Is there any easy proof of Brouwer's fixed point theorem?

Question

I am taking a course in game theory. For proving the Nash equilibrium we require Brouwer's fixed point theorem. But I have not taken a topology course so I am finding the proof difficult to understand. You may explain the same Brouwer's in little easy way.

Answer 1

Topological Spaces From Distance To Nighborhood by Gerard Buskes and Arnoud van Rooij provide a nice 'intuitive proof' (Section~4.21). It is not exact, but it has plenty of diagrams and explanations.

Answer 2

I think so. Hirsch's proof, which is in Guillemin and Pollack.

It goes (something like) this:

Suppose $p:D^2\to D^2$ doesn't have a fixed point. Define $r:D^2\to S^1$ by $r(x)$ is the point on the segment from $x$ to $p(x)$ on $S^1$. $r$ would be a retraction of the disk onto the circle. This is impossible.

Answer 3

Aigner gives an elementary proof for the case $n = 2$ in Proofs from the BOOK.

But if I remember correctly you need Brouwer's fixed point theorem for arbitrary $n$ in the proof of Nash.

saulspatz has a source for this (and the general case) in his comment to the question.

Answer 4

You could have a look at John Milnor's take. His methods are all quite elementary (the most difficult prerequisite needed here is the well-known-but-rarely-proved change of variables theorem). However, in my opinion this proof is indeed quite mysterious and John himself seems to agree with that.

Answer 5

Of course "easy" has a large degree of subjectivity to it. But there is a famous proof due to Emanuel Sperner that was (and still is) striking for minimizing the amount of topology needed.

These course notes of Jacob Fox give an exposition using Sperner's Lemma. The document starts from scratch and proves Brouwer's Theorem (for all $n$) in just under 2.5 pages.