# Proving the Brouwer Fixed Point Theorem

Brouwer's Fixed Point Theorem: Let $$f : D^{n+1} \to D^{n+1}$$ be a continuous map, then $$f$$ has a fixed point.

The proof goes something like this:

Proof: Suppose that $$f : D^{n+1} \to D^{n+1}$$ is a continuous map and suppose that $$f$$ has no fixed point. Since $$f$$ has no fixed point we have that $$f(x) \neq x$$ for all $$x \in D^{n+1}$$. Define a function $$g : D^{n+1} \to S^n$$ in the following way. For any $$x \in D^{n+1}$$ let $$g(x)$$ be the (unique) point on $$S^{n}$$ at which the ray from $$x$$ to $$f(x)$$ intersects $$S^n$$. Then it's usually left to the reader to check that $$g$$ is well-defined, continuous and is a retract (since $$g(x) = x$$ for all $$x \in S^n$$).

But then since $$g$$ is a retract it follows that $$S^n$$ is a retract of $$D^{n+1}$$ a contradiction since homology theory tells us otherwise. $$\square$$

Now my question is the following, what exactly is a closed form formula of $$g$$ (if it exists)? I guess it would involve quite some use of the metric on $$\mathbb{R}^{n+1}$$.

• Well, your question is understandable but not very rigourous. The proof you have been given tells you that such function $g$ does not exists, so why one should expect to find a closed form ? – nicomezi Nov 13 '18 at 11:08

Yes, it exists. Find a $$t\in[0,\infty)$$ such that $$\bigl\lVert x+t\bigl(f(x)-x\bigr)\bigr\rVert=1$$ and then put $$g(x)=x+t\bigl(f(x)-x\bigr)$$. In order to find such a $$t$$ you do\begin{align}\bigl\lVert x+t\bigl(f(x)-x\bigr)\bigr\rVert=1&\iff\bigl\lVert x+t\bigl(f(x)-x\bigr)\bigr\rVert^2=1\\&\iff\lVert x\rVert^2+2t\bigl\langle x,f(x)-x)\bigr\rangle+t^2\bigl\lVert f(x)-x\bigr\rVert^2=1\\&\iff t^2\bigl\lVert f(x)-x\bigr\rVert^2+2t\bigl\langle x,f(x)-x)\bigr\rangle+\lVert x\rVert^2-1=0\\&\iff t=\frac{-\bigl\langle x,f(x)-x)\bigr\rangle+\sqrt{\bigl\langle x,f(x)-x)\bigr\rangle^2-\bigl\lVert f(x)-x\bigr\rVert^2\bigl(\lVert x\rVert^2-1\bigr)}}{\bigl\lVert f(x)-x\bigr\rVert^2},\end{align}since you are taking the positive root.