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Math 205B - Topology\\
\emph{ }\\
Dr. Baez\\
\emph{ }\\
January 12, 2007\\
\emph{ }\\
Christopher Walker\\
\emph{ }\\
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\begin{question}\rm{
A subset $A$ of $\mathbb{R}^n$ is said to be \textbf{\textit{star convex}} if for some point $a_0$ of $A$, all the line segments joining $a_0$ to other points in $A$ lie in $A$.
\begin{enumerate}
\item Find a star convex set that is not convex.
\item Show that if $A$ is star convex, $A$ is simply connected.
\end{enumerate}
}
\end{question}
\begin{proof}\emph{ }\\
\begin{enumerate}
\item Consider the subset $A$ of $\mathbb{R}^2$ bounded by the parametric polar curve
$$\{(r(t), \theta(t))\mid r(t)=sin(5t)+1.2, \theta(t)=t, 0\leq t<2\pi\}.$$
(See attached graph.) This subset is star convex if we let $a_0$ be the origin (and even looks like a star!). This is because the line segment from the origin to any other point in $A$ is contained in $A$. This is not convex since the line segment between $(1,0)$ and $(0,1)$ is not contained in $A$.
\item Assume $A$ is star convex. We need to show that $A$ is path connected and that $\pi_1(A, a_0)$ is trivial for the point $a_0$ in the definition of star convex. Let $a_1,a_2\in A$. We can create a path from $a_1$ and $a_2$ by first traveling along the line segment from $a_1$ to $a_0$, then along the line segment from $a_0$ to $a_2$. This path is contained entirely in $A$ by definition of star convex, and so $A$ is path connected.\\
\indent Next, consider a loop $g$ at $a_0$. Define $H:I\times I\to A$ as the straight line path-homotopy $H(x,t)=ta_0 + (1-t)g(x)$. We have that $H$ is continuous, $H(x,0)=g(x)$, and $H(x,1)=a_0$. Thus $H$ is a path-homotopy from $g$ to the constant loop at $a_0$. This gives us that $\pi_1(A,a_0)=\{0\}$ (the trivial group), and so $A$ is simply connected.
\end{enumerate}
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\includegraphics[scale=0.5]{starconvex.jpg}
\end{proof}
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\begin{question}\rm{
Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$; let $\beta$ be a path in $X$ from $x_1$ to $x_2$. Show that if $\gamma=\alpha *\beta$, then $\hat{\gamma}=\hat{\beta} \circ \hat{\alpha}$.
}
\end{question}
\begin{proof} Let $[f]\in \pi_1(X, x_0)$. We show that $\hat{\gamma}([f])=\hat{\beta} \circ \hat{\alpha}([f])$.
$$\begin{array}{rll}
\hat{\gamma}([f]) & = [\bar{\gamma}]*[f]*[\gamma] & \\
& = [\overline{\alpha*\beta}]*[f]*[\alpha*\beta] & \\
& = [\bar{\beta}*\bar{\alpha}]*[f]*[\alpha*\beta] & \\
& = [\bar{\beta}]*[\bar{\alpha}]*[f]*[\alpha]*[\beta] & \\
& = [\bar{\beta}]*\hat{\alpha}([f])*[\beta] & \\
& = \hat{\beta}(\hat{\alpha}([f]) & \\
& = \hat{\beta} \circ \hat{\alpha}([f]) & \\
\end{array}$$
Therefore $\hat{\gamma}=\hat{\beta} \circ \hat{\alpha}$.
\end{proof}
\newpage
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\begin{question}\rm{
Let $x_0$ and $x_1$ be points of the path-connected space $X$. Show that $\pi_1(X,x_0)$ is abelian if and only if for every pair $\alpha$ and $\beta$ of paths from $x_0$ to $x_1$, we have $\hat{\alpha}=\hat{\beta}$.
}
\end{question}
\begin{proof} $(\Rightarrow)$\\
Assume $\pi_1(X,x_0)$ is abelian. We need to show for $g\in \pi_1(X,x_0)$ that $\hat{\alpha}([g])=\hat{\beta}([g])$. Recall that by $\bar{\beta}$ we mean the inverse path to $\beta$.
$$\begin{array}{rll}
\hat{\alpha}([g]) & = [\bar{\alpha}]*[g]*[\alpha] & \\
& = [\bar{\alpha}]*[g]*[\alpha]*[\bar{\beta}]*[\beta] & \emph{\rm{Since }}[\bar{\beta}]*[\beta]\emph{\rm{ is in the identity class of }}\pi_1(X,x_0)\\
& = [\bar{\alpha}]*[g]*([\alpha]*[\bar{\beta}])*[\beta] & \\
& = [\bar{\alpha}]*([\alpha]*[\bar{\beta}])*[g]*[\beta]& \emph{\rm{Since }} [\alpha]*[\bar{\beta}]\in \pi_1(X,x_0) \emph{\rm{ and }}\pi_1(X,x_0)\emph{\rm{ is abelian.}} \\
& = ([\bar{\alpha}]*[\alpha])*[\bar{\beta}]*[g]*[\beta] & \\
& = [\bar{\beta}]*[g]*[\beta] & \\
& = \hat{\beta}([g]) & \\
\end{array}$$
Thus $\hat{\alpha}=\hat{\beta}$.\\
\\
$(\Leftarrow)$\\
Now assume that for any two paths $\alpha$ and $\beta$ from $x_0$ and $x_1$, We have that $\hat{\alpha}=\hat{\beta}$. Now take $f,g\in\pi_1(X,x_0)$. Let $\alpha$ be any path from $x_0$ to $x_1$ and define $\beta=g* \alpha$ which is a path from $x_0$ to $x_1$. Since $\hat{\alpha}=\hat{\beta}$ we get that:
$$\begin{array}{rll}
\hat{\alpha}([g]*[f]) & = \hat{\beta}([g]*[f]) & \\
& = [\overline{g*\alpha}]*[g]*[f]*[g*\alpha] & \\
& = [\bar{\alpha}*\bar{g}]*[g]*[f]*[g*\alpha] & \\
& = [\bar{\alpha}]*[\bar{g}]*[g]*[f]*[g]*[\alpha] & \\
& = [\bar{\alpha}]*[f]*[g]*[\alpha] & \\
& = \hat{\alpha}([f]*[g]) & \\
\end{array}$$
This gives us that $\hat{\alpha}([g]*[f])=\hat{\alpha}([f]*[g])$. But $\hat{\alpha}$ is an isomorphism, so $[g]*[f]=[f]*[g]$. Therefore $\pi_1(X,x_0)$ is abelian.
\end{proof}
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\begin{question}\rm{
Let $A\subset X$; suppose $r:X\to A$ is a continuous map such that $r(a)=a$ for each point $a\in A$. (The map $r$ is called a \textbf{\textit{retraction}} of $X$ onto $A$.) If $a_0\in A$, show that
\begin{equation}\nonumber
r_*;\pi_1(X,a_0)\to \pi_1(A,a_0)
\end{equation}
is surjective.
}
\end{question}
\begin{proof}
To show that $r_*;\pi_1(X,a_0)\to \pi_1(A,a_0)$ is surjective, we will show that for $[f]\in\pi_1(A,a_0)$ there exist $[g]\in\pi_1(X,a_0)$ such that $r_*([g])=[f]$. This will follow from the fact that since $A\subset X$, then $f$ is also a loop in $X$ so consider the equivalence class $[f]\in\pi_1(X,a_0)$. With this and the fact that $r(a)=a$ for all $a\in A$, we have that $r_*([f]) = [r\circ f]= [f]$, and thus $r_*$ is surjective.
\end{proof}
\newpage
\textbf{Question.} Which capital english letters are simply connected? Which ones have $\pi_1=\mathbb{Z}$?\\
\\
For simply connected we need all loops at a give point to be path-homotopic to the constant loop at that point. One way of looking at this is that the letter must not have any ``holes''. The set of simply connected capital english letters is
$$\{C,E,F,G,H,I,J,K,L,M,N,S,T,U,V,W,X,Y,Z\}$$
Which leaves the remaining letters as possible candidates for $\pi_1=\mathbb{Z}$ (Since any simply connected set will have $\pi_1=\{0\}$, the trivial group). Of these the following are the capital english letters with $\pi_1=\mathbb{Z}$.
$$\{A,D,O,P,Q,R\}$$
And since $B$ has two ``holes'' it will have a different fundamental group.
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