I want to show the following statements are equivalent :
- There is no odd function [i.e f(-x) = -f(x) ] from $S^n \to S^{n-1}$
- Every odd function from $S^{n-1} \to S^{n-1}$ is not nullhomotopic.
I tried showing that if an odd function is nullhomotopic it can be extended to the cone and by gluing two such cones (upper and lower hemisphere) it can be extended to $S^n$. Can I say this extended map is odd? And what about the other way?
Both of these theorems can be found in the excellent book "Using the Borsuk-Ulam Theorem".
The key insight, though, is that the projection $\pi : (x_1, \ldots, x_{n+1}) \mapsto (x_1, \ldots, x_n)$ is a homeomorphism of the upper hemisphere of $S^n$ with the ball $B^n$. You've already made a similar observation (extending a map to the cone, and thus to either hemisphere), so I don't feel too bad showing you how to use this observation to solve the problem.
Recall $f : S^{n-1} \to S^{n-1}$ is nullhomotopic exactly when it can be extended to a map $\tilde{f} : B^n \to S^{n-1}$ agreeing with $f$ on $\partial B^n = S^{n-1}$.
First say you have an odd function $f : S^n \to S^{n-1}$. Define a map $g : D^n \to S^{n-1}$ by $g = f \circ \pi^{-1}$.
Do you see why $g \restriction S^{n-1}$ is odd? Since $g \restriction S^{n-1}$ can be extended to $g$, then, we see $g \restriction S^{n-1}$ is the desired odd nullhomotopic map.
Now we run the argument in reverse!
Say we have an odd nullhomotopic map $g : S^{n-1} \to S^{n-1}$. Then, by nullhomotopicness, $g$ extends to a map $\tilde{g} : B^n \to S^{n-1}$. Thus $f = \tilde{g} \circ \pi$ is a map from the upper hemisphere of $S^n$ to $S^{n-1}$ which is odd on the boundary.
Of course, there's only one way to extend this to the lower hemisphere if we want our map to be odd everywhere: defining $f(-x) = -f(x)$ does the job.
Do you see why this map $f$ is well defined and continuous?
I hope this helps ^_^