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\begin{center}
Math 205B - Topology\\
\emph{ }\\
Dr. Baez\\
\emph{ }\\
February 2, 2007\\
\emph{ }\\
Christopher Walker\\
\emph{ }\\
\end{center}
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\begin{exercise}\rm{
Show that if $A$ is a retract of $B^2$, then every continuous map $f:A\to A$ has a fixed point.
}
\end{exercise}
\begin{proof}
Let $r:B^2\to A$ be a retraction, and let $i:A\to B^2$ be the inclusion map. Consider the map $g=i\circ f\circ r :B^2\to B^2$. This is continuous, since it is the composition of continuous maps. Thus by the Brouwer Fixed Point Theorem, there exist $y\in B^2$ such that $g(y)=y$. This gives us that $y=f(r(y))$. But this means $y\in A$, and so $y=r(y)$. Therefore $f(y)=y$, and so $f$ has a fixed point.
\end{proof}
\newpage
\begin{exercise}\rm{
Show that if $h:S^1\to S^1$ is nulhomotopic, then $h$ has a fixed point and $h$ maps some point $x$ to its antipode $-x$.
}
\end{exercise}
\begin{proof}
Since $h:S^1\to S^1$ is nulhomotopic, then by Theorem 55.3, $h$ can be extended to a map $k:B^2\to S^1$. If $i: S^1\to S^1$ is the inclusion map, then $f=i\circ k$ is a continuous map from $B^2$ to itself. Hence by the Brouwer Fixed Point Theorem, there exist $z\in B^2$ with $z=f(z)=i(k(z))$. This gives us that $k(z)=z$, which tells us that $z\in S^1$, and so $z=k(z)=h(z)$. Thus $z$ is a fixed point of $h$.\\
\par Consider the antipode map $a:S^1\to S^1$ given by $a(x)=-x$ and the homotopy $F(x,t)=e^{i\pi t}h(x)$. $F$ is continuous since it is the product of continous maps, $F(x,0)=h(x)$, and $F(x,1)=-h(x)=a\circ h(x)$. Thus we have $h\simeq a\circ h$ and so $a\circ h$ is nulhomotopic. Now by the first part above $a\circ h$ has a fixed point. This means there exist $x\in S^1$ such that $x=a(h(x))$, or $x=-h(x)$. Thus $h(x)=-x$ and so $h$ maps some point to its antipode.
\end{proof}
\newpage
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\begin{exercise}
\rm{
Suppose that you are given the fact that for each $n$, there is no retraction $r:B^{n+1}\to S^n$. Prove the following:
\begin{enumerate}
\item The identity map $\iota:S^n\to S^n$ is not nulhomotopic.
\item The inclusion map $j:S^n\to \mathbb{R}^{n+1}-\textbf{0}$ is not nulhomotopic.
\item Every non-vanishing vector field on $B^{n+1}$ point directly outward at some point of $S^n$, and directly inward at some point of $S^n$.
\item Every continuous map $f:B^{n+1}\to B^{n+1}$ has a fixed point.
\item Every $n+1\times n+1$ matrix with positive real entries has a positive eigenvalue.
\item If $h:S^n\to S^n$ is nulhomotopic, then $h$ has a fixed point and $h$ maps some point $x$ to its antipode $-x$.
\end{enumerate}
}
\end{exercise}
\begin{proof} Parts (a), (b), (c), and (d) were proven previously, so we will assume they are true.\\
\emph{ }\\
(e)
Let $A$ be a $n+1\times n+1$ matrix with positive real entries, and let $T:\mathbb{R}^{n+1}\to \mathbb{R}^{n+1}$ be the linear transformation given by $T(x)=Ax$. Now consider the region $B=\{x\in \mathbb{R}^{n+1}\mid \norm{x}=1, \emph{\rm{ and }} x_i\geq 0 \emph{\rm{ for }} i=1,...,n+1\}$. $B$ is homeomorphic to $B^{n+1}$ and so all the result above about $B^{n+1}$ hold for this new region $B$. Specifically we will use the generalization of the Brouwer Fixed Point Theorem; part (d).\\
\indent Let $T^*(x)=\frac{T(x)}{\norm{T(x)}}$. $T^*$ is defined on $B$ since for any $x\in B, \norm{x}=1$, so $\norm{T(x)}\neq 0$ because $A$ has all positive entries. Thus $T^*$ is a continuous map from $B$ to itself, so by part (d) $T^*$ has a fixed point, say $x_0$. This gives us that $T^*(x_0)=x_0$ or $x_0=\frac{T(x_0)}{\norm{T(x_0)}}$. It then follows that $Ax_0=\norm{T(x_0)}x_0$, and so $\norm{T(x_0)}$ is a positive eigenvalue of $A$.\\
\emph{ }\\
(f)
We have previously shown that we can generalize part (i) and (ii) of Theorem 55.3, so we will use that here.
Since $h:S^n\to S^n$ is nulhomotopic, then by the generalization of Theorem 55.3, $h$ can be extended to a map $k:B^{n+1}\to S^n$. Let $i: S^n\to S^n$ be the inclusion map. Then $f=i\circ k$ is a continuous map from $B^{n+1}$ to itself. Hence by the part (d) (the generalized Brouwer Fixed Point Theroem), there exist $z\in B^{n+1}$ with $z=f(z)=i(k(z))$. This gives us that $k(z)=z$, which tells us that $z\in S^n$, and so $z=k(z)=h(z)$. Thus $z$ is a fixed point of $h$.\\
\par Again, consider the antipode map $a:S^n\to S^n$ given by $a(x)=-x$.
Since $h$ is nulhomotopic so is $a\circ h$, since if $F: S^n \times I \to S^n$ is a nulhomotopy for $h$, $a \circ F$ is a nulhomotopy for $a \circ h$.
Now by the first part above $a\circ h$ has a fixed point. This means there exist $x\in S^n$ such that $x=a(h(x))$, or $x=-h(x)$. Thus $h(x)=-x$ and so $h$ maps some point to its antipode.
\end{proof}
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