Wikipedia:Reference desk/Archives/Mathematics/2012 June 20

= June 20 =

Homotopy
Let $$\pi : T \twoheadrightarrow B$$ be a fibre bundle. For each $$b \in B$$ let $$F_b$$ denote the fibre over b. Consider the map $$ f : T \to T$$ given by $$f(t) := F_{\pi(t)}$$. Under what conditions is $$f$$ homotopic to the identity $$\text{id}_T : T \to T$$. I'm pretty sure that I need to consider a sequence of homotopy groups, but I'm not entirely sure. Any suggestions? — Fly by Night  ( talk )  23:04, 20 June 2012 (UTC)


 * I think there is a mistake in your definition of f. Fπ(t) is a subset of T, not an element in T. Rckrone (talk) 05:27, 21 June 2012 (UTC)


 * Exactly: a point goes to its fibre. Is there a problem with such a multi-valued function? — Fly by Night  ( talk )  07:08, 21 June 2012 (UTC)
 * It's not a map from T to T. It's map from T to the powerset of T.--2601:9:1300:5B:21B:63FF:FEA7:CABD (talk) 07:12, 21 June 2012 (UTC)
 * ...which is a problem for constructing a homotopy to the identity. I don't know what it means for two maps with different target spaces to be homotopic. Rckrone (talk) 16:33, 21 June 2012 (UTC)


 * I wouldn't go as far as to say it's a map into the power space. It's a one-to-many map into T and a surjective map into T/F. — Fly by Night  ( talk )  18:06, 21 June 2012 (UTC)
 * I suppose I've just given myself an extra condition: I suppose I need it to be a vector bundle with a connected base space. Although I would like to relax the conditions as much as possible. — Fly by Night  ( talk )  18:08, 21 June 2012 (UTC)