Spheres in different dimension are not homotopy equivalent

Question

Is there a way to prove that $\textbf{S}^n$ and $\textbf{S}^m$ are not homotopy equivalent if $n\neq m$ without using the machinery of homology or higher homotopy groups?

Answer 1

Just for fun, here's an answer that doesn't use the machinery you're talking about, at the expense of using crazier machinery. Suppose you had a homotopy equivalence $S^n \stackrel{\sim}{\to} S^{n+k}$ for $k > 0$. By smashing this with $S^{km}$, you get homotopy equivalences $S^{n+km} \stackrel{\sim}{\to} S^{n + (k+1)m}$ for each $k$. Composing these gives you a sequence $$ S^n \stackrel{\sim}{\to} S^{n+k} \stackrel{\sim}{\to} S^{n+2k} \stackrel{\sim}{\to} \dotsb $$ and taking the colimit gives you a weak equivalence $S^n \stackrel{\sim}{\to} S^\infty$. But $S^\infty$ is contractible. (And as with Andreas's answer, you'd then have to prove that spheres aren't contractible.)

Answer 2

I can prove, without mentioning groups, that if spheres $S^m$ and $S^n$ with $m<n$ were homotopy equivalent, then they'd be contractible, but I don't immediately see a group-free proof that spheres aren't contractible.

Suppose $f:S^m\to S^n$ and $g:S^n\to S^m$ were inverses of each other up to homotopy. By cellular or simplicial approximation, you can deform $f$ so that it's not surjective (as $m<n$), and then you can deform it to a constant by pushing away from a point not in the image of $f$. So $f$ is homotopic to a constant; therefore so are $f\circ g$ and $g\circ f$, which are homotopic to the identity maps of the two spheres. Therefore, these spheres are contractible,

Answer 3

Suppose $S^{n} \simeq S^{n+d}$. Notice that $S^{n}$ is a subspace of $S^{n+d}$. So there exists a deformation retraction $H:S^{n+d} \times I \to S^n$ such that $H((x_1, ..., x_{n+1}, x_{n+2}, ..., x_{n+d+1}),0)=(x_1, ..., x_{n+1}, x_{n+2}, ..., x_{n+d+1})$ and $H((x_1, ..., x_{n+1}, x_{n+2}, ..., x_{n+d+1}),0)=(x_1, ..., x_{n+1}, 0, ..., 0)$.

Restrict $H$ to $n+1$th and $n+2$ coordinate to get $F$. Notice $F$ is still continuous since $H$ is. $F$ defines a deformation retraction from a circle to a an interval. $F((x_{n+1}, x_{n+2}),0)=(x_{n+1}, x_{n+2})$ and $F((x_{n+1}, x_{n+2}),1)=(x_{n+1}, 0)$. But this is a contradiction and hence $S^{n}$ cannot be homotopy equivalent to $S^{n+d}$ as we assumed. $\blacksquare$