What are the homotopy groups of $\mathbb{R}^2\setminus S^1$ = the plane without the circle?

Question

I’m reading “Essential Topology” by Martin Crossley. One of the questions in the exercises is to compute the homotopy groups of $\mathbb{R}^2 - S^1$. Intuitively the space seems homotopic to a circle with a point at its center. So my gut tells me that the homotopy groups are just $\pi_n(S^1)$.

Answer 1

As pointed out in the comments the space $\mathbb{R}^2\setminus S^1$ is not connected, so its homotopy groups will depend on your particular choice of basepoint.

Now $\mathbb{R}^2\setminus S^1$ has two path components. Namely the open unit disc $E^2$ and the points $X=\{x\in\mathbb{R}^2\mid \|x\|>1\}$. You should know (or prove) that if a space is path connected then any choice of a basepoint will yield isomorphic homotopy groups for it (although the isomorphisms are not canonical). Thus to consider the homotopy groups of $\mathbb{R}^2\setminus S^1$ we really need to consider (and it will suffice to do so) both groups

$$\pi_n(E^2;(0,0)),\qquad \pi_n(X;(2,0))$$

for $n\geq 0$.

Here you have that $E^2\simeq\ast$ is contractible by the convex homotopy

$$(x,t)\mapsto (1-t)x,\qquad x\in E^2,\,t\in I.$$

Hence

$$\pi_n(E^2;(0,0))\cong\{0\}$$

is the trivial group for all $n\geq1$ (we have already noted that it is path connected).

Finally you have already observed that $X$ deformation retracts onto the circle $S^1(2)$ of radius $2$ by the homotopy

$$(x,t)\mapsto (1-t)x+2t\frac{x}{\|x\|},\qquad x\in X,\,t\in I,$$

so in particular there is a homotopy equivalence $X\simeq S^1(2)$. Since $S^1(2)$ is homeomorphic to the standard circle $S^1$ and homotopy groups are invariant under homotopy equivalence we have that

$$\pi_n(X;(2,0))\cong\pi_n(S^1;(1,0))$$

for all $n\geq 0$.