N-sphere is simply connected for n greater than 1

Statement

Suppose $n$ is a natural number greater than 1. Then, the $n$-sphere $S^{n}$ is a Simply connected space (?). In other words, it is a Path-connected space (?) and its Fundamental group (?) is the trivial group.

Related facts

Stronger facts

• n-sphere is (n-1)-connected

Facts used

1. Union of two simply connected open subsets with path-connected intersection is simply connected, which is a corollary of the Seifert-van Kampen theorem.
2. Suspension of path-connected space is simply connected (this actually follows from the previous fact, but the explanation it offers is in some ways more direct and intuitive).

Proof

Proof using fact (1)

We try to show that the $n$-sphere $S^{n}$ satisfies the conditions for fact (1). Denote by $p$ and $q$ any two antipodal (diametrically opposite) points of $S^{n}$. Then, define $U=S^{n}\setminus \{p\}$ and $V$ as $S^{n}\setminus \{q\}$. We obtain $U\cap V=S^{n}\setminus \{p,q\}$. We find that:

No.	Assertion	Uses $n>1$?	Justification
1	Both $U$ and $V$ are open	No	Since $S^{n}$ is a manifold, points are closed, so their complements are open subsets.
2	Both $U$ and $V$ are simply connected	No	In fact, via the stereographic projection relative to the point $p$, $U$ is homeomorphic to $\mathbb {R} ^{n}$, which is contractible and hence simply connected. Similarly, via the stereographic projection relative to the point $q$, $V$ is homeomorphic to $\mathbb {R} ^{n}$, which is contractible, and hence simply connected.
3	The intersection $U\cap V$ is nonempty and path-connected	Yes	In fact, it is homeomorphic to $S^{n-1}\times (0,1)$. Since $n>1$, $n-1\geq 1$, so $S^{n-1}$ is path-connected, hence $S^{n-1}\times (0,1)$ is path-connected.

Thus, we see that the conditions to apply fact (1) are met, and we obtain that $S^{n}$ has trivial fundamental group.

Proof using fact (2)

For this, we note that $S^{n}$ is the suspension of $S^{n-1}$. For $n>1$, $n-1\geq 1$, so $S^{n-1}$ is a path-connected space. Hence, by fact (2), $S^{n}$ is simply connected.

Note that this proof is actually the same as the proof using fact (1), but is stated in a different conceptual language that perhaps sheds more light.