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\begin{document}
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\begin{center}
Math 205B - Topology\\
\emph{ }\\
Dr. Baez\\
\emph{ }\\
January 19, 2007\\
\emph{ }\\
Christopher Walker\\
\emph{ }\\
\end{center}
\addtocounter{section}{53}
\begin{theorem}
The map $p:\mathbb{R}\to S^1$ given by the equation
$$p(x)=(cos(2\pi x), sin (2\pi x))$$
is a covering map
\end{theorem}
\begin{proof}
First $p$ is continuous since it is the product for continuous functions.
Consider the following open subsets of $S^{'}$.
$$\begin{array}{l}
U_1=\{(cos(2\pi x), sin (2\pi x))\mid x\in(-\frac{1}{4},\frac{1}{4})\}\\
U_2=\{(cos(2\pi x), sin (2\pi x))\mid x\in(\frac{1}{4},\frac{3}{4})\}\\
U_3=\{(cos(2\pi x), sin (2\pi x))\mid x\in(0,\frac{1}{2})\}\\
U_4=\{(cos(2\pi x), sin (2\pi x))\mid x\in(\frac{1}{2},1)\}\\
\end{array}$$
These open set totally cover $S^{'}$ so we only need to show that each one is evenly covered. Consider the slices of $\mathbb{R}$ to be.
$$\begin{array}{l}
V_{n_1}=(n-\frac{1}{4}, n+\frac{1}{4})\\
V_{n_2}=(n+\frac{1}{4}, n+\frac{3}{4})\\
V_{n_3}=(n, n+\frac{1}{2})\\
V_{n_4}=(n-\frac{1}{2}, n)\\
\end{array}$$
Where the $V_{n_i}$'s corresponds to $U_i$. These evenly cover there corresponding $U_i$ since at least one of $cos(2\pi x)$ or $sin (2\pi x)$ is monotone on these intervals, The end points of the $V_{n_i}$'s map to the end points of $U_i$ and are thus surjective by the Intermediate Value Theorem, and thus a homeomorphism.
\end{proof}
\newpage
\addtocounter{definition}{2}
\begin{theorem}
If $p: E\to B$ and $p':E^{'}\to B^{'}$ are covering maps, then
$$p\times p' :E\times E^{'}\to B\times B^{'}$$
is a covering map.
\end{theorem}
\begin{proof}
Let $b\in B$ and $b'\in B^{'}$, and let $U$ and $U^{'}$ be the evenly covered neighborhoods of $b$ and $b'$. Also, let $\{V_\alpha\}$ and $\{V_\beta^{'}\}$ be the slices of $p^{-1}(U)$ and $p^{-1}(U^{'})$. Consider the inverse image of $U\times U^{'}$ under $p\times p'$.
$$\begin{array}{rll}
(p\times p')^{-1}(U\times U^{'} & = p^{-1}(U)\times p^{-1}(U^{'}) & \\
& = \cup V_\alpha \times \cup V_\beta^{'} & \\
& = \cup ( V_\alpha \times V_\beta^{'}) & \\
\end{array}$$
Since each of the families $\{V_\alpha\}$ and $\{V_\beta^{'}\}$ are disjoint and open, then the family $\{V_\alpha \times V_\beta^{'}\}$ is also disjoint and open, and since $p$ and $p'$ are homeomorphisms from the respective slices to the evenly covered neighborhoods, then $p\times p'$ is a homeomorphism from $V_\alpha \times V_\beta^{'}$ to $U\times U^{'}$, and so $p\times p'$ is a covering map.
\end{proof}
\newpage
\addtocounter{definition}{-3}
\begin{exercise}\rm{
Let $Y$ have the discrete topology. Show that if $p: X\times Y\to X$ is projection on to the first coordinate, then $p$ is a covering map.}
\end{exercise}
\begin{proof}
First, $p$ is continuous since for any open set $U\in X$, $p^{-1}(U) = \{ U\times V\mid V\subseteq Y\}$. Since $Y$ has the discrete topology, then all $V\subseteq Y$ are open, and so $p^{-1}(U)$ is open since it is a union of open set, thus $p$ is continuous. $p$ is surjective since it is a projection. \\
We will show that $X$ itself is evenly covered to prove $p$ is a covering map. We can write $p^{-1}(X)$ as the disjoint union of open sets $V_y=X\times \{y\}$ for all $y\in Y$. $p$ restricted to $V_y$ is already surjective and continuous, so we only need to show injective and $p^{-1}$ (the canonical embedding) is continuous for each $V_y$ to get our homeomorphism. Let $(x_1, y), (x_2, y)\in X\times Y$. With this, $p(x_1, y_1)= p(x_2, y_2)$ implies $x_1=x_2$, which gives us $(x_1, y)= (x_2, y)$. Finally for any open set $U\times W\subseteq X\times Y$, we have the pre-image to be the open set $U$, so $p^{-1}$ is continuous. Thus $p$ restricted to $V_y$ is a homeomorphism, so $p$ is a covering map.
\end{proof}
\newpage
\addtocounter{definition}{1}
\begin{exercise}\rm{
Let $p: E\to B$ be a covering map; Let $B$ be connected. Show that if $p^{-1}(b_0)$ has $k$ elements for some $b_0\in B$, then $p^{-1}(b)$ has $k$ elements for every $b\in B$. In such a case, $E$ is called a \textit{\textbf{k-fold covering}} of $B$.
}
\end{exercise}
\begin{proof}
Since $B$ is connected than the only subsets of $B$ that are both open and closed are $B$ and $\emptyset$. We will show that the set $A_k=\{b\in B\mid p^{-1}(b) \emph{\rm{has \emph{k} elements}}\}$ is non-empty, open, and closed, and thus equal to $B$.\\
First, $A\neq\emptyset$ since $b_0\in A$. Next since $p$ is a covering map, then for the each point $b\in B$, there exist a neighborhood $U_b$ that is evenly covered. If $p^{-1}(b)$ contains $k$ elements, then there are $k$ slices of $E$ that are each homeomorphic to $U_b$, so each element of $U_b$ has $k$ elements in its pre-image, thus $U_b$ is contained in $A_k$. This gives us that $A_k$ is the union of these evenly covered neighborhoods, so $A_k$ is open.\\
$B$ is the disjoint union of a family of sets $\{A_j\}$ where each $A_j$ is defined the same way as $A_k$ above. Since each of these is open by the argument above, we have that $B-A_k$ is open, and so $A_k$ is closed. Therefore $B=A_k$.
\end{proof}
\newpage
\addtocounter{definition}{1}
\begin{exercise}\rm{
Show that the map of Example 3 is a covering map. Generalize to the map $p(z)=z^n$.
}
\end{exercise}
\begin{proof}
We will generalize since this proof holds for $n=2$. Let $p(z)=z^n$. $p$ is easily seen to be continuous and surjective. Consider the points of $S^{'}$ to be complex numbers of the form $z=e^{\theta i}$. We will split this into two cases. \\
Case 1: $z=1$ \\
For $z=1$ let $U=S^{'}-\{-1\}$. We will consider the disjoint family of sets in the domain to be $V_j=\{e^{\theta i}\mid \frac{2(j-1)\pi}{n}-\frac{\pi}{n}<\frac{2(j-1)\pi}{n}<\frac{2(j-1)\pi}{n}+\frac{\pi}{n}\}$ for $j=1,...,n$. $p$ restricted to $V_j$ is injective since $z^n$ is monotone over an interval smaller than $\frac{2\pi}{n}$. It is surjective since the endpoint each map to -1 in $S^{'}$, so by the Intermediate Value Theorem each point has a pre-image. The inverse $p^{'}=z^{\frac{1}{n}}$ is also continuous, so $V_j\im U$ for each $j$.\\
Case 2: $z\neq1$\\
For $z\neq 1$ let $U=S^{'}-{1}$. For this we have that $V_j=\{e^{\theta i}\mid \frac{2(j-1)\pi}{n}<\theta <\frac{2j\pi}{n}\}$ for $j=1,...,n$. These $V_j$'s are homeomorphic to $U$ by the same argument above.
\end{proof}
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