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Let $\omega,\tau:[0,1]\rightarrow[0,1]\times[0,1]$ be paths in the square such that $\omega(0)=(0,0)$, $\omega(1)=(1,1)$, $\tau(0)=(1,0)$, and $\tau(1)=(0,1)$. Show that $\omega(s)=\tau(t)$ for some $s,t\in[0,1]$. Hint: use Brouwer fixed point theorem.

I know that to use the hinted theorem, I need to consider that the unit square is homeomorphic to the 2-dimensional disc $D^2$. I also need to define a function $f:[0,1]\times[0,1]\rightarrow[0,1]\times[0,1]$ so that the fixed point corresponds to the point where $\omega(s)=\tau(t)$.

mrose
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1 Answers1

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The following proof is a modification of some results from the paper The Jordan Curve Theorem Via the Brouwer Fixed Point Theorem by Ryuji Maehara (The American Mathematical Monthly vol. 91/10, 1984, pp. 641--643).

Note that by applying the transformation $t\mapsto 2t-1$ we may consider the interval $[-1,1]$ instead of $[0,1]$ and continuous functions $\omega,\tau\colon[-1,1]\to[-1,1]\times[-1,1]=:I$ such that $\omega(-1)=(-1,-1)$, $\omega(1)=(1,1)$, $\tau(-1)=(1,-1)$, $\tau(1)=(-1,1)$. If $\omega(s)\not=\tau(t)$ for all $s,t\in[-1,1]$ the continuous function $N$, $N(s,t) :=\Vert\omega(s)-\tau(t)\Vert$, where $\Vert(u,v)\Vert:=\max\{\vert u\vert,\vert v\vert\}$ is different from $0$ for all $s,t$. Thus $F\colon I\to I$, $F(s,t):=\frac{\tau(t)-\omega(s)}{N(s,t)}$ is well-defined and continuous. Then by Brouwer's fixed point theorem there would exist $(u,v)\in I$ such that $F(u,v)=(u,v)$. Since $\Vert F(u,v)\Vert=1$ by definition of $F$ this implies $\vert u\vert=1$ or $\vert v\vert=1$. Let $\omega=(\omega_1,\omega_2)$, $\tau=(\tau_1,\tau_2)$ with continuous functions $\omega_i,\tau_j\colon[-1,1]\to[-1,1]$. We have four cases.

  1. $u=-1$. Then $\tau_1(u)=-1$ and $-1=\frac{1-\omega_1(v)}{N(-1,v)}$ implying $1-\omega_1(v)<0$ or $\omega_1(v)>1$, a contradiction.
  2. $u=1$. Then $\tau_1(u)=-1$ and therefore (similar as above) $-1-\omega(v)>0$, again a contradiction.
  3. $v=-1$. Then $\omega_2(v)=-1$ and finally $\tau_2(u)+1<0$, again a contradiction.
  4. $v=1$. Then $\omega_2(v)=1$ which implies the contadiction $\tau_2(u)-1>0$.

Thus there are $s,t$ such that $N(s,t)=0$, as desired.

Jens Schwaiger
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