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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 $\mathbb{R}^{n+1}$.

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  • $\begingroup$ 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 ? $\endgroup$ – nicomezi Nov 13 '18 at 11:08
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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.

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