Timeline for Can the fundamental group of any manifold be realized as the fund grp of a finite space?
Current License: CC BY-SA 2.5
11 events
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| Nov 10, 2010 at 17:17 | history | edited | Abhishek Parab | CC BY-SA 2.5 |
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| Nov 10, 2010 at 17:15 | comment | added | Abhishek Parab | Yes, got it! Sorry.. | |
| Nov 10, 2010 at 17:06 | vote | accept | Abhishek Parab | ||
| Nov 10, 2010 at 17:04 | comment | added | Ryan Reich | Well, the fundamental group of a 1-skeleton is free. The 2-skeleton provides the relations, as you have demonstrated. | |
| Nov 10, 2010 at 16:59 | comment | added | Abhishek Parab | Sorry I had to be away. Ryan is right, the points on S^1 with positive (resp negative) Y-coordinate are all identified. The space is like, {1, -1, N, S}. @Ryan, I believe it is the 1-skeleton. A loop inside a disk (disk in R^2) can be homotoped to a loop on the boundary S^1. Am I committing some obvious blunder here? | |
| Nov 10, 2010 at 15:58 | comment | added | Ryan Reich | It looks like he means that a is the northern hemisphere (collapsed to a single point), c is the southern hemisphere (likewise collapsed), and b and d are 1 and -1. Then the four sets he mentioned are the two open hemispheres and the complements of the two other points, which are indeed (with $a \cup b$ and $\emptyset$) the saturated open sets for the quotient of $S^1$ collapsing each hemisphere to different points. | |
| Nov 10, 2010 at 15:05 | comment | added | Jim Conant | I never paused to consider whether $\pi_1(X)$ could be nonzero for non-Hausdorff finite spaces. Is there a simple way of describing your space $X$ as a quotient of $S^1$? You mentioned identifying the two hemispheres in some way!? | |
| Nov 10, 2010 at 13:22 | comment | added | Ryan Reich | You mean the 2-skeleton, right? | |
| Nov 10, 2010 at 13:14 | history | edited | Abhishek Parab | CC BY-SA 2.5 |
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| Nov 10, 2010 at 13:01 | answer | added | Dan Petersen | timeline score: 68 | |
| Nov 10, 2010 at 12:52 | history | asked | Abhishek Parab | CC BY-SA 2.5 |