Talk:Bundle map

A new article - sorry if it is a bit rough. I lost my first attempt by closing my web browser by mistake, so I was less patient than usual... Geometry guy 02:19, 11 February 2007 (UTC)[reply]

I'm glad I'm not the only one that happens to ;) Fropuff 02:22, 11 February 2007 (UTC)[reply]

:) And thanks for tidying up some of the rough edges. Geometry guy 02:25, 11 February 2007 (UTC)[reply]

Must bundle maps always be continuous?

I was following along with the lecture series "Lectures on the Geometric Anatomy of Theoretical Physics", given by Dr. Frederic P. Schuller and freely available on YouTube. In Lecture 6, when he introduces bundle maps (about 50 minutes in), he does not include the requirement that they are continuous. I also don't see that condition included here http://mathworld.wolfram.com/BundleMap.html. I have been unable to find other information on the topic in my brief attempt at research (it doesn't help that the article lists no references), and was wondering if someone can clarify for me. Did Dr. Schuller simply forget to say the word continuous? Or is there some reasoning for intentionally leaving it out? Are both definitions used, and it simply depends on the author? — Preceding unsigned comment added by 2605:6000:2A80:6600:4D4D:A585:85E5:F50F (talk) 01:02, 28 June 2017 (UTC)[reply]

The Induced Map?

"In other words, $\varphi$ is fiber-preserving, and $f$ is the induced map on the space of fibers of $E$ : since $\pi _{E}$ is surjective, $f$ is uniquely determined by $\varphi$ ."

This sentence is confusing. It's not clear what is meant by "the induced map on the space of fibers of $E$ ". First off, I think it should say "the induced map from $M\to N$ ." Second, what exactly is the induced map? I can see that IF $\varphi$ is fiber-preserving THEN there exists a unique induced map $f$ . But, as written, the definition of a bundle map does not have "fiber-preserving" as a strict condition. Rather, from the given definition, it is a straightforward theorem/corollary that bundle maps are fiber preserving.

I think the paragraph could be reworked:

"The map $\varphi$ is fiber-preserving, that is, if $x\in \pi _{E}^{-1}(m)$ Then $\varphi (x)\in \pi _{F}^{-1}(f(m))$ . A bundle map $\varphi$ is said to be a bundle map covering $f$ . If only a map $\varphi :E\to F$ is given which is both continuous and fiber-preserving, then, because $\pi _{E}$ is surjective, a map $f_{\varphi }:M\to N$ is uniquely induced by $\varphi$ such that $\varphi$ is a bundle map covering $f_{\varphi }$ ."

Twistar48 (talk) 19:56, 24 March 2024 (UTC)[reply]