If $h:S^n\rightarrow S^n$ is null-homotopic prove that $h$ maps some point $x$ to its antipode $-x$.
I'm not sure how to proceed with this problem. I understand that null-homotopic means that $h$ is homotopic to a constant map $g:S^n\rightarrow S^n$. Unfortunately I haven't found a way to get a handle on the antipodal points. Any insights would help greatly.
We can show that if a map does not satisfy the property mentioned above it is homotopic to the identity map of the sphere on itself or at (least if you don't want to do homotopy theory here) has degree one and thus is not null-homotopic.
First let us show that if a map $g$ has no fixed points it is homotopic to antipodal one. The homotopy between $g$ and antipodal map $\sigma$ is explicitly given by the formula:
$$ (x,t)\mapsto \frac{(1-t)g(x)-tx}{|(1-t)g(x)-tx|}. $$
Since the antipodal map has degree $(-1)^{n+1},$ where $n$ is a dimension of a sphere, we get that any map without fixed points has the same degree.
Now consider the map $\sigma\circ f$, it has no fixed points and thus is homotopic to antipodal map. Now using multipliactivity of the degree we see that $\deg(f)=(-1)^{2n+2}=1.$ Thus it is not null-homotopic.