Why are the fundamental groups $\pi_{1}(S^3)$ and $\pi_{1}(S^2)$ trivial?

Why are the fundamental groups $\pi_{1}(S^3)$ and $\pi_{1}(S^2)$ trivial? Is there a straightforward proof of this?

Solution 1:

There is a straightforward proof that for all $n\geq 2$, $\pi_1(S^n))$ is trivial. Let $n\geq 2$ be an integer, and let $c$ be a loop (based at $N=(1,0,\dots,0)$).

1. Show that $c$ is homotopic to a map $c'$ that misses a point in $S^n$, say $p\in S^n\setminus\lbrace N\rbrace$.
2. Since $S^n\setminus\lbrace p\rbrace$ is homeomorphic to $\Bbb R^n$, which is contractible, and $c'$ is, by construction, entirely contained in $S^n\setminus\lbrace p\rbrace$ , you can homotope it to the point $N$ inside $S^n\setminus\lbrace p\rbrace$.

To show the first claim, use the open cover of the sphere by the two open sets $U=S^n\setminus\lbrace N\rbrace$ and $V=S^n\setminus\lbrace -N\rbrace$. By continuity of $c$, and and the $\epsilon$ lemma of Lebesgue applied to the open cover $\big(c^{-1}(U),c^{-1}(V)\big)$ of $[0,1]$, there exists an integer $m$ such that $$\text{for all }k\in\lbrace 0,\dots,m-1\rbrace,\;c\left(\left[\frac{k}{n},\frac{k+1}{n}\right]\right)\subset U\text{ or }V$$ Now you can easily deform $c$ relative to endpoints on each interval $\left[\frac{k}{n},\frac{k+1}{n}\right]$ to a map that is "a straight line connecting $c\left(\frac{k}{n}\right)$ to $c\left(\frac{k+1}{n}\right)$" inside $U$ or $V$, whichever contains $c\left(\left[\frac{k}{n},\frac{k+1}{n}\right]\right)$ (they may both contain it). This will be $c'$, and it clearly misses most of the sphere.

Another proof goes as follows. You probably haven't studied it yet, but one theorem of Whitehead states that any map between (pointed) CW complexes is homotopic to a cellular map. Since the circle $S^1$ is a one dimensional CW complex, and the spheres of dimension $n\geq 2$ have a CW decomposition with only one $0$ cell and one $n$ cell, the theorem implies that a loop $c:S^1\to S^n$ is homotopic (relative to basepoints) to a map $c:S^1\to \mathrm{sk}_1(S^n)=\mathrm{pt}$, where $\mathrm{sk}_1(S^n)$ (the $1$ skeleton of $S^n$ with the above cellular decomposition) is a point, i.e., $c$ is homotopic to the constant path.

The only good thing about this last argument is that it generalizes to show that the homotopy groups $\pi_k(S^n)$ are trivial for $1\leq k<n$.