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 [tex]\mathbb{R}^n[/tex] to itself. Then there is a fixed point [tex]x\in G[/tex] such that [tex]f(x)=x[/tex].

**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 [tex]r(x)[/tex] 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 [tex]r(x)=x[/tex] 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 [tex]\pi_n(S^n)\ne 0[/tex] (which is a tremendous task to prove).

By a simplicial approximation theorem there is a triangulation of G and a map [tex]r^*\colon G\to G[/tex] which is piecewise linear and which is homotopic to *r* by a homotopy relative to [tex]\partial G[/tex]. So [tex]r^*[/tex] 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 [tex]r=r^*[/tex]

So what do we have: a triangulated convex closed bounded set G and a map [tex]r\colon G\to \partial G[/tex] 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 [tex]\partial G[/tex]. Let us now choose a point x from [tex]\partial G[/tex] 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 [tex](r\restriction s)^{-1}\{x\}[/tex] 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 [tex](r\restriction s)[/tex] maps the corners to the corners and is linear we may assume for a while (changing the co-ordinate system) that

[tex]s=\{(x_0,\dots,x_n)\in [0,1]^{n+1}\mid x_0+\cdots + x_n = 1\},[/tex]

[tex]b=\{(x_0,\dots,x_{n-1})\in [0,1]^n\mid x_0+\cdots + x_{n-1}=1\}[/tex]

and

[tex](r\restriction s)(x_0,\dots,x_n)=(x_0,\dots,x_{n-2},1-x_0-\cdots-x_{n-2}).[/tex]

In this view we have

[tex]x=(\frac{1}{n},\dots,\frac{1}{n})[/tex]

being the barycenter of b and so

[tex](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}\}[/tex]

which proves the claim.

It follows that [tex]r^{-1}\{x\}[/tex] 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 [tex]\partial G[/tex], in which case it is a face of just one n-simplex. Because for each s, the set [tex](r\restriction s)^{-1}\{x\}[/tex] consist of at most one line segment, the polygon can never cross itself, and so must continue until it meets [tex]\partial G[/tex] again, this time at a different point [tex]y\ne x[/tex]. 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. [tex]\square[/tex]

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