Brouwer fixed point theorem
From Academic Kids

In mathematics, the Brouwer fixed point theorem states that every continuous function from the closed unit ball D^{ n} to itself has a fixed point. In this theorem, n is any positive integer, and the closed unit ball is the set of all points in Euclidean nspace R^{n} which are at distance at most 1 from the origin.
The theorem has several "real world" illustrations. One common informal version is "you can't comb a hairy ball smooth"; another one works as follows: take two equal size sheets of graph paper with coordinate systems on them, lay one flat on the table and crumple up (but don't rip) the other one and place it any way you like on top of the first. Then there will be at least one point of the crumpled sheet that lies exactly on top of the corresponding point (i.e. the point with the same coordinates) of the flat sheet. This is a consequence of the n = 2 case of Brouwer's theorem applied to the continuous map that assigns to the coordinates of every point of the crumpled sheet the coordinates of the point of the flat sheet right beneath it.
The Brouwer fixed point theorem was one of the early achievements of algebraic topology, and is the basis of more general fixed point theorems which are important in functional analysis. The case n = 3 was proved by L. E. J. Brouwer in 1909. Jacques Hadamard proved the general case in 1910, and Brouwer found a different proof in 1912. Since it must have an essentially nonconstructive proof, it ran contrary to Brouwer's intuitionist ideals.
Proof outline
A full proof of the theorem would be too long to reproduce here, but the following paragraph outlines a proof omitting the difficult part. It is hoped that this will at least give some idea why the theorem might be expected to be true. Note that the boundary of D^{ n} is S^{ n1}, the (n1)sphere.
Suppose f : D^{ n} → D^{ n} is a continuous function that has no fixed point. The idea is to show that this leads to a contradiction. For each x in D^{ n}, consider the straight line that passes through f(x) and x. There is only one such line, because f(x) ≠ x. Following this line from f(x) through x leads to a point on S^{ n1}. Call this point g(x). This gives us a continuous function g : D^{ n} → S^{ n1}. This is a special type of continuous function known as a retraction: every point of the codomain (in this case S^{ n1}) is a fixed point of the function.
Intuitively it seems unlikely that there could be a retraction of D^{ n} onto S^{ n1}, and in the case n = 1 it is obviously impossible because S^{ 0} isn't even connected. The case n=2 takes more thought, but can be proven by using basic arguments involving the fundamental groups. For n > 2, however, proving the impossibility of the retraction is considerably more difficult. One way is to make use of homology groups: it can be shown that H_{n1}(D^{ n}) is trivial while H_{n1}(S^{ n1}) is infinite cyclic. This shows that the retraction is impossible, because a retraction cannot increase the size of homology groups.
There is also an almost elementary combinatorial proof. Its main step consists in establishing Sperner's lemma in n dimensions.
A quite different proof can be given based on the game of Hex. The basic theorem about Hex is that no game can end in a draw. This is equivalent to the Brouwer fixed point theorem for dimension 2. By considering ndimensional versions of Hex, one can prove that in general that Brouwer's theorem is equivalent to the "no draw" theorem for Hex.
Generalizations
 Lefschetz fixedpoint theorem
 For a number of generalizations of the Brouwer fixed point theorem to infinite dimensions, see fixed point theorems in infinitedimensional spaces.
External link
 Brouwer's Fixed Point Theorem for Triangles (http://www.cuttheknot.org/do_you_know/poincare.shtml#brouwertheorem) (based on Sperner's lemma)it:Teorema del punto fisso di Brouwer