Prove that $S^n$ is $(n-1)$-connected. I heard someone mention that since $S^n$ is $(n-1)$-connected (i.e. $\pi_k (S^n) = 0$ for $k<n$), $\pi_2 (S^3)=0$. However, I can't seem to imagine how this is the case. The person who said it made it seem trivial, so I've come to the conclusion that this is probably a stupid question, but are there any easy/intuitive ways to understand this?
 A: Continuous maps are homotopic to smooth maps. Smooth maps $S^k \to S^n$, for $k<n$, are for dimensional reasons not surjective due to Sard's theorem. By removing a point from $S^n$ you get $\mathbb{R}^n$, and thus such maps are homotopic to a constant.
A: It suffices to show that any map $S^n\to S^{n+k}$ is nullhomotopic for any $k\geq 1$. Give $S^n$ and $S^{n-k}$ the a CW structure consisting of only one $0$-cell and one $n$-cell (or $n-k$ cell respectfully). Use cellular approximation to obtain a cellular map $S^n\to S^{n+k}$. By definition, for any CW complex $X$, we have $X^{k-1}\subset X^k$, so by definition of cellular maps the image $S^n$ must be contained in the $n$-skeleton of $S^{n+k}$. But this is just a point (namely the $0$-skeleton of $S^{n+k}$).
The proof can also be cast in terms of Whitney's approximation and Sard's theorem (as the other answers indicated).
A: The only way I am aware of for proving this is the Freudenthal suspension theorem, which is not very hard.
A: The argument goes that when $0\lt i\lt n $, any continuous map from $S^i $ to $S^n $ is homotopy equivalent to a non-surjective map.  Then since $S^n\setminus\{\text {pt}\}\cong\mathbb R^n $ is contractible,  the map is homtopically trivial.
To make the first part rigorous,  see the cellular approximation theorem.  This is somewhat difficult because of the existence of space-filling curves.  When $i=1$ and $n=2$, there's the intuition that a rubber band wrapped around a sphere can be deformed to a point.
