# Even More On BFPT

I wouldn’t mind writing this unless I haven’t already posted two different proofs of Brouwer’s Fixed Point Theorem. Namely a friend of mine and a reader of this blog, exposed me to another way of proving it.

This proof has the disadvantage that it uses approximation by piecewise linear maps, although it retains the elementarity.

Theorem (BFPT) Let f be a continuous map from a bounded convex closed subset G of $$\mathbb{R}^n$$ to itself. Then there is a fixed point $$x\in G$$ such that $$f(x)=x$$.

Proof Let us make a counter assumption that there are no such fixed points. First let us define the retraction r as follows. Given a point x, draw a line segment starting at f(x), going through x and ending at the boundary of G. Then set $$r(x)$$ to be that point on the boundary: This is actually the same map which is used in the standard proof of BFPT using algebraic topology. Namely r is a retraction: it holds that $$r(x)=x$$ for all x on the boundary and r is continuous (exercise) which implies that the n:th homotopy goups of G and its boundary must coincide which is nonsense as soon as you know that $$\pi_n(S^n)\ne 0$$ (which is a tremendous task to prove).

By a simplicial approximation theorem there is a triangulation of G and a map $$r^*\colon G\to G$$ which is piecewise linear and which is homotopic to r by a homotopy relative to $$\partial G$$. So $$r^*$$ is also a retraction. I will skip this part of the proof here; it is technical and elementary and can be found in any introductory book to Algebraic Topology (say C. R. F. Maunder: Algebraic Topology, from which this hole post is initializing). From now on denote $$r=r^*$$

So what do we have: a triangulated convex closed bounded set G and a map $$r\colon G\to \partial G$$ which is linear restricted to each simplex of the triangulation. So it maps the corners of the n-simplices of the triangulation of G to the corners of n-1-simplices of the triangulation of the boundary $$\partial G$$. Let us now choose a point x from $$\partial G$$ which is in the barycenter of the simplex b of the triangulation of the boundary (in the barycenter = in the middle). We claim that given any simplex s in the triangulation of G, the inverse image $$(r\restriction s)^{-1}\{x\}$$ is either empty or a line which passes from one face of s to another through the interior of s. This is easy to see: since $$(r\restriction s)$$ maps the corners to the corners and is linear we may assume for a while (changing the co-ordinate system) that

$$s=\{(x_0,\dots,x_n)\in [0,1]^{n+1}\mid x_0+\cdots + x_n = 1\},$$
$$b=\{(x_0,\dots,x_{n-1})\in [0,1]^n\mid x_0+\cdots + x_{n-1}=1\}$$
and
$$(r\restriction s)(x_0,\dots,x_n)=(x_0,\dots,x_{n-2},1-x_0-\cdots-x_{n-2}).$$
In this view we have
$$x=(\frac{1}{n},\dots,\frac{1}{n})$$
being the barycenter of b and so
$$(r\restriction s)^{-1}\{x\}=\{(\frac{1}{n},\dots,\frac{1}{n},x,y)\in [0,1]^{n+1}\mid x+y=\frac{1}{n}\}$$
which proves the claim.

It follows that $$r^{-1}\{x\}$$ is a polygon consisting of line segments which starts at x, each segment joining on to the next one at a point in the interior of some (n-1)-simplex: this is because each (n-1)-simplex is a face of exactly two n-simplexes unless it is in $$\partial G$$, in which case it is a face of just one n-simplex. Because for each s, the set $$(r\restriction s)^{-1}\{x\}$$ consist of at most one line segment, the polygon can never cross itself, and so must continue until it meets $$\partial G$$ again, this time at a different point $$y\ne x$$. But by the assumption r(y)=x which contradicts the property of the retraction r that r(x)=x for all x in the boundary. $$\square$$
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