Calculating the fundamental group of $X \setminus \mathbb S^1$, for $n\ge 3$

Question

1. $X = \mathbb S^n$

I think I got this one very roughly. Since we had that $\mathbb S^n\setminus\{N\}\cong\mathbb R^n$ by stereographic projection, we can suppose that $N$ is sent to the point of infinity in $\mathbb R^n$ and say that $\mathbb S^n\setminus\mathbb S^1\cong \mathbb R^n\setminus\{$straight line$\}(\cong\mathbb R^n\setminus\{$axis$\})$. Since $\Pi_1(\mathbb R^n\setminus\{$axis$\})\cong \Pi_1(\mathbb S^{n-2})$ (all of this has been proved beforehand, I just don't want to make my proof any longer), we have that \begin{equation*} \Pi_1(\mathbb S^n\setminus\mathbb S^1) = \begin{cases} \mathbb Z & \quad \text{if $n=3$} \\ \{0\} & \quad \text{if $n>3$} \end{cases} \end{equation*} The different cases rub me the wrong way, so I don't know whether it is correct or not.

2. X = $\mathbb R^n$

In page 46 of Hatcher's Algebraic Topology, we see an example of an application of Seifert-Van Kampen Theorem where they get the fundamental group of $\mathbb R^3\setminus\mathbb S^1$, which is $\mathbb Z$. In this case I haven't finished the proof, since I have no idea how to move on from here. Maybe $\mathbb R^n\setminus\mathbb S^1$ is homotopic to $\mathbb R^n\setminus\{$straight line$\}$, and we get the same result as before? Or maybe we get that it deformation retracts to $\mathbb S^{n-1}\vee \mathbb S^1$ (or is homotopic, at the very least).

Could anyone, please, help me out?

Answer 1

Ok, I think I found the answer to the 2nd case.

If we reason similarly to the 1st case, we have that if we have that $\mathbb R^n\cong \mathbb S^n\setminus\{N\}$, and locate $N\in\mathbb S^1$, then we have that $\mathbb R^n\setminus \mathbb S^1 \cong \mathbb S^n\setminus\mathbb S^1 (*)$, hence $\Pi_1(\mathbb R^n\setminus \mathbb S^1) \cong \Pi_1(\mathbb S^n\setminus\mathbb S^1)$. We computed in 1. the latter fundamental group, so by the isomorphism, \begin{equation*} \Pi_1(\mathbb R^n\setminus\mathbb S^1) = \begin{cases} \mathbb Z & \quad \text{if $n=3$} \\ \{0\} & \quad \text{if $n>3$} \end{cases} \end{equation*}

I don't know if $(*)$ is very precise. Is this answer correct?

P.S.: Thanks to Tabes Bridges for confirming that 1. was correct.