Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recently, I was asked to calculate the fundamental group of the space $X= \{a,b,c,d\}$ with open sets generated by $\{ a, c, abc, acd \}$.

Turns out, $\pi_1(X)\cong \mathbb Z$ and in fact, $X$ is the quotient of $S^1$ (with the northern and southern hemispheres identified). But the result was not so easy to prove and this motivates the questions:

  • Is the fundamental group of every connected manifold (other restrictions / generalizations on the manifold are welcome) the fundamental group of a finite space? (Of course, it would not be Hausdorff). (I observe that there are many redundant points on a manifold where homotopy-equivalent loops need not traverse)

  • Is calculating $\pi_1$ of such finite spaces easier than for the given space? (If yes, this gives a method to calculate fundamental groups of many familiar spaces)

Perhaps the fact that -- $\pi_1$ of any CW complex just depends on its 1-skeleton [edit:2 skeleton]-- may be helpful.

share|improve this question
    
Note: The formulas with braces must be surrounded with backticks. –  Guillaume Brunerie Nov 10 '10 at 12:59
    
Thanks, fixed the braces. –  Abhishek Parab Nov 10 '10 at 13:15
5  
You mean the 2-skeleton, right? –  Ryan Reich Nov 10 '10 at 13:22
1  
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!? –  Jim Conant Nov 10 '10 at 15:05
2  
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. –  Ryan Reich Nov 10 '10 at 15:58

1 Answer 1

up vote 53 down vote accepted

In fact, there is the following theorem: Every finite CW complex is weakly homotopy equivalent to a finite topological space, and vice versa.

For simplicial complexes, this theorem is realized by mapping a complex to its face poset, and using the correspondence between finite posets and finite topological spaces. In the other direction, one maps a poset to its order complex.

In general it is not easy to compute homotopy groups of a finite topological space. I know that there are some techniques in Jonathan Barmak's Ph.D. thesis.

share|improve this answer
10  
+1: This is such a striking result that I am almost upset that no one told me about it in the algebraic topology classes I took in my youth. –  Pete L. Clark Nov 10 '10 at 15:25
5  
You went to Chicago, right? It must have been before Peter May started to love this theorem; he taught a summer course on it in 2003. –  Ryan Reich Nov 10 '10 at 15:54
    
This result is fantastic! (This being my first course in Alg Topology, I didn't know weak homotopy equivalence definition, but looked up.) So now may I ask, how one calculates the fund group of a CW complex 'efficiently'? Unfortunately, Barmak's thesis seems to be in French. –  Abhishek Parab Nov 10 '10 at 17:05
3  
Jonathan Barmak's thesis is definitely not in French. :) The abstract and introduction are in Spanish, though. As for how to compute the fundamental group of a concretely given CW complex, I can only give two tips: (i) use van Kampen wherever possible; (ii) think of the 1-cells as generators and 2-cells as relations. –  Dan Petersen Nov 10 '10 at 17:16
3  
May has some notes on his webpage: math.uchicago.edu/~may/MISCMaster.html –  Dan Petersen Nov 11 '10 at 16:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.