Timeline for Could there be any homotopy group without "Lebesgue Number Lemma"?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 29 at 13:24 | vote | accept | user51223 | ||
| Jun 27 at 22:05 | comment | added | Johannes Ebert | The Lebesgue covering lemma is basically the statement that continuous functions on compact metric spaces are uniformly continuous. In that disguise, the lemma is a key ingredient for most of the analytical tools you mentioned (integration, approximation of continuous by smooth functions and surely also in Sard's theorem). The standard proof of the $\pi_1$-Hurewicz theorem is combinatorial, but the computation of $H_1 (S^1)$ relies on excision hence also on the covering lemma. | |
| May 7 at 7:27 | comment | added | user51223 | @TimothyChow Thank you so much for the clarifications. | |
| May 6 at 12:47 | comment | added | Timothy Chow | @user51223 The "synthetic" version of $\pi_1(S^1)=\mathbb{Z}$ is not exactly the same assertion as the classical assertion that $\pi_1(S^1)=\mathbb{Z}$. In particular, it can't be proved in the same way. Everything in homotopy type theory is homotopy invariant, and the covering space map $\mathbb{R}\mapsto S^1$, which plays such an important role in the classical proof that $\pi_1(S^1)=\mathbb{Z}$, is homotopic to a constant function, and therefore useless in homotopy type theory. Also, point-set topology (the Lebesgue number lemma) plays no role in homotopy type theory. | |
| May 6 at 12:14 | comment | added | Andy Putman | By the way, you might be interested in the following very short proof of SvK (due to Grothendieck): www3.nd.edu/~andyp/notes/SeifertVanKampen.pdf | |
| May 6 at 12:12 | comment | added | Andy Putman | @user51223: Again, probably the Lebesgue number lemma is needed for the most general form of SvK, but in many (most?) specific examples you can avoid it. | |
| May 6 at 12:05 | comment | added | user51223 | @AndyPutman No, thee is not specific computation that I am after it. But, I always got the impression that there is “essentially” one way to compute the fundamental group of the circle, to start with, which is a kind of rigidity and the motivation for homotopy type theory!!? Perhaps, I am putting a sequence of vague statement together to make a deduction. However, having said these, I think for a class of important examples the Seifert-van Kampen theorem would be an imortant tool to compute $\pi_1$for which I think the Lebeshue number lemma is needed! | |
| May 6 at 11:44 | comment | added | Andy Putman | @user51223: that’s a little vague. If you literally want to know whether you need Lebesgue numbers to prove the most general version of the lifting theorem, then probably the answer is “yes” (though I’m not sure there’s a way to make that precise). But for most specific computations, you can find devices that bypass that. Is there a specific computation you are interested in? | |
| May 6 at 3:15 | comment | added | user51223 | Thank for the wonderful comments, and your answer @AndyPutman. I feel the mention of $S^1$ in my question was somehow a mistake. The general question was about homotopy groups whose computations, in general, rely on covering space theory if we exclude pathological cases! That part of the theory, the construction of liftings, as far as I know depends on the Lebesgue number lemma. Could that be avoided too? | |
| May 5 at 11:44 | comment | added | Andy Putman | @DenisT: I’m not sure exactly what you’re getting at. Are you claiming that it needs Lebesgue numbers? The proof I had in mind (which is the standard proof) can be found in these notes: staff.ustc.edu.cn/~wangzuoq/Courses/21F-Manifolds/Notes/…. Can you point to where in that proof the issue lies? | |
| May 5 at 6:49 | comment | added | Denis T | Existence of mollifiers obviously depends on finiteness of covering dimension. | |
| May 4 at 22:04 | comment | added | Z. M | @NoahSchweber It is called the Whitney approximation theorem. | |
| May 4 at 19:17 | history | edited | Andy Putman | CC BY-SA 4.0 |
Actually, simplicial approximation does use the Lebesgue Number Lemma, so I removed that approach.
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| May 4 at 19:10 | history | edited | Andy Putman | CC BY-SA 4.0 |
added 351 characters in body
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| May 4 at 19:04 | comment | added | Noah Schweber | That second paragraph is indeed a nice fact worth knowing on its own! | |
| May 4 at 18:56 | history | answered | Andy Putman | CC BY-SA 4.0 |