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\begin{document}
\begin{center}
{\Large Exercises from February 9th} \\[.3em]
{\small 18.904 Spring 2011}
\end{center}
\vskip 1ex
These are exercises from JJ's and Kyle's talks. Solutions are on the following page.
\begin{exercise}[JJ]
\label{jj}
Let $D$ be the open unit disc and let $f: D \rightarrow D$ be a continuous map. Does the Brouwer Fixed Point Theorem hold in this context? That is, must $f(D)$ have a fixed point?
\end{exercise}
\begin{exercise}[Kyle]
\label{kyle}
Let $f,g$ be homotopic loops in $S^1$. Show that $\tilde{f}(1)=\tilde{g}(1)$, where $\tilde{f}$ and $\tilde{g}$ denote the lifts to
maps $\R \to S^1$ taking $0$ to the basepoint of $S^1$.
\end{exercise}
\newpage
\par\noindent{\it Solution to Exercise~\ref{jj}}:
The theorem does not hold in this case, because we cannot find a retract from the open disc to the 1-sphere, a step that is key
to our proof of the theorem.
More specifically, consider the continuous map that shrinks the distance between a point in $D$ and the border of $D$ by half:
\begin{displaymath}
f(x, y) = \left( \tfrac{1}{2} (\sqrt{1-y^2}+x), y \right).
\end{displaymath}
The map $f$ does not fix any points in the open disc.
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{kyle}}:
Let $F:I\times I\to S^1$ be a homotopy from $f$ to $g$ (that is,
$F(t,0)=f(t)$, $F(t,1)=g$, and $F(0,s)=F(1,s)=x_0$). For $s\in I$,
let $h(s)$ be a path of the homotopy (where $h(s)(t)=F(t,s)$).
Since $F$ is continuous and $I\times I$ is compact, let $\delta$ be
the Lebesgue number of the open cover $\{F\inv(U_1),F\inv(U_2)\}$,
where $U_1$ and $U_2$ are as before. Let $\ell$ denote the
operation which lifts a loop to a path in $\R$ (so $\ell
f=\tilde{f}$).
Claim: if $s_1,s_2\in I$ are such that
$\abs{s_1-s_2}<\frac{\delta}{2}$, then $\ell(h(s_1))(1)$ and
$\ell(h(s_2))(1)$ are in the same path component of $p\inv(U_1)$ or
$p\inv(U_2)$. We see that $h(s_1)(t)$ and $h(s_2)(t)$ are always
both in $U_1$ or in $U_2$ for all $t\in I$ since
$\abs{s_1-s_2}<\delta$. Let $0=t_0