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Brouwer's fixed point theorem:

Every continuous function $f$ from a convex compact subset $K$ of a Euclidean space to $K$ itself has a fixed point.

I am wondering why the word "convex" is in there. It seems to me that it is necessary and sufficient for $K$ to have no holes, which is a weaker condition than convexity.

Necessary: if $K$ has a hole, then the continuous mapping that simply rotates points around this hole has no fixed point.

Sufficient: A unit disk in any number of dimensions has a Brouwer fixed point. Any compact, hole-less set $K$ is homeomorphic to the unit disk in some number of dimensions. If we map $K$ to the unit disk $D$ with a homeomorphism $h$, then consider the function from $D$ to $D$ given by $h \circ f \circ h^{-1}$. This is a continuous function from the unit disk to itself (which is convex and compact), so it has a Brouwer fixed point $x$. Then $h(f(h^{-1}(x))) = x$, so $f(h^{-1}(x)) = h^{-1}(x)$, so $h^{-1}(x)$ is a Brouwer fixed point of $K$.

What's wrong with this argument?

GMB
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    Most probably it's just a way to sidestep the problem of defining exactly what "no holes" means in a rigorous way that both works and makes sense for all dimensions. (Which is an interesting question, but it's nice to be able to state and use the fixed point theorem without answering it). – hmakholm left over Monica Mar 07 '13 at 18:13
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    How about "homeomorphic to a unit disk?" – GMB Mar 07 '13 at 18:15
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    It lacks formality. How do you define 'hole-less'? (Your definition is sidestepping, as Henning mentioned.) – copper.hat Mar 07 '13 at 18:15
  • x @LevDub: Okay, but then your preferred wording is so trivially equivalent to the one you don't like that the difference is not worth caring about, since both premise and conclusion are clearly invariant under homeomorphism. (Except that you'd incur an obligation to prove that convex compacts subsets of $\mathbb R^n$ are homeomorphic to a closed ball in some $\mathbb R^k$). – hmakholm left over Monica Mar 07 '13 at 18:19
  • Okay. So essentially, Brouwer is stated as is because it's simpler, and we trust mathematicians to use their noggins and realize that you get homeomorphisms for free. That's reasonable, I guess. Thanks! – GMB Mar 07 '13 at 18:22

3 Answers3

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The spirit of your question is correct -- the hypothesis of convexity is unnecessary, and indeed any compact subset of Euclidean space without "holes" has the fixed point property. In particular:

Theorem. Let $K$ be any compact, locally contractible subspace of $\mathbb{R}^n$ whose reduced homology groups are all trivial. Then every continuous map $f\colon K\to K$ has a fixed point.

Here "locally contractible" and "all reduced homology groups are trivial" are a precise formulation of what it means for a space to not have "holes". This theorem was proven by Lefschetz, and it follows from the famous Lefschetz fixed point theorem. Specifically, the Lefschetz fixed point theorem is usually stated only for simplicial complexes, but it is known that every compact, locally contractible subset of $\mathbb{R}^n$ is a retract of a simplicial complex, from which the above theorem follows.

By the way, the reasoning you gave is not correct, because a compact, hole-less set $K$ need not be homeomorphic to a disk. For a simple example, the union of finitely many line segments in $\mathbb{R}^2$ meeting at a point is compact and "hole-less" (in the sense of the theorem above), but is not homeomorphic to a disk in any dimension. The above theorem says that any map from such a space to itself must have a fixed point.

Jim Belk
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It is just that a convex compact set is homeomorphic to the unit ball. And the fixed point theorem works for those spaces.

Damien L
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You better also demand connectedness in your definition of "holeless" because otherwise you can take two circles (compact) and have a continuous function which just switches them. Convex implies path connected, so that's why Brouwer doesn't apply.

Dylan Yott
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