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\begin{document}
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\begin{center}
Math 205B - Topology\\
\emph{ }\\
Dr. Baez\\
\emph{ }\\
February 16, 2007\\
\emph{ }\\
Christopher Walker\\
\emph{ }\\
\end{center}

\addtocounter{section}{58}
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\begin{exercise}\rm{
(Pointed version.) 
Show that given a collection $\mathcal{C}$ of pointed spaces, 
the relation of homotopy equivalence is an equivalence relation 
on $\mathcal{C}$.
}
\end{exercise}

\begin{proof}
Recall that two pointed spaces $(X,x_0)$ and $(Y,y_0)$ are homotopy
equivalent if there exist maps $f\colon (X,x_0)\to Y$ and $g\colon
(Y,y_0)\to (X,x_0)$ where $f\circ g\simeq 1_Y$ and $g\circ f\simeq
1_X$.  Here the relation `$\simeq$' is the appropriate concept of
`homotopic' for pointed maps: that is, $h \simeq h'$ if that there
exists a {\it pointed} homotopy from $h$ to $h'$.  We need to check
that homotopy equivalence is reflexive, symmetric, and transitive.
\begin{itemize}

\item Reflexive - for all $X\in \mathcal{C}$, $1_X\colon (X,x_0)\to
(X,x_0)$ gives us that $1_X\circ 1_X=1_X$, so $X$ is homotopy
equivalent to $X$.

\item Symmetric - Let $X, Y\in \mathcal{C}$. If $X$ is homotopy
equivalent to $Y$, then by definition of homotopy equivalence, $Y$ is
homotopy equivalent to $X$. This is because the definition is symmetrical.

\item Transitive - Let $X, Y, Z\in \mathcal{C}$ where $X$ is homotopy
equivalent to $Y$, and $Y$ is homotopy equivalent to $Z$. This implies
there exist $f\colon X\to Y$ and $g \colon Y\to X$ with $f\circ
g\simeq 1_Y$ and $g\circ f\simeq 1_X$. Also, there exist $h\colon Y\to Z$
and $k\colon Z\to Y$ with $h\circ k\simeq 1_Z$ and $k\circ h\simeq 1_Y$. Now
consider $h\circ f\colon X\to Z$ and $g\circ k\colon Z\to X$. For each step, we
use the fact that composition preserves the relation $\simeq$.
First
$$\begin{array}{rl}
h\circ f\circ g\circ k & \simeq h\circ 1_Y\circ k\\
                       & \simeq h\circ k\\
                       & \simeq 1_Z\\
\end{array}$$
Similarly we have the other direction.
$$\begin{array}{rl}
g\circ k\circ h\circ f & \simeq g\circ 1_Y \circ f\\
                       & \simeq g\circ f\\
                       & 1_X\\
\end{array}$$
Thus $X$ is homotopy equivalent to $Z$.
\end{itemize}
Therefore homotopy equivalence is an equivalence relation.                       
\end{proof}

\newpage
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\begin{exercise}\rm{
Show that a retract of a contractible space is contractible
}
\end{exercise}

\begin{proof}
Let $A$ be a retract of $X$, and let $X$ be contractible. This means
$1_X$ is homotopic to a constant map, say $f(x)=x_0$. Let
$H\colon X\times I\to X$ be the homotopy from $1_X$ to $f$ with
$H(x,0)=x$ and $H(x,1)=f(x)=x_0$. If $r\colon X\to A$ is the retraction of
$X$ to $A$, then we can consider the homotopy $r\circ H|_A\colon A\times
I\to A$. This is continuous since it is the composition of continuous
maps. Also $r\circ H|_A(x,0)=r(x)=x$ , since $x\in A$, and $r\circ
H|_A(x,1)=r(f(x))=r(x_0)$. This gives us that $1_A$ is homotopic to
the constant map $g(x)=r(x_0)$, and so $A$ is contractible.
\end{proof}

\newpage
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\begin{exercise}\rm{
\begin{enumerate}
\item Show that $\mathbb{R}^1$ and $\mathbb{R}^n$ are not homeomorphic if $n>1$.
\item Show that $\mathbb{R}^2$ and $\mathbb{R}^n$ are not homeomorphic if $n>2$.
\end{enumerate}
}
\end{exercise}

\begin{proof}\emph{ }\\
\begin{enumerate}
\item We will assume that exist a map $h\colon \mathbb{R}^1\to \mathbb{R}^n$ that is a homeomorphism. We will then consider what happens If we remove $0$ in $\mathbb{R}^1$ and its image $h(0)$ from $\mathbb{R}^n$. $\R^1-0$ is a disconnected space, but $\R^n-h(0)$ is connected. Removal of a point and its image always preserves homeomorphism, thus $\R^1-0$ and $\R^n-h(0)$ are homeomorphic. But connectedness is invariant under homeomorphism, so we have a contradiction.
\item We will use a similar technique for this part. Assume that there exist a map $h\colon \R^2\to R^n$ that is a homeomorphism. We will again remove the origin from $\R^2$ and the image $h(0,0)$ from $\R^n$. Removal of a point and its image always preserves homeomorphism, thus $\R^1-(0,0)$ and $\R^n-\{h(0,0)\}$ are homeomorphic. We now have that $\R^2-\{(0,0)\}$ is homeomorphic to $S^1$, and $\R^n-\{h(0,0)\}$ is homeomorphic to $S^{n-1}$. $S^1$ is not simply connected, however $S^{n-1}$ is simply connected for $n>2$. Thus since simply connectedness is invariant under homeomorphism, then $\R^1-(0,0)$ and $\R^n-\{h(0,0)\}$ are not homeomorphic, a contradiction.
\end{enumerate}
\end{proof}

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