# 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.

## 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 $\R^n$, which is contractible and hence simply connected. Similarly, via the stereographic projection relative to the point $q$, $V$ is homeomorphic to $\R^n$, 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 $n > 1$, $n - 1 \ge 1$, 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 $n > 1$, $n - 1 \ge 1$, 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.