# Is there a simple proof of Borsuk-Ulam, given Brouwer?

(Brouwer) Any continuous function from a convex compact subset K of a Euclidian space to itself has a fixed point.

Given this lemma, is there a simple proof of:

(Borsuk-Ulam) Any continuous function $f \, : \, S^n \to R^n$ (where $S^n$ is the $n$-sphere) has a point $x$ for which $f(x) = f(-x)$.

?

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I don't think so. –  Carsten Schultz Jan 9 at 11:31

It appears that Borsuk-Ulam is strictly harder than Brouwer. Quote from Using the Borsuk-Ulam Theorem : Lectures on Topological Methods in Combinatorics and Geometry:

"It is instructive to compare this with the Brouwer fixed point theorem (...). The statement of the Borsuk–Ulam theorem sounds similar (and actually, it easily implies the Brouwer theorem; see below). But it involves an extra ingredient besides the topology of the considered spaces: a certain symmetry of these spaces, namely, the symmetry given by the mapping $x \mapsto −x$ (which is often called the antipodality on $S^n$ and on $\Bbb R^n$)."

The exact strenght of Brouwer is known: is equivalent to WKL$_0$ over RCA$_0$ (Simpson, Subsystems of Second Order Arithmetic) but Borsuk-Ulam does not appear in the index of Subsystems...

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This doesn't completely answer the question, just a very special case.

Let $S_+^n$ be the (closed) upper hemisphere of $S^n$ and $S_-^n$ the lower hemisphere. Suppose our map $f:S^n\to\mathbb R^n$ besides being continuous, also has the properties

• $f(S^n_+) = f(S^n_-) = B^n$,
• the restriction $f_-:S^n_-\to B^n$ is a homeomorphism.

(Here $f_-:S_-^n\to B^n$ is defined by $f_-(x)=f(x)$. Define $f_+:S_+^n\to B^n$ analogously.)

Next, define the antipodal map $\phi:S^n_-\to S^n_+$ by $\phi(x)=-x$. This gives us a well-defined continuous map $f_+\circ\phi\circ f_-^{-1}:B^n\to B^n$, which has a fixed point by Brouwer's theorem. More specifically, this means that there exists an $x\in B^n$ such that $$f(\phi(f_-^{-1}(x)))=x=f(f_-^{-1}(x)).$$ Writing $y = f_-^{-1}(x)$, this means precisely that $f(-y)=f(y)$, proving Borsuk-Ulam in this special case.

If anyone sees a way to generalize this, I'd be interested to know.

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