The number of retracts of $\mathbb{R}^2 \setminus \{(0,0)\}$ up to homeomorphic We know that $\mathbb{S}^1$ is a (deformation) retract of $\mathbb{R}^2 - \{ (0,0)\}$. Obviously, the number of retracts of $\mathbb{R}^2\setminus \{ (0,0)\}$ equals to 1 up to homotopy equivalence. My question is that:  
How many retracts are there up to homeomorphic for  $\mathbb{R}^2 \setminus \{ (0,0)\}$?
 A: This is more a comment than a answer. Clearly there are no more that $2^{\Bbb R}$ retracts. Here is a way to build $2^{\Bbb N}$ different retracts. I don't know if there are $2^{\Bbb R}$ retracts or not.
Consider $$R = S^1 \cup \{(0,y) : y \geq 1\} \cup \{(x,y) : x \in (-1,1) , y \in \Bbb N_{>1} \}$$
If you make a picture it is clear that $\Bbb R^2 \backslash \{(0,0)\}$ retracts on $R$. Now, $R$ is concretely a circle with a line and an infinity of crossing. If $(x_1,x_2, \dots, x_n, \dots ) \in 2^{\Bbb N}$, we can modify $R$ by thickening the $n$-th segment if $x_n = 1$ and do nothing else. We call this space $R_x$, concretly $R \cap \{ n < y < n+1 \} = \{(x,y) : x \in [-1/2,1/2], n < y < n+1\}$ if $x_n=1$ and $R_x \cap \{ n < y < n+1 \} = R \cap \{n<y<n+1\}$ else. In particular $R_0 = R$. Again a picture shows that $\Bbb R^2 \backslash \{(0,0)\}$ retracts on $R_x$. 
Now, clearly if $f : R_{x} \to R_{x'}$ is an homeomorphism, $f(0,n) = (0,n)$ for all $n \geq 1$. In particular, $f$ restricts to an homeomorphism between every "strip" of $R_x$ and $R_{x'}$. Since $[0,1]$ is not homeomorphic to $[0,1] \times [-1/2,1/2]$ it follows that any sequence $x = \{x_1, \dots, x_n, \dots \}$ defines a retract of $\Bbb R^2 \backslash \{0\}$, with all a different topological type. 
A: There are infinitely many. For instance, given any natural number $n>2$, take $n$ unit discs and glue them together (so that each disc touches two other discs, each at a single point) into a ring around the origin. None of these rings are homeomorphic.
