non-trivial finite fundamental group I am a physics student and I want to know about finite fundamental group of subspace of $\mathbb{R}^3 $. Is there a subspace of $\mathbb{R}^3$ with nontrivial finite fundamental group?
I am just familiar with $\pi_1(RP^n)$ for n>1 and $\pi_1 (SO(n))  $ for n>2,  which have nontrivial finite fundamental group and $RP^2$ can not be embedded into $\mathbb{R}^3 $.
Thanks.
 A: The answer to the stated question is "we do not know if there are subsets of $R^3$ with nontrivial finite fundamental groups." It is also unknown if there are subsets of $R^3$ whose fundamental groups contain nontrivial elements of finite order. The latter was discussed in this Mathoverflow question. Below, a group $G$ is called torsion-free if it contains no nontrivial elements of finite order.
My personal take is that the standard invariants of algebraic topology such as fundamental groups, homotopy groups, singular homology groups are designed to work with "nice" spaces and should not be used for arbitrary topological spaces (instead one should use "Chech versions" of the standard invariants). What constitutes "nice" is a bit vague but includes manifolds, simplicial complexes, more generally, cell-complexes.  All these spaces share the property that they are locally contractible. Examples of topological spaces (even among compact subsets of $R^n$) which are not nice are the Cantor set,  Warsaw circle,
Hawaiian earring (and many more). The list of "nice spaces" might include something more general, spaces which are locally path-connected and semi-locally simply connected. The latter combination of conditions appears frequently in the theory of covering spaces (if you read any introductory algebraic topology book you will almost surely encounter these notions). Informally speaking, these two conditions for a topological space $X$ mean:
(a) Any two points in $X$ which are sufficiently close can be connected by a path in $X$ close to these points.
(b) Every sufficiently small loop in $X$ is null-homotopic in $X$.
The examples of "not-nice" spaces I gave above violate one of these two properties.
Here is what I can prove:
Theorem. Suppose that $X\subset R^3$ is a closed subset which is both locally path-connected and semilocally simply connected. Then $G=\pi_1(X,x)$ is torsion-free.
Sketch of the proof. Let $c: S^1\to X$ be a loop in $X$ (based at $x\in X$) representing an element of finite order in $G$. Consider a system of open neighborhoods $U_i$ of $X$ in $R^3$, i.e. open subsets of $R^3$ containing $X$, whose intersection equals $X$:
$$
\bigcap_{i\ge 1} U_i=X. 
$$
As I explained in my answer here, fundamental groups of open subsets of $R^3$ are torsion-free. Hence, for each $i$, the group $\pi_1(U_i, x)$ is torsion-free. Since $c$ represents a finite order element of $\pi_1(X,x)$, it also represents a finite order element of $\pi_1(U_i, x)$. Hence, this element of $\pi_1(U_i, x)$ is trivial.
Therefore, for each $i$ there exists a (continuous) extension $f_i: D^2\to U_i$ of $c: S^1\to X$, where $D^2$ is the unit disk with the boundary $S^1$. With a bit more thought, one sees that the images $f_i(D^2)$ can be chosen to lie in a fixed compact subset $K\subset R^3$ (independent of $i$), say, any closed round ball containing $c(S^1)$ would work. Now, using the assumption that $X$ is both locally path-connected and semilocally simply connected, for all sufficiently large $i$, one can "push" $f_i$ into $X$, i.e. find a continuous map $g_i: D^2\to X$ "close" to $f_i$ such that the restriction of $g_i$ to $S^1$ equals $c$. I omit the proof, it imitates the "standard" argument due to Borsuk that locally contractible finite-dimensional compact spaces are absolute neighborhood retracts. (Take a sufficiently fine triangulation of $D^2$. For vertices $v$ of the triangulation define $g_i(v)$ to be a point in $X$ closest to $f_i(v)$. Then use local path-connectedness to $X$ to extend $g_i$ to the edges, then use semilocal simple connectivity of $X$ to extend $g_i$ to the faces of the triangulation.) I do not need infinitely many $g_i$'s, just one, call it $g$. Thus, we found a continuous map $g: D^2\to X$ extending $c: S^1\to X$, hence, $c$ represents the trivial element of $G=\pi_1(X)$. In other words, $G$ is torsion-free. qed
