# Topological Content of the Kakutani Fixed Point Theorem

Reading about general equilibria, the Kakutani fixed point theorem seems to be a central tool. It states (following Wikipedia)

For $S \subset \mathbb{R}^n$, non-empty, compact and convex, and $\phi : S \to 2^S$, if $\phi$ has a closed graph and, for all $s \in S$, $\phi(s)$ is nonempty and convex, then there exists an $x \in S$ such that $x \in \phi(x)$.

It is often called a generalization of the Brouwer fixed point theorem, but I'm not sure I exactly see that as Brouwer holds for any subset homeomorphic to the closed ball, and plenty of those are nonconvex. Regardless, the proofs that I've found on the internet usually proceed by reducing to a simplex. Then, for every subdivision of that simplex into smaller simplices of side length $1/n$, you can generate a function to $S$ (not $2^S$) that agrees with $\phi$ on the vertices. Brouwer that thing, and you get a series of fixed points for each sized subdivision. Pass to a convergent subsequence, take the limit and you prove that you get a fixed point in the sense of the theorem by appealing to convexity of the target sets. Or something like that; I might have missed a detail or three.

Yuck. Maybe I'm betraying my anti-analysis prejudices, but compared to the proof of the Brouwer fixed point theorem using homology, I'm left a little unsatisfied. So, the question is, is there a way to think about the Kakutani fixed point theorem topologically? Perhaps not, given the role convexity seems to play in the proof, but then you could ask, like Brouwer for convex sets, is Kakutani a special case of a more topological theorem?

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Re. convexity: Economists tend to think of/be taught the Brouwer Theorem as being a statement about compact convex sets in $\mathbb{R}^n$ - then they don't need to be taught about homeomorphisms! –  Tom Smith Apr 23 '10 at 10:06
Not that this is an answer to your question (which is an interesting one) but you can hide the nasty details of the reduction to the Brouwer fixed point theorem by using the Michael selection theorem which gives a continuous map $f\colon S \to S$ with $f(x) \in \phi(x)$ for all $x$ to which you then can directly apply the Brouwer theorem. –  Torsten Ekedahl Apr 24 '10 at 9:03
@TorstenEkedahl No, you can't. The Michael selection theorem requires the correspondence to be lower hemicontinuous, not upper hemicontinuous. –  Michael Greinecker Jul 2 '13 at 11:26

## 2 Answers

Here is one suggestion of a topological result. Just as for the Brouwer fixed point theorem formulating it in the right generality seems tricky so I will stick to the case of a finite polyhedron $X$. An $h$-correspondence $f\colon X \multimap X$ will to me be a closed subset $\Gamma\subseteq X\times X$ such that the fibres of the projection $p\colon\Gamma\rightarrow X$ on the first factor are contractible. This implies by (Lacher: Cell-like mappings, I, Pacif. Journ. of Math., 30:3, 1969) that for every open set $U\subseteq X$ that the map $p^{-1}(U) \rightarrow U$ is a (proper) homotopy equivalence. In particular, which is actually easier to prove, $p$ induces an isomorphism in cohomology $p^\ast\colon H^\ast(X,\mathbb Z) \rightarrow H^\ast(X,\mathbb Z)$ and we define $f^\ast$ to be $(p^\ast)^{-1}q^\ast$, where $q\colon \Gamma\rightarrow X$ is the projection on the second factor.

The statement is then that if $f$ is an $h$-correspondence whose alternating trace on cohomology is different from $0$, then $f$ has a fixed point, i.e., an $x\in X$ such that $x\in q(p^{-1}(x))$. If $f$ is one-valued this is the usual (consequence of) Lefschetz fixed point formula and I will reduce to that case (arguably the reduction is not less messy than the reduction in the Kakutani case). Assume therefore that $f$ has no fixed point. That means that $\Gamma$ is disjoint from the diagonal $\Delta\subseteq X\times X$ and as both are compact they have non-zero distance (as it is somewhat simpler to argue in terms of distances I assume we have fixed a metric on $X$). I claim now that there is a continuous map $g\colon X \rightarrow \Gamma$ such that $p(g(x))$ has arbitrarily small distance to $x$ and such that $pg$ is homotopic to the identity. Assuming that we get that $g^\ast$ is an inverse to $p^\ast$ and hence that $(qg)^\ast=f^\ast$ and $qg$ has fixed point. However, the graph of $qg$ is arbitrarily close to $\Gamma$ and hence we may arrange it so that it is disjoint from $\Delta$ and hence has no fixed points which is a contradiction.

The construction of $g$ is along the lines of Lemma 2.3 of Lacher: First one replaces the given triangulation of $X$ by a an iterated barycentric subdivision to make stars of simplices small and then one uses that if $U$ is such a star, then $p^{-1}(U) \rightarrow U$ is a homotopy equivalence to construct $g$ on cells by induction of the skeleta of $X$. As stars contract nicely to their simplices we get the homotopy.

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This might be off topic, but I thought I'd throw it out there anyway: the conclusion of the Kakutani theorem resembles that of Lawvere's fixed point theorem (see his Diagonal arguments and cartesian closed categories), which seems to be more general. Would it be possible that the topological conditions in Kakutani's theorem enable an application of Lawvere's?

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Here is what appears to be a related thread: mathoverflow.net/questions/136478/… –  Ricardo Andrade Oct 29 '13 at 2:23