. In the PL category the map $r$ has a section given by Alexander trick, as mentioned in the comment above. I do not know the answer in the smooth category, but I suspect it should be "no".
Sep 7, 2011 at 21:56	history	edited	Oscar Randal-Williams		tags
Sep 7, 2011 at 21:55	answer	added	Oscar Randal-Williams		timeline score: 14
Sep 7, 2011 at 19:43	comment	added	Peter Samuelson		Could someone give a reference for Johannes's comment?
Sep 7, 2011 at 18:38	comment	added	Johannes Ebert		It is true if you replace smooth by piecewise linear (Alexander trick).
Sep 7, 2011 at 17:17	comment	added	user16750		orientation preserving Diffeomorphism ($S^k$)
Sep 7, 2011 at 15:22	comment	added	Oscar Randal-Williams		With which structure group?
Sep 7, 2011 at 15:19	answer	added	Igor Rivin		timeline score: 3
Sep 7, 2011 at 15:17	history	edited	Neil Strickland	CC BY-SA 3.0	spelling, capitalisation, grammar in title
Sep 7, 2011 at 15:00	history	asked	user16750	CC BY-SA 3.0