# Prove Without Using Borsuk-Ulam

I am teaching a summer course and a student asked me a question he found online. It asks

"Show that there is not a continuous mapping $f:S^n\longrightarrow S^1$ with $f(-x)=-f(x)$ for all $x$."

The easy answer is that this would contradict the Borsuk-Ulam Theorem if we view $S^1$ as embedded in $\mathbb{R}^n$. Is there perhaps another way to prove this using degrees of maps with $S^1\subseteq S^n$? Or, can we pass to a map from $\mathbb{R}P^n$ to itself? Or is there at least some solution that doesn't require a big theorem?

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For $n=2$ this result in turn implies Borsuk-Ulam (as Wikipedia demonstrates). And I don't know if the $n=2$ case of Borsuk-Ulam has a substantially easier proof than the general one. –  user31373 Jul 17 '12 at 23:04

I don't know exactly what you consider a big theorem. Embedding $S^1\subset S^n$ as intersection with a plane through the origin, the restriction to this $S^1$ would have odd degree (does that require a big theorem?), while the embedded $S^1$ can be continuously contracted to a point. The degree of the restriction cannot change under the deformation, but $0$ is not odd.