Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking for a proof that Brouwer's fixed point theorem implies Sperner's lemma. All proof I've found just prove that there must be at least one completely colored n-simplex, not that there must be an odd number of them.

Thanks in advance!

share|improve this question
here is going the other way math.mit.edu/~fox/MAT307-lecture03.pdf –  john mangual Jul 29 '12 at 17:47
Do you ave any reason to suspect there is such a proof? –  Michael Greinecker Jul 29 '12 at 17:53
There is a common statement that the two theorems is equivalent, and the proof is easy if you just want to prove the existence of at least on such simplex. Different sources names different definitions of the theorem. Any information about the relationship is welcomed. –  user36773 Jul 29 '12 at 18:13
The classic proof tat Brouwer implies the existence of a completely labeled subsimplex is due to Yoseloff in the 1974 paper "Topologic Proofs of some Combinatorial Theorems". I don't think there is a proof that Brouwer implies the full theorem. One certainly doesn't need it to prove the fixed point theorem. There is a paper by Cuong Le Van in which he uses topological degree theory to prove Sperner's lemma in full generality. I tink he also discusses the relation to the fixed point theorem. –  Michael Greinecker Jul 29 '12 at 19:25
Thanks for all your helpful comments! –  user36773 Jul 29 '12 at 21:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.