Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It seems that the fundamental group of any subset of $\mathbb{R^2}$ will not have an element of finite order. Though the 3-dimensional version is an open problem I couldn't immediately see why it is true in the 2-dimensional case. Please shed some light on this.

share|improve this question
2  
It's a corollary of a theorem of Eda's. See the reference in the question from this thread: mathoverflow.net/questions/4478/… –  Ryan Budney May 1 '11 at 21:15
    
Thanks! but it seems to have no reference, he just mentioned it right? –  Dinesh May 1 '11 at 22:14
    
In defining Fundamental group we assume it to be path connected..does "any subset" makes sense here? –  Dinesh May 1 '11 at 22:28
3  
Eda, K. Fundamental group of subsets of the plane. Topology and its Applications Volume 84, Issues 1-3, 24 April 1998, Pages 283-306 –  Ryan Budney May 1 '11 at 22:28
    
Regarding your path-connected concern, usually in a question like this one simply replaces "fundamental group of $X$" by "fundamental group of all components of $X$", or restricts to path-connected sets, etc. –  Ryan Budney May 1 '11 at 22:29

1 Answer 1

up vote 11 down vote accepted

I don't have a copy of the 1998 paper by Eda from which the result apparently follows. However, the question can be answered using results from the paper The fundamental groups of subsets of closed surfaces inject into their first shape groups by Fischer & Zastrow, which is freely available on the arXiv. They show that the fundamental group of a compact proper subset of a closed surface is locally free (among other things). A group is called locally free if every finitely generated subgroup is free, which implies that it is torsion-free (the subgroup generated by an element $g\not=1$ is free, hence infinite, so $g$ does not have finite order). From this, we can show the following.

1) If $X$ is a proper subset of a closed surface and $x\in X$, then $\pi_1(X,x)$ is torsion-free.

In particular, $\mathbb{R}^2$ is homeomorphic to the sphere $S^2$ with a point removed, so every $X\subseteq\mathbb{R}^2$ is homemorphic to a proper subset of $S^2$ and $\pi_1(X,x)$ is torsion-free for $x\in X$.

To show that (1) does indeed follow from the stated result, consider a closed curve $\gamma\colon[0,1]\to X$ with $\gamma(0)=\gamma(1)=x$ such that $[\gamma]\in\pi_1(X,x)$ has finite order, so that $[\gamma]^{n}=1$ for some $n\ge1$. It needs to be shown that $[\gamma]=1$. First, writing $\gamma^{n}=\gamma\ast\cdots\ast\gamma$ for the n-fold product of $\gamma$, we have that $\gamma^{n}$ is null-homotopic in $X$. Letting $H\colon[0,1]^2\to X$ be a null-homotopy for $\gamma^{(n)}$ then $Y\equiv{\rm Im}(H)\subseteq X$ is compact and $\gamma^n$ in null-homotopic in $Y$. By the result from the paper, $\pi_1(Y,x)$ is torsion-free, so $\gamma$ is null-homotopic in $Y$ and, hence, in $X$.


Update: Reading through the paper, it is apparent that the following is true:

If $x\in X\subseteq\mathbb{R}^2$ and $\gamma\colon[0,1]\to X$ is a curve with $\gamma(0)=\gamma(1)=x$ which is not null-homotopic in $X$, then there is a finite $F\subseteq\mathbb{R}^2\setminus X$ such that $\gamma$ is not null-homotopic in $\mathbb{R}^2\setminus F$.

This implies the result asked for, since if $\gamma$ is not null-homotopic in $X$, then it is not null-homotopic in $\mathbb{R}^2\setminus F$ for some finite set $F\subseteq\mathbb{R}^2\setminus X$. However, the fundamental group of $\mathbb{R}^2\setminus F$ is freely generated by $\vert F\vert$ elements, so is torsion free. Therefore, $\gamma^n$ is not null-homotopic in $X\subseteq\mathbb{R}^2\setminus F$.

To prove the quoted result, let $X^\prime\subseteq X$ be the union of $\gamma([0,1])$ and the bounded connected components of $\mathbb{R}^2\setminus\gamma([0,1])$ which are subsets of $X$. Also let $U_1,U_2,\ldots$ be the bounded connected components of $\mathbb{R}^2\setminus\gamma([0,1])$ which do not lie entirely in $X$. If we choose $r > 0$ such that $\gamma([0,1])\subseteq B_r(0)$ and let $B_{r_i}(z_i)\subseteq U_i$ be open balls about points $z_i\not\in X$ then the Riemann mapping theorem can be used to show that $X^{\prime\prime}=\bar B_r(0)\setminus\bigcup_iB_{r_i}(z_i)$ retracts onto $X^\prime$. So, $\gamma$ is not null-homotopic in $X^{\prime\prime}$.

It then needs to be shown that $\gamma$ is not null-homotopic in $X_n=\bar B_r(0)\setminus\bigcup_{i=1}^nB_{r_i}(z_i)$ for a finite n. This is the difficult part, and is established in the appendix of the linked paper by using the fact that sets such as $X^{\prime\prime}$ are homeomorphic to something like a Sierpinski carpet (at least, the Sierpinski carpet is the most complex that it can get). A lengthy argument is used, counting crossings of horizontal and vertical strips in the square to show that $\gamma$ cannot be null-homotopic in $X_n$ for every $n$. Then, setting $F=\{z_1,z_2,\ldots,z_n\}$, $\mathbb{R}^2\setminus F$ retracts onto $X_n$, so $\gamma$ is not null-homotopic in $\mathbb{R}^2\setminus F$.


I'll also give a quick explanation of where the result stated in the paper comes from, although I haven't worked through all of the details (yet).

The idea is to embed the fundamental group $\pi_1(X,x)$ into another group $\check{\pi}_1(X,x)$ called the first shape homotopy group, which is calculated as an inverse limit of fundamental groups of simplicial complexes. If $\mathcal{U}$ is a finite open cover of $X$ (or, just locally finite if $X$ is not compact) then you can define the abstract simplicial complex $N(\mathcal{U})$ consisting of subsets $\Delta\subseteq\mathcal{U}$ for which $\bigcap\Delta\not=\emptyset$. Then you can define natural homomorphisms of fundamental groups $\pi_1(X,x)\to\pi_1(N(\mathcal{U}),\{U\})$. Taking the inverse limit over refinements of the partition gives the first shape group and a natural homomorphism $\pi_1(X,x)\to\check{\pi}_1(X,x)$. In general, this does not have to be either injective or surjective. However, for subsets of a closed surface then it is injective, as stated in the title of the paper. It can fail to be injective in three dimensions, and an example of this failure is provided in the paper.

In the case where $X$ is compact, then it can be written as the intersection of a sequence of piecewise-linear submanifolds of the surface with nonempty boundary, which each have finitely generated and free fundamental groups. The first shape group can then be written as the inverse limit of a sequence of finitely generated and free groups. You can show that subgroups of inverse limits of free groups are locally free.

share|improve this answer
    
Thanks George! Though I could not completely grasp the solution I understood the sketch. I am just wondering how did you see that the paper you cited is relevant for this question. Have you read the paper solely for the purpose of answering this? –  Dinesh May 9 '11 at 6:45
    
@Dinesh: Yes, I just tried searching with google for the paper by Eda and came across this one instead. It's clear from the abstract that it is relevant. Also, the result that you only need to use a finite number of points in the complement of the set was something that came to mind when I first looked at your question, but it was only looking through this paper that I found out that this approach works. I skimmed through the paper rather than reading it in detail. The main body doesn't contain anything particularly new, and the real work is in the appendix. –  George Lowther May 9 '11 at 17:21

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.