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Is there a way to show that $S^2$ does not have the same homotopy type of a point without using homology groups?

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You mean, without the theory of degree? – 1015 Mar 23 '13 at 12:27
What do you already know about algebraic topology? – Martin Brandenburg Mar 23 '13 at 13:14
up vote 0 down vote accepted

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.

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Your proof in the first paragraph is just a reformulation of the claim ... – Martin Brandenburg Mar 23 '13 at 13:13

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.

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I think that assuming $\mathbb{R}^3\setminus \{0\}$ is not contractible rather defeats the point. The two standard proofs of Brouwer's fixed point theorem use homology and higher homotopy groups of the sphere (which itself is a calculation normally reliant on the homology of the sphere). – Dan Rust Mar 23 '13 at 13:08
I am looking now at a proof in Hatcher's 'Algebraic Topology' for the $D^2 \to D^2$ case of Brouwer's, and it only uses the non-triviality of the fundamental group of $S^1$. I'm hoping that is trivial enough to be assumed. – fixedp Mar 23 '13 at 13:14

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 $S^2$.

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