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

Let $\backslash pi\; :\; T\; \backslash twoheadrightarrow\; B$ be a fibre bundle. For each $b\; \backslash in\; B$ let $F\_b$ denote the fibre over b. Consider the map $f\; :\; T\; \backslash to\; T$ given by $f(t)\; :=\; F\_\{\backslash pi(t)\}$. Under what conditions is $f$ homotopic to the identity $\backslash text\{id\}\_T\; :\; T\; \backslash to\; T$. 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. F_π(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)