Details about the Brouwer fixed-point theorem

Question

I read about the construction of a function $G : D_n → ∂D_n$ for an $n$-dimensional disk $D_n$ using $F(p)$ and $p$ in the proof of Brouwer fixed-point theorem, yet I don't understand how to construct such functions.

I am wondering if someone could give me a more explicit example of $G$, for example, when $n = 2$ and $p \in D_2$? Thanks in advance!

Answer 1

Are you referring to the proof attributed to Hirsch, in which we map the point $p$ to $G(p)\in\partial D$ given by extending the segment through $p$ and $F(p)$ until it hits the boundary circle? If so I think that's about it. Of course, here we use the fact that two points determine a line. And that line has to encounter the boundary at some stage. So that will be $G(p)$.

Now it's easy to see that $G$ is continuous, and is a retraction of the disk onto the boundary. But that is proved elsewhere not to be possible (that part might use a little algebraic topology, say. That's the way I remember: $S^1$ and $D^2$ have different homology. There's probably a differential geometry way, but my cohomology has gotten rusty. There's probably a more basic way too).

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I guess if you want more explicit, we could form the vector $p+t(F(p)-p)$, and then choose $t$ so that it has norm $1$.

Answer 2

If $F(x,y)=(F_1(x,y),F_2(x,y))$ then you need to find the $t\leq 0$ so that:

$$(x,y)+t(F_1(x,y)-x,F_2(x,y)-y)$$ is on the boundary.

This requires $$\left(tF_1+(1-t)x\right)^2+\left(tF_2+(1-t)y\right)^2=1$$

You get a quadratic equation $$at^2+bt +c =0\\a=(F_1-x)^2+(F_2-y)^2\\ b=2x(F_1-x)+2y(F_2-y)\\ c=x^2+y^2-1$$ Assuming $F(x,y)\neq (x,y),$ we have $a>0.$ We also have $c\leq 0.$ A little more geometry gives you that $b<0$ when $x^2+y^2=1.$ (Short argument: The sign of $b$ is the same as the sign cosine of the angle between $(x,y)$ and $(F(x,y)-(x,y))$ and that can’t be positive when $(x,y)$ is on the boundary and $F(x,y)$ is in the unit disk.)

So the non-positive root of the quadratic is:

$$t(x,y)=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

We can see that $t$ is continuous since $a,b,c$ are continuous and $a\neq 0.$

Then $$G(x,y)=(1-t(x,y))(x,y)+t(x,y)F(x,y)$$ is continuous and $|G(x,y)|=1$ for all $(x,y),$ by our construction.

When $x^2+y^2=1,$ we see that $t(x,y)=0$ (since $c=0$ and $b<0$) and thus $G(x,y)=(x,y).$