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?