TITLE: Topological Content of the Kakutani Fixed Point Theorem
QUESTION [14 upvotes]: 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?

REPLY [2 votes]: I'm still a student so a bit hesitant to try to make a contribution here, but as an economist quite enamored by topology, I've wondered similar things.  The below is based on an old attempt at a solution to an exercise in a rather challenging book (reference at end).  
Consider first a result on coincidences of certain mappings.  

Theorem Let $X$ be a finite dimensional compact Hausdorff space, $S$ a compact, convex subset of Euclidean space.  Let $f,g: X\to S$ be continuous maps, with $f$ a closed, surjective map with $\tilde{H}^q(f^{-1}(s))=0$ for all $q\ge 0$ and $s \in S$ (where $\tilde{H}^*$ denotes reduced Cech cohomology with coefficients in some module $G$).  Then there exists a coincidence point $f(x) = g(x)$.  

Proof: Define $\partial S$ as the boundary of $S$ and $\partial X = f^{-1}(\partial S)$.  Consider the two long exact sequences in relative cohomology for the pairs $(X, \partial X)$ and $(S, \partial S)$.  By the Vietoris-Begle Theorem and the Five Lemma, $f^*$ acting on the relative groups is an isomorphism, and in particular, is non-zero, given $\tilde{H}^{n-1}(S,\partial S; G) \neq 0$.  
Now, if there were no $x \in X$ for which $f(x) = g(x)$ then using the ray from $g(x)$ through $f(x)$ we can define a function $\bar{f}: X \to \partial S$ that is homotopic to $f$, and which agrees with $f$ restricted to $\partial X$.  But $\bar{f}^*$ is zero on the relative groups because $\bar{f}$ takes X onto $\partial S$, hence $f^*$ is zero, contradicting the above.  QED
Now, consider an upper-hemicontinuous correspondence $\phi: S\to S$.  Let the graph of this correspondence play the role of $X$ in the above, the projection to the domain $f$, and the projection to the codomain $g$. Call $\phi$ "acyclic valued" if the projection to the domain obeys the conditions on $f$ above.  Then coincidence points of $f$ and $g$ are precisely those fixed points $s^* \in \phi(s^*)$.
If you are interested, the above proof is set as an exercise in Repeated Games by Mertens, Sorin, and Zamir (p. 54-55), as well as a number of extensions.