fundamental group of $X=\{(x_0,x_1,\dots, x_n)\in S^{n}| x_{0}^2 +x_{1}^2 <1 \}$ for $n\geq 3$

Question

I have trouble finding this fundamental group. I know that $X$ can also be seen as $S^n$ minus the circumference $x_0 ^2+ x_{1}^2=1$, and I guess that the fundamental group is trivial, because $S^n$ is a variety of dimension 3 while I am subtracting something of dimesion 1, but I don't see ho to prove it. (I am not sure about the variety thing, we have yet to study them properly, it is more an intuition)

Answer 1

First of all, note that the complement of a dimension-1 thingy inside a dimension-3 thingy can be quite complicated (search for "knot complement"). But for this specific problem, there's a lovely proof that generalizes. For $0\le k<n$, define $$ X_{n,k} = \big\{ (x_0,\dots,x_n) \in S^n\colon x_0^2+\cdots+x_k^2 < 1 \big\} $$ (so that you are asking about $X_{n,1}$). Then we can show that the following spaces are all homeomorphic: $$ X_{n,k} \times \Bbb R,\quad \Bbb R^{n+1} \setminus \Bbb R^{k+1} ,\quad (\Bbb R^{n-k} \setminus \{0\}) \times \Bbb R^{k+1},\quad S^{n-k-1}\times \Bbb R^{k+2}. $$ (By $\Bbb R^{n+1} \setminus \Bbb R^{k+1}$ we refer to the obvious inclusion of these Euclidean spaces: $\{ (x_0,\dots,x_n) \} \setminus \{ x_0,\dots,x_k,0,\dots,0) \}$.) In particular, since $\pi_1(A\times B) \cong \pi_1(A)\times \pi_1(B)$ and $\pi_1(\Bbb R)$ is trivial, the fact that $X_{n,k} \times \Bbb R$ is homeomorphic to $S^{n-k-1}\times \Bbb R^{k+2}$ shows that $\pi_1(X_{n,k} \cong \pi_1(S^{n-k-1})$. In particular, $\pi_1(X_{3,1}) \cong \Bbb Z$ while $\pi_1(X_{n,1}) \cong \{0\}$ for $n\ge4$.

The key here is that $S_n\times\Bbb R_{>0}$ is homeomorphic to $\Bbb R^{n+1}\setminus\{0\}$: this is the familiar map $(x_0,\dots,x_n) \times \lambda \mapsto \lambda(x_0,\dots,x_n)$ that splits the quotient map from $\Bbb R^{n+1}\setminus\{0\}$ onto $S_n$. And of course, $\Bbb R_{>0}$ is homeomorphic to $\Bbb R$ itself (via $\exp$ and $\log$, for example). This shows that the third and fourth spaces above are homeomorphic. But the same method (adding a "radial coordinate" $\lambda$) shows that $X_{n,k} \times \Bbb R_{>0}$ is homeomorphic to $\Bbb R^{n+1} \setminus \Bbb R^{k+1}$, which is the first pair of spaces above. And the middle two spaces are obviously homeomorphic by "factoring out" the common $k+1$ coordinates.

Another way of phrasing this argument is that the spaces $X_{n,k}$, $\Bbb R^{n+1} \setminus \Bbb R^{k+1}$, $\Bbb R^{n-k} \setminus \{0\}$, and $S^{n-k-1}$ are all homotopy equivalent.