# 1.08 Brouwer's fixed point theorem

## Video

Below the video you will find accompanying notes and some pre-class questions.

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## Notes

### Brouwer's fixed point theorem

*(0.30)* Let \(F\colon D^2\to D^2\) be a continuous map, where
\(D^2=\{(x,y)\in\mathbf{R}^2\ :\ x^2+y^2\leq 1\}\) is the
2-dimensional disc. Then there exists a point \(x\in D^2\) such that
\(F(x)=x\) (a *fixed point*).

*(1.40)* Assume, for a contradiction, that \(F(x)\neq x\) for all
\(x\in D^2\). Then we can define a map \(G\colon D^2\to\partial
D^2\) (where \(\partial D^2\) denotes the boundary circle) as
follows: consider the ray starting at \(F(x)\), passing through
\(x\) and let \(G(x)\) to be the unique point of intersection of
this ray with \(\partial D^2\).

*(3.45)* If there were a fixed point, we would not know which ray to
draw (as \(F(x)=x\)), so this map is only well-defined because we
have assumed there are no fixed points.

*(3.58)* We claim that:

- \(G\) is continuous; this is intuitively clear, but requires proof (below).
- \(G(x)=x\) if \(x\in\partial D^2\); this is clear, because if \(x\in\partial D^2\), no matter where \(F(x)\) is, the ray from \(F(x)\) to \(x\) intersects the boundary at \(x\), so \(G(x)=x\).

*(5.25)* Given these two claims, we get a contradiction as
follows. Let \(i\colon\partial D^2\to D^2\) be the inclusion map of
the boundary; the composition \(G\circ i\) equals the identity map
on \(\partial D^2\) by claim 2. This implies
\[id=G_*\circ i_*\colon\pi_1(\partial D^2)\to\pi_1(\partial D^2).\]
We have \(\pi_1(\partial D^2)=\mathbf{Z}\), and this is saying that
the identity map on the integers factors through the map
\(i_*\colon\pi_1(\partial D^2)\to\pi_1(D^2)=0\):

*(7.30)* This implies that \(G_*\circ i_*=0\). This is impossible, as
it implies \(n=G_*(i_*(n))=G_*(0)=0\) for all
\(n\in\mathbf{Z}\). Therefore we have a contradiction, and we have
proved Brouwer's fixed point theorem, modulo claim 1.

*(9.20)* It remains to prove claim 1 (that \(G\) is continuous). We
can write \(G\) as a composition \(H\circ j\) where:

- \(j\colon D^2\to D^2\times D^2\setminus\{(x,x)\ :\ x\in D^2\}\) is the map \(j(x)=(x,F(x))\).
*(11.30)*\(H\colon D^2\times D^2\setminus\{(x,x)\ :\ x\in D^2\}\) is the map defined as follows. Given two distinct points \(x\neq y\) in \(D^2\), let \(H(x,y)\) denote the point where the ray from \(y\) through \(x\) intersects \(\partial D^2\).

*(12.25)* By definition, we have \(G(x)=H(x,F(x))=H(j(x))\). By the
properties of the product topology, the map \(j\) is continuous (more
generally, if \(p\) and \(q\) are continuous then \((p,q)\) is
continuous). It therefore suffices to show that \(H\) is continuous.

*(13.25)* We can write \(H\) explicitly in coordinates. The ray from
\(y\) through \(x\) is given in parametric form by \(y+t(x-y)\). The
condition that the ray meets the boundary of the disc is
\[|y+t(x-y)|^2=1,\] which is a quadratic equation in \(t\):
\[t^2|x-y|^2+2t(x-y)\cdot y+|y|^2-1=0.\] (Note that the leading term
is never zero since \(x\neq y\)). This has solution
\[t_\pm=\frac{-2(x-y)\cdot y\pm\sqrt{4((x-y)\cdot
y)^2-4|x-y|^2(|y|^2-1)}}{2|x-y|^2}\] We are only interested in
solutions with \(t\geq 0\) as the ray only points in one
direction. Therefore the point \(H(x,y)\) is \(y+t_+(x-y)\), which is
now expressed purely in terms of rational functions and surds, all of
which are continuous.

## Pre-class questions

- Brouwer's fixed point theorem tells us that continuous maps between 2-discs have fixed points. Is the same true for maps between 2-dimensional annuli? (An annulus is \(S^1\times[0,1]\)).
- Brouwer's fixed point theorem also holds for maps \(F\colon D^n\to D^n\) where \(D^n\) is the \(n\)-dimensional disc; can the proof above be adapted to cover this case, or are new ideas required?

## Navigation

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**1.09 Homotopy equivalence**. - Index of all lectures.