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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?

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    $\begingroup$ You must assume your complex is finite; otherwise consider a translation of the real line. $\endgroup$
    – Tom Church
    Commented Jan 25, 2013 at 2:20
  • $\begingroup$ Indeed, the complex must be finite. Thanks. $\endgroup$
    – Ami Paz
    Commented Jan 25, 2013 at 16:43
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    $\begingroup$ @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 ) $\endgroup$
    – Michał Kukieła
    Commented Jul 4, 2013 at 22:18

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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 $f$.

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  • $\begingroup$ 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
    Commented Jan 25, 2013 at 18:04
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    $\begingroup$ I think that ENR, euclidean neighborhood retract, is the key phrase. $\endgroup$
    – Tom Goodwillie
    Commented 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
    Commented Jan 26, 2013 at 22:39
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    $\begingroup$ 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 $C$. $\endgroup$
    – Tom Goodwillie
    Commented Jan 27, 2013 at 0:02
  • $\begingroup$ @Tom, that's a nice construction! $\endgroup$
    – Greg Friedman
    Commented Feb 8, 2013 at 8:11

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