(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 $f(x) = f(-x)$.


  • $\begingroup$ I don't think so. $\endgroup$ – Carsten S Jan 9 '14 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...

UPDATE: a "elementary" proof of Borsuk-Ulam.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.