§ Brouwer's fixed point theorem

§ General statement:

Given an nD simplex that has been subdivided, and a function that maps vertices of the subdivision to vertices of the simplex such that the function on the boundary of the simplex maps it to endpoints of the boundary, we will always have a subdivided simplex with all vertices of the original simplex.

§ 1D

  • Given a line with endpoints a,ba, b and points in between, we will always have an occurrence of abab on the line.
  • Can prove something slightly stronger: there will always be an odd number of abab on the line.

§ 2D

  • Given a triangle labelled abcabc and a subdivision of it, there will be a smaller triangle labelled abcabc.
  • Consider all smaller triangles.
  • Call a side with bcbc a door.
  • How many doors can a triangle have? It can have 0 doors if it is labelled aaaaaa, or abbabb, or some such.
  • It can have 1 door if it is:
/ \
  • It can have two doors if it is:
// \
  • We can't have three doors. So triangles can have 0, 1, or 2 doors.
  • If we find a triangle with one door, we are done, since it will have abcabc.
  • Now start from the bottom of the triangle where we have the side bcbc. Here we will find at least one edge bcbc.
  • Walk along the triangle, entering any triangle with a door.
  • If that's the only door of the triangle, we are done.
  • If not, then the triangle has two doors. Exit the current triangle through the other door (the door we did not enter from). This will take us to another triangle.
  • See that we cannot terminate the walk by exiting from sides ABAB or ACAC, for such a side will be of the form abab. Then to have a door, we will get a bcbc, so the triangle must be abcabc, ie a triangle we are looking for!
  • So if we ever escape the simplex, we must escape from the bottom side BCBC. This removes an even number of bcbc edges from BCBC. But we know there are an odd number of bcbc from BCBC, so we must find a triangle ABCABC eventually.