Reading Lovasz's lecture notes on evasive graph properties, I encountered the following extension of Brouwer's fixed point theorem:

Any continues map from a contractible [finite] simplicial complex to itself has a fixed point.

Lovasz refers this to Lefschetz. Indeed, it seems Lefschetz fixed-point theorem guarantees the existence of a fixed point under some conditions. But I could not find a simple proof for Lefschetz theorem, while Brouwer's theorem has an elementary proof using Sperner's Lemma.

My questions are:

  1. How should one interpret the conditions of Lefschetz fixed-point theorem, and why do they hold for a contractible complex? (A good reference will be appreciated.)

  2. Is there an easy way to prove the above statement, for the restricted case of a collapsible complex, and a simplicial, bijective map, without using homology?

  • 2
    $\begingroup$ You must assume your complex is finite; otherwise consider a translation of the real line. $\endgroup$
    – Tom Church
    Jan 25, 2013 at 2:20
  • $\begingroup$ Indeed, the complex must be finite. Thanks. $\endgroup$
    – Ami Paz
    Jan 25, 2013 at 16:43
  • 1
    $\begingroup$ @Tom Church: In fact you do not need finiteness, a weaker condition of being rayless suffices. See the paper: V. Okhezin, "On the fixed-point theory for non-compact maps and spaces. I", Topological Methods in Nonlinear Analysis 5 (1995), 83-100 ( tmna.ncu.pl/files/v05n1-05.pdf ) $\endgroup$ Jul 4, 2013 at 22:18

1 Answer 1


You have to assume that your complex is finite. Then the Lefschetz Fixed Point Theorem definitely says that $f$ must have a fixed point if the (homologically defined) Lefschetz number of $f$ is not zero. If the complex is contractible, then the Lefschetz number of $f$ must be $1$.

This fact about contractible complexes reduces to the Brouwer Theorem if you use the following fact: any finite contractible complex is a retract of $D^n$ for some $n$. That is, there exist continuous maps $i:X\to D^n$ and $r: D^n\to X$ such that $r\circ i$ is the identity. For continuous $f:X\to X$, any fixed point of $i\circ f\circ r:D^n\to D^n$ gives you a fixed point of $f$.

  • $\begingroup$ Thanks. I assume the contractility of $X$ is used to prove that such $D^n$ and $r$ exist. Could you give some guidelines for the proof of this fact? $\endgroup$
    – Ami Paz
    Jan 25, 2013 at 18:04
  • 2
    $\begingroup$ I think that ENR, euclidean neighborhood retract, is the key phrase. $\endgroup$ Jan 25, 2013 at 18:18
  • $\begingroup$ @Tom, I read about ENR, and indeed any finite complex is an ENR. But this holds even if the complex is not contractible; where do the contractibility goes into the picture? $\endgroup$
    – Ami Paz
    Jan 26, 2013 at 22:39
  • 1
    $\begingroup$ Embed $X$ in $\mathbb R^n$. It's a retarct of a nbhd $N$. In $N$ is a smaller nbhd $K$ that is a union of cubes with edges parallel to the axes. $X$ is also a retarct of $K$. Now let $C$ be a cube containing $K$, and extend the retraction $K\to X$ to a map $C\to X$ (cell by cell) to get that $X$ is a retract of $C$. $\endgroup$ Jan 27, 2013 at 0:02
  • $\begingroup$ @Tom, that's a nice construction! $\endgroup$ Feb 8, 2013 at 8:11

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.