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\vspace{-2em}

Trying to prove \nameref{brouwer}. We have shown
that it is equivalent to \nameref{no_retrac_thm}.

Next step is to establish equivalence with a
weaker cousin of no \nameref{no_retrac_thm}.

\begin{flashcard}[no-retrac-equiv-lemma]
\begin{lemma*}[Cousin of No Retraction Theorem]
  \nameref{no_retrac_thm} is equivalent to the
  following: divide $\partial \ol{D}$ into 3 equal
  arcs.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/9f08b06a7e3b41f5.png}
  \end{center}
  In polars:
  \begin{align*}
    I
    &= \left\{(1, \theta) : 0 \le \theta \le
    \frac{2\pi}{3}\right\} \\
    J
    &= \left\{(1, \theta) : \frac{2\pi}{3} \le \theta \le
    \frac{4\pi}{3}\right\} \\
    K
    &= \left\{(1, \theta) : \frac{4\pi}{3} \le \theta \le
    2\pi\right\}
  \end{align*}
  Then there does not exist $g : \ol{D} \to
  \bound D$ continuous such that $g(I) \subseteq
  I$, $g(J) \subseteq J$, $g(K) \subseteq K$.
\end{lemma*}

\begin{proof}
  \cloze{If this cousin is true, then
  \nameref{no_retrac_thm} follows at once.

  On the other hand if the cousin is false, and
  such a $g$ exists, then if $T$ is rotation about
  $0$ through $\pi$ then $T \circ g$ has no fixed
  point.}
\end{proof}
\end{flashcard}

The cousin has a triangle version.

\begin{theorem*}
  If $\ol{\triangle}$ is an equilateral triangle
  (closed) with sides $\tilde{I}, \tilde{J},
  \tilde{K}$, then there does not exist $G :
  \ol{\triangle} \to \ol{\bound \triangle}$ such
  that $G(\tilde{I}) \subseteq \tilde{I}$,
  $G(\tilde{J}) = \tilde{J}$, $G(\tilde{K}) =
  \tilde{K}$.
\end{theorem*}

\vspace{-1em}
This is equivalent to the previous result by
homeomoprhism (use a homeomorphism $H : \ol{D} \to
\ol{\triangle}$ with $H(I) = \tilde{I}$, $H(J) =
\tilde{J}$, $H(K) = \tilde{K}$).

\begin{flashcard}[triangle-retrac-equiv-sets-thm]
\begin{theorem*}
  The following two statements about an
  equilateral $\ol{\triangle}$ with sides $I, J,
  K$ (note sides include end point vertices) are
  equivalent:
  \begin{enumerate}[(i)]
    \item \cloze{There does not exist $h : \ol{\triangle}
      \to \bound D$ continuous with $h(I)
      \subseteq I$, $h(J) \subseteq J$, $h(K)
      \subseteq K$.}
    \item \cloze{There does not exist $A, B, C$ closed,
      $A, B, C \subseteq \ol{\triangle}$ such that
      $A \cup B \cup C = \ol{\triangle}$, $A
      \supseteq I$, $B \supseteq J$, $C \supseteq
      K$ and $A \cap B \cap C =
      \emptyset$.}
  \end{enumerate}
\end{theorem*}

\begin{proof}
  \cloze{If we could find $h : \ol{\triangle} \to \bound
  \triangle$ with $h(I) \subseteq I$, $h(J)
  \subseteq J$, $h(K) \subseteq K$. Then let
  \begin{align*}
    A
    &= h^{-1}(I) \\
    B
    &= h^{-1}(J) \\
    C
    &= h^{-1}(K)
  \end{align*}
  Then
  \[ A \cap B \cap C = h^{-1}(I \cap J \cap K) =
  h^{-1}(\emptyset) .\]

  For the other direction, suppose conversely that
  we have $A, B, C$ closed such that $A \supseteq
  I$, $B \supseteq J$, $C \supseteq K$, $A \cup B
  \cup C = \triangle$, $A \cap B \cap C =
  \emptyset$. We look at the triangle in $\RR^3$
  given by
  \[ \{(x, y, z) : x + y + z = 1, x, y, z \ge 0\}
  .\]
  Now look at $d(\bf{x}, A)$, $d(\bf{x}, B)$,
  $d(\bf{x}, C)$. Remember that
  \[ d(\bf{x}, E) = \inf_{\bf{E} \in E} \|\bf{x} -
  \bf{e}\| ,\]
  and if $E$ is closed, then $d(\bf{x}, E) = 0$
  if and only if $\bf{x} \in E$. We also remarked
  before that $\bf{x} \mapsto d(\bf{x}, E)$ is
  continuous. So
  \[ x \mapsto d(x, A), \qquad x \mapsto d(x, B),
  \qquad x \mapsto d(x, C) \]
  are continuous. So
  \[ x \mapsto d(x, A) + d(x, B) + d(x, C) \]
  is continuous. Since $A \cap B \cap C =
  \emptyset$, we have for each $x$ that at least
  one of the distances is positive. Thus
  \[ d(\bf{x}, A) + d(\bf{x}, B) + d(\bf{x}, C) >
  0 .\]
  Thus the function $F$ given by
  \[ \bf{x} \mapsto \left( \frac{d(x, A)}{d(x, A)
  + d(x, B) + d(x, C)}, \frac{d(x, B)}{d(x, A) +
  d(x, B) + d(x, C)}, \frac{d(x, C)}{d(x, A) +
  d(x, B) + d(x, C)} \right) \]
  is well defined and continuous. Let the
  components of $F(x)$ be denoted by $F_i(x)$ for
  $i = 1, 2, 3$. Note $F_j(x) \ge 0$, and $F_1(x)
  + F_2(x) + F_3(x) = 1$. So $F$ maps
  $\ol{\triangle}$ to $\ol{\triangle}$
  continuously. If $\bf{x} \in I$ then
  $F_1(\bf{x}) = 0$, so $F(A) \subseteq I$.
  Similarly for the others.}
\end{proof}
\end{flashcard}

Thus we have changed out problem to a colouring
problem.

We attack this by looking at a finite problem.
Take an equilateral triangle and cut it up by
using $n$ equally spaced lines parallel to each of
the sides.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/198a1290a2e042f5.png}
\end{center}
If you colour the vertices obeying the following
rule by triangle $ABC$:
\begin{itemize}
  \item All vertices on $AB$ except $B$ are red
  \item All vertices on $BC$ except $C$ are blue.
  \item All vertices on $CA$ except $A$ are green.
\end{itemize}
Remaining vertices are your choice. Then there is
at least one small triangle with all 3 colours.

This is known as Sperner's Lemma.