Is there a subset of $\mathbb R^2$ that has fundamental group $\mathbb Z^2$? I know the fundamental group of the torus is $\mathbb Z^2$. I am wondering is it possible to find a subset of the plane such that the fundamental group of this subspace is $\mathbb Z^2$? I can't seem to come up with an example.
 A: The following is likely to be in the literature, but it is easier to prove than to find a reference. 
Proposition.  If $A\subset R^2$ is a nonempty subset then for every $a\in A$ the fundamental group $\pi_1(A,a)$ is locally free. 
(Recall that a group $G$ is called locally free if every finitely generated subgroup of $G$ is free. For instance, the additive group of rational numbers is locally free, but not free.) 
I first give an intuitive explanation and then a formal proof. Our intuition comes from connected polygonal subsets $A\subset R^2$, i.e. finite planar simplicial subcomplexes.  Every such subcomplex is homotopy-equivalent to a finite graph: just keep collapsing 2-dimensional simplices $s$ in $A$ which have one "free edge" i.e. an edge $e$ such that $e$ is contained in a unique 2-dimensional simplex in $A$, namely, $s$. After finitely many such collapses we obtain a finite connected subgraph $B\subset A$. It is a standard exercise in algebraic topology (using van Kampen's theorem) to show that the fundamental group of a finite connected graph is free. The trouble with this intuition is that there are (even compact) subsets in $R^2$ which are quite different from simplicial complexes, e.g. the topologists's sine curve or the Lakes of Wada continuum, which tend to defy our intuition of planar sets. Below is a formal proof of the proposition. 
Proof. First, I need a bit of group theory. 
Definition. The rank, $rank(G)$, of a (finitely generated) group $G$ is the minimal number of generators needed to generate $G$. (Note that $G$ need not be free!) 
It is clear that if $Q$ is a quotient group of $G$ then $rank(Q)\le rank(G)$. What's less obvious is the following 
Lemma. 1 $rank(F_n)=n$ for every $n$. 
(This lemma follows by considering the abelianization of $F_n$, which is ${\mathbb Z}^n$; alternatively, it follows from Lemma 2 below.)  
Lemma 2. If $\phi: F_n\to F_m$ is a non-injective epimorphism of free groups then $rank(F_m)< rank(F_n)$. 
Lemma 2 follows from the Hopfian property of free groups of finite rank. 
Given this, let's prove the proposition. The proof is by induction on rank of finitely generated subgroups of $\pi_1(A,a)$. There is nothing to prove for subgroups of rank $0$ (they are trivial), so assume that the claim holds for all subgroups of rank $<n$. 
The key fact that we need is the following theorem due to 
Fischer and Zastrow mentioned by George  Lowther in his answer to this question:
Theorem. If $K\subset R^2$ is a compact subset, then the fundamental group $\pi_1(K,x)$ is locally free for every $x\in K$.  
Now, suppose that $G$ is a rank $n$ subgroup of $\pi_1(A,a)$. I claim that $G$ is free. Let $g_1,...,g_n$ be generators of $G$, they are represented by loops $L_1,...,L_n$ in $A$ based at $a$. If the elements $g_1,...,g_n$ do not satisfy any nontrivial relation in $\pi_1(A,a)$, then $G$ is free and we are done. Suppose, hence,  that there exists $R$, a reduced word in the alphabet $g_i^{\pm 1}$ such that $R$ represents the identity element in $\pi_1(A,a)$. The word $R$ corresponds to a continuous map $f: D^2\to A$ (such that the restriction of $f$ to the boundary of $D$ is a certain concatenation of the loops $L_i$ and their inverses which is given by the word $R$). The union of the images of the loops $L_i$ and the map $f$ is a compact path-connected subset $C\subset A\subset R^2$. Let 
$h_1,...,h_n\in \pi_1(C,a)$ denote the elements represented by the loops $L_1,...,L_n$. 
By the above theorem, $\pi_1(C,a)$ is locally free, hence, the subgroup $H$ of $\pi_1(C,a)$ generated by  $h_1,...,h_n$, is free. We then obtain a non-injective epimorphism
$$
F_n\to H
$$
sending the free generators to $h_1,...,h_n$. By Lemma 2 above, $rank(H)< n$. Hence, by the induction assumption, the image of $H$ in $\pi_1(A,a)$ 
(under the homomorphism induced by the inclusion map $(C,a)\to (A,a)$), which is our subgroup $G$, is free. qed
Now, this proposition immediately implies that for every planar subset $A\subset R^2$, $\pi_1(A,a)$ cannot contain ${\mathbb Z}^2$, since ${\mathbb Z}^2$ is finitely-generated (actually, 2-generated) and is not free. 
Edit. An example of a path-connected (even compact and locally path-connected)) planar subset with locally free but not free fundamental group is given by Hawaiian Earring. I do not know, however, if, ${\mathbb Q}$ is isomorphic to the fundamental group of a planar set.  
