Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 2 at 9:30 | comment | added | Sam Nead | Surely this proof is circular… | |
| Jun 6, 2011 at 19:00 | comment | added | Steven Gubkin | We really only need that the map from H_1 to \pi_1 is the abelianization, which doesn't seem to use \pi^1(S^1) - see hatcher section 2.A. | |
| Jun 5, 2011 at 17:19 | comment | added | Jeff Strom | I don't think you can prove Hurewicz without knowing $\pi_1(S^1)$. | |
| Dec 8, 2010 at 18:27 | comment | added | Johannes Ebert | There is a proof of $\pi_1 (C^{\times})=Z$, using elementary complex analysis, which is so elementary and simple that even the standard covering theory proof looks like a nuke to me. | |
| Oct 23, 2010 at 13:57 | comment | added | Dan Ramras | On the other hand, it's perfectly natural to use the Eckmann-Hilton argument in order to show that the degree map is a homomorphism. I don't know a reference that uses E-H here, but that's how I did it in class this semester. | |
| Oct 23, 2010 at 13:52 | comment | added | Dan Ramras | I'm surprised people don't like this. It's not the most extreme example here, but imagine this scenario: you see a colleague in the hall and ask what he's teaching today. "I'm introducing the fundamental group." And you ask if he'll compute $\pi_1(S1)$. "Well, not today. I'll define the fundamental group, but before I can compute $\pi_1(S1)$ I'll have to set up singular cohomology (long exact sequences, excision, all that) and then once I've explained simplicial homology we can get back to $\pi_1(S1)$. It'll be a month or two." I bet you'd worry about your colleague's sanity. | |
| Oct 19, 2010 at 4:49 | comment | added | HJRW | Hmm, I'm torn. I'm very happy to have learned this proof, which I think is rather elegant. So +1 for that. But I don't think it's significantly harder work than the regular proof, so it certainly doens't seem like a nuke. So -1 for that. Let's leave it at +0. | |
| Oct 18, 2010 at 21:38 | comment | added | Arend Bayer | Somehow this proof doesn't feel like a pointless nuke to me, rather it does explain from some perspective what's going on. The usual proof, i.e. proving the path-lifting property for the covering $\mathbb{R} \to S^1$, is also a bit of work, and breaking up loops into paths on $\mathbb{R}$ feels unsatisfying. It also seems odd to never mention that $[s \mapsto e^{2 \pi n s}] * [s \mapsto e^{2 \pi m s}] = [s \mapsto e^{2 \pi (m+n) s}]$ has something to do with a group structure on $S^1$... | |
| Oct 18, 2010 at 19:40 | comment | added | Christian Blatter | "The fundamental group of the circle is $\mathbb Z$" In my opinion this is one of the most consequential theorems in all of mathematics. So I don't mind if people suggest proofs coming from very different sources. | |
| Oct 18, 2010 at 19:34 | comment | added | Steven Gubkin | Yes, but the fundamental group of the circle is so basic that it is usually the first thing computed in any algebraic topology course or book. Isn't using Hurewicz and the equivalence of singular and simplicial homology a nuke for this problem? If not, I do not see why any of the other answers work | |
| Oct 18, 2010 at 17:51 | comment | added | Mariano Suárez-Álvarez | But this is not a nuke in the sense that a lot of homology theory is designed to be able to carry out precisely this kind of things. | |
| Oct 18, 2010 at 17:11 | comment | added | Steven Gubkin | I don't see what you mean. The homology will not generally give you complete information about the fundamental group. The power here is that homology is easy to compute and in some instances you can get information about the homotopy groups through homology using Hurewicz. | |
| Oct 18, 2010 at 16:27 | comment | added | darij grinberg | This is not a nuke. For $H_1$, the whole homology theory could be an exercise for the reader. | |
| Oct 18, 2010 at 15:24 | history | edited | Darsh Ranjan | CC BY-SA 2.5 |
corrected spelling
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| Oct 18, 2010 at 15:17 | history | answered | Steven Gubkin | CC BY-SA 2.5 |