Brouwer fixed-point theorem

In the language of spheres
Any continuous map from a disc to itself must have a fixed point. In other words, for any natural number $$n$$, if $$D^n$$ denotes the spherical disc in $$\R^n$$, any continuous map $$f:D^n \to D^n$$ must have a point $$x$$ such that $$f(x) = x$$.

In the language of simplices
Any continuous map from the standard $$n$$-simplex, to itself has a fixed point.

Case $$n = 1$$
This says that any continuous map from the closed unit interval $$[0,1]$$ to itself has a fixed point. This particular case is often proved as a consequence of the intermediate value theorem for continuous real-valued functions. Specifically, if $$f:[0,1] \to [0,1]$$ is the function, then the function $$g(x) := f(x) - x$$ crosses over from a non-positive to a nonnegative function and hence must be zero for some intermediate value of $$x$$.

Facts used

 * 1) uses::No-retraction theorem: This states that there does not exist a continuous retraction from $$D^n$$ to $$S^{n-1}$$, i.e., there is no continuous map from $$D^n$$ to $$S^{n-1} = \partial D^n$$ that restricts to the identity map on $$S^{n-1}$$.

Proof
The Brouwer fixed-point theorem follows easily from the no-retraction theorem. Suppose $$f:D^n \to D^n$$ is a continuous map with no fixed points. Define a map $$g:D^n \to S^{n-1}$$, that sends $$x \in D^n$$ to the unique point on $$S^{n-1}$$ that is colllinear with $$x$$ and $$f(x)$$ in such a way that $$x$$ lies between that point and $$f(x)$$. We can see that:


 * Since $$f(x)$$ is never equal to $$x$$, and $$x$$ is inside the unit disc, $$g$$ is well-defined throughout $$D^n$$
 * $$g$$ is continuous
 * $$g$$ is a retraction because it fixes every point on $$S^{n-1}$$