# Brouwer fixed point theorem for smooth maps (Milnor)

This is Lemma 6 in page 14 of Milnor's Topology from the Differentiable Viewpoint.

Lemma 6. Any smooth map $$g:D^n \to D^n$$ has a fixed point.

Proof. Suppose $$g$$ has no fixed point. For $$x \in D^n$$, put $$u(x)=\frac{x-g(x)}{|x-g(x)|}, t(x)=-x \cdot u(x)+\sqrt{1-x \cdot x+(x\cdot u(x))^2 }$$ and define $$f(x)=x+t(x)u(x)$$. Then $$f$$ is a smooth retraction of $$D^n$$ onto $$S^{n-1}$$, contradicting the above lemma.

There is a figure (below) in the book. But I can't see how the formula of $$f$$ come from. Also, I can't see that $$t$$ and $$f$$ are well-defined. How do I have derive the formula?

We're trying to create a retraction that maps points $$g(x)$$ to points $$f(x)$$. The idea is that since there are no points such that $$g(x) = x$$, we can always find a line from $$g(x)$$ to $$x$$, and extend this line to meet the boundary at $$f(x)$$.

Intuitively, such a process ought to be continuous: wiggling the point $$x_0 \in D^n$$ should wiggle $$g(x_0)$$ by a small amount (since $$g$$ is smooth), and should hence wiggle $$f(x_0)$$ by a small amount (since $$f$$ is a line that's joining two smoothly wiggling points, the wiggle in the endpoint of $$f$$ as it touches the disk should be small).

So here, $$u(x)$$ is the normalized direction vector from $$g(x)$$ to $$x$$: we take the direction $$g(x) - x$$ and then normalize with $$|g(x) - x|$$ to create $$u(x)$$.

Next, we need to extend the line starting at $$x$$, pointing in direction $$u(x)$$, with magnitude that will take it to the surface of the disk $$D^n$$.

Notice that such a vector will be of the form $$\texttt{start} + \texttt{dir} \cdot \texttt{length}$$, where:

• $$\texttt{start}$$ is the starting point: $$x$$.
• $$\texttt{dir}$$ is the direction vector to move in: $$u(x)$$
• $$\texttt{length}$$ is the distance to move: $$t(x)$$

So that gives us the formula for $$f(x) \equiv x + u(x)t(x)$$

To derive the formula for $$t(x)$$, notice that $$f(x)$$ is a point on the sphere, so we need $$||f(x)||^2 = 1$$. From this, we get:

\begin{align*} ||f(x)||^2 &= 1 \\ ||x + t u||^2 &= 1 \\ ||x||^2 + ||tu||^2 + 2(x \cdot t u) &= 1 \\ ||x||^2 + t^2||u^2|| + 2 (x \cdot u) t - 1 &= 0 \\ t^2||u^2|| + 2 (x \cdot u) t - 1 + ||x||^2 &= 0 \\ t^2 \cdot 1 + 2 (x \cdot u) t - 1 + ||x||^2 &= 0 \qquad \text{||u|| = 1 since u is unit vector}\\ \end{align*}

Solving the quadratic for $$t^2$$ with $$a = 1, b = 2 (x \cdot u), c = ||x^2 - 1||$$, we arrive at the desired expression for $$t$$.

The norm of $$f(x)$$ is $$1$$: $$|f(x)|=|x+t(x)u(x)|=1\implies x\cdot x+2t(x)x\cdot u(x)+(t(x))^2=1$$. Now apply the quadratic formula to get the expression for $$t(x)$$.

You have the expression then for $$t(x)$$ as a function of $$x$$, and it and $$f$$ are well-defined.