Show that there is no homeomorphism $f:\mathbb{R}^2\setminus\mathbb{S}^1\rightarrow \mathbb{R}^2\setminus\mathbb{S}^1$ such that $f(0, 0) = (2, 0)$. I proceed as follows (everything has its usual topology):
Suppose there is such a homeomorphism $f$. Then the restriction of $f$ to $\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$ is a homeomorphism whose image is a connected component of $\mathbb{R}^2\setminus\mathbb{S}^1$. As $f(0,0)=(2,0)$, its image must be $\{(x,y)\in\mathbb{R}^2:x^2+y^2>1\}$. However, this is not possible, as those spaces have different fundamental group:
- The unit disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$ is convex, and therefore it has trivial fundamental group.
- The annulus $\{(x,y)\in\mathbb{R}^2:x^2+y^2>1\}$ has as a deformation retract the sphere centered at the origin of radius $2$, which has fundamental group $\mathbb{Z}$, and so does the annulus.
Is everything correct?