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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)[reply]

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

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)[reply]

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)[reply]

...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)[reply]

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)[reply]

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)[reply]