$S^2$ and a point are not homotopy equivalent Is there a way to show that $S^2$ does not have the same homotopy type of a point without using homology groups?
 A: If $S^2$ were homotopy equivalent to a point, then every vector bundle over $S^2$ would be trivial.  But, by the hairy ball theorem, the tangent bundle of $S^2$ is nontrivial.
The hairy ball theorem can be proven via Brouwer's fixed point theorem for maps $f:D^2\rightarrow D^2$, which doesn't require the knowledge of homology groups of $S^2$.  
A: If you are allowed to assume $\mathbb{R}^3-\{0\}$ is not contractible, it's not too difficult to write a homotopy equivalence between that and $S^2$.
Alternatively, you can assume otherwise and use Brouwer's fixed point theorem to derive a contradiction.
A: Note that if $S^2\simeq \{*\}$, then every map $f\colon X\rightarrow S^2$ is homotopic to the constant map. This is clearly not the case though for the identity map $\mbox{id}\colon S^2\rightarrow S^2$ and so it can not be the case that the sphere $S^2$ and the point $\{*\}$ are homotopy equivalent.

I should add that the identity map on the sphere is only clearly not homotopic to the constant map when using intuition. To prove this rigorously, I would think that using the degree of the map is the simplest way, but then this is essentially a homology argument. It might be possible to prove that if $f$ is homotopic to the identity map then $f$ is surjective without using a homology argument. Whether this then fits your criterion is debatable.
