Homology and Fundamental group of $\mathbb{R}^4\setminus S^1$

Question

I was attempting to help someone with this problem and realized I could not solve it myself.

Let $S^1=\{(x,y,0,0):x^2+y^2=1\}$ be the unit circle in $\mathbb{R}^4$ and consider $M=\mathbb{R}^4 \setminus S^1 $. Compute $\pi_1(M)$ and $H_*(M)$.

It should have trivial fundamental group and therefore, trivial first homology group, which we can prove by showing that any loop which "passes through" the circle can slide past it and become trivial. Also, it is path connected so $H_0(M)\cong \mathbb{Z}$. But I do not have any good idea of how to deformation retract this space to something simpler we can work with and find the homology groups for $n\geq 2$. Any help would be appreciated.

Answer 1

One way is: $S^{n-1}\backslash S^{k-1}$ is homotopy equivalent to $S^{n-k-1}$, which is quite easy to see by $S^{n-1}\subset\mathbb R^n$, $S^{k-1}=S^{n-1}\cap\mathbb R^{k}$, $S^{n-k-1}=S^{n-1}\cap\mathbb (\mathbb R^{k})^\perp$. Similarly we get also that $S^{n-1}\backslash(\{*\}\cup S^{k-1})$ ($*$ a point not on $S^{k-1}$) is homotopy equivalent to $S^{n-k-1}\vee S^{n-2}$, the additional $S^{n-2}$ is around $*$. In your case ($\mathbb R^4=S^4\backslash\{*\}$) we have $n=5$ and $k=2$, i.e. the result is $S^2\vee S^3$.