Fundamental group of $S^3$ with finitely many points removed? I had a question about the fundamental group if the $3$-sphere with finitely many points removed.
Since $\pi_1(\mathbb{S}^{n-1}) \cong \pi_1({\mathbb{R}}^n\setminus\{0\})$, I thought $\pi_1(\mathbb{S}^{3}\setminus\{p_1,..., p_{n}\}) \cong \pi_1({\mathbb{R}}^4\setminus\{0, x_1,..., x_{n-1}\})$.
And since ${\mathbb{R}}^4\setminus\{0, x_1,..., x_{n-1}\}$ is simply connected $\pi_1({\mathbb{R}}^4\setminus\{0, x_1,..., x_{n-1}\}) = \{e\}$.
Therefore $\pi_1(\mathbb{S}^{n-1})= \{e\}$.
Is this argument valid? 
Thanks in advance for any help :).
 A: Your final answer is only coincidentally correct. The reasoning is wrong.
The major problem with your reasoning is that you take the homotopy equivalence $S^{n-1}\simeq\mathbb{R}^n\backslash\{0\}$. But $X\simeq Y$ doesn't imply that $X\backslash\{x_0\}\simeq Y\backslash\{y_0\}$ for some $x_0\in X$ and $y_0\in Y$. For example $X=\mathbb{R}^2$ is homotopy equivalent to $Y=\mathbb{R}$ but there is no pair of points $x_0\in X,y_0\in Y$ making $X\backslash\{x_0\}$ homotopy equivalent to $Y\backslash\{y_0\}$. In fact $X\backslash A$ is not homotopy equivalent to $Y\backslash B$ for any two nonempty finite subsets, even of different size (simply because the former is always connected while the latter is always disconnected). Even more: their fundamental groups (regardless of the choice of base points) are always different.
So you need some other topological invariant that on one hand induces the same fundamental group and on the other hand is preserved when removing points. And one such invariant is obviously homeomorphism. Therefore lets slightly modify your idea: use the stereographic projection to show that $S^{n-1}\backslash\{x_0\}\cong\mathbb{R}^{n-1}$ (note the different "$\cong$" sign, i.e. homeomorphic). And so instead of gaining points you lose them, and you get the correct formula
$$\pi_m(S^n\backslash\{x_0, x_1,\ldots, x_k\})\cong\pi_m(\mathbb{R}^n\backslash\{y_1,\ldots, y_k\})$$
(here "$\cong$" means "group isomorphism"). The relationship between $x$ and $y$ points is that $\{y_i\}_{i=1}^k$ is the image of $\{x_i\}_{i=1}^k$ under the stereographic projection based at $x_0$ (side note: with this observation it can be easily generalized to any difference $S^n\backslash A$, even when $A$ is infinite). But since $\mathbb{R}^n\backslash A$ is homeomorphic to $\mathbb{R}^n\backslash B$ for any two finite subsets $A,B$ of the same size (which I leave as an interesting exercise) then we can choose $\{y_i\}_{i=1}^k$ points arbitrarly.
Finally for $m=1$ and $n\geq 3$ the formula yields the trivial group. By Seifert-van Kampen for example, or by noticing that $\mathbb{R}^n\backslash\{y_1,\ldots, y_k\}$ is homotopy equivalent to the wedge sum of spheres of dimension $n-1$.
