% vim: tw=50
% 25/01/2023 11AM

\begin{enumerate}[(i)]
    \setcounter{enumi}{7}
    \vspace{-0.5\baselineskip}
    \item[] The torus $T^2$ as $[0, 1]^2 / \sim$.
        $P = [0, 1] \times [0, 1]$.
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/366be6bc9ca211ed.png}
        \end{center}
        If $p \in \operatorname{interior}(P)$
        we pick $\delta > 0$ small so that
        $\ol{B_\delta(p)}$ lies in the
        $\operatorname{interior}(P)$. Now
        argue as before: the quotient map is
        injective on $\ol{B_\delta(P)}$ and
        homeomorphism on its interior. If $p
        \in \operatorname{edge}(P)$:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/6e27ba869ca211ed.png}
        \end{center}
        Say $p = (0, y_0) \sim (1, y_0) = p'$.
        Take $\delta$ small enough so half discs
        of radius $\delta$ as shown don't meet the
        vertices. Define a map as
        \[ (x, y) \mapsto
        \begin{cases}
            (x, y - y_0) & \text{on $U$} \\
            (x - 1, y - y_0) & \text{on $V$}
        \end{cases} \]
        Recall: If $X = A \cup B$, $A$, $B$ closed
        and $f \colon A \to Y$ and $g \colon B \to
        Y$ are continuous, and $g |_{A \cap B} =
        g|_{A \cap B}$ then they define a
        continuous map on $X$.

        \myskip
        $f_u$ and $f_V$ are
        continuous on $U, V \subset [0, 1]^2$,
        they induce continuous maps on $q_U$ and
        $q_V$. ($q \colon [0, 1]^2 \to [0, 1]^2 /
        \sim$). In $T^2$, $\half$ discs, $q_U$ and
        $q_V$ overlap, but our maps agree as they
        are compatible with the equivalence
        relation. So $f_U$ and $f_V$ ``glue'' to
        give a continuous map on an open
        neighbourhood of $[p] \in T^2$ to $\RR^2$.
        Now the ``usual argument'' (pass to closed
        disc, use Topological Inverse Function
        Theorem pass back to interior) shows that
        if $[p] \in T^2$ lies in
        $\operatorname{edge}(P)$, it has a
        neighbourhood homeomorphic to a disc.

        \myskip
        Finally at a vertex of $[0, 1]^2$
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/201475309ca411ed.png}
        \end{center}
        and analogously we get that a vertex has a
        neighbourhood homeomorphic to a sic and
        $[0, 1]^2 / \sim$ is a \emph{topological
        surface}.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/b6bb7f929ca411ed.png}
        \end{center}
        For a general polygon: \emph{similar
        idea}. Same situation as $T^2$ for
        interior points and points on edges. How
        about vertices?
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/1312e4249ca511ed.png}
        \end{center}
        If $v \in \operatorname{vertex}(P)$ has
        $k$ vertices in its equivalence class,
        there exist $k$ sectors in $P$. Any sector
        can be identified with our favourite
        sector (for example a quarter circle is
        homemorphic to a semi circle). In ($*$) we
        get an open disc neighbourhood of $v$ (red
        dots) via
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/5f364d969ca511ed.png}
        \end{center}
        If we have $k = 1$, we must have either
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/7ed8bf629ca511ed.png}
        \end{center}
        But this quotient space is homemorphic to
        $D^2 \subset \RR^2$. These open
        neighbourhoods of points in $P / \sim$
        show that $P$ is locally homeomorphic to a
        disc.

        \myskip
        Exercise: Check that $P / \sim$ is
        Hausdorff and second countable.
    \item Connected sums: Given topological
        surfaces $\Sigma_1$ and $\Sigma_2$ we can
        remove an open disc from each and glue the
        resulting boundary circles
        \begin{center}
            \includegraphics[width=0.8\linewidth]
            {images/0fe3a0d49ca711ed.png}
        \end{center}
        Explicitly take $\Sigma_1 \setminus D_1
        \ci \Sigma_2 \setminus D_2$ and impose a
        quotient relation $\theta \in \partial D_1
        \sim \theta \in \partial D_2$ where
        $\theta$ parametrisation. $S' = \partial
        D_i$, $\partial D_i$ is the boundary of
        $\partial D_i$. The result $\Sigma_1 \#
        \Sigma_2$ is called the connected sum of
        $\Sigma_1$ and $\Sigma_2$. In principle
        this depends on many choices and it takes
        some effort to prove that it is
        \emph{well-defined}.

        \begin{lemma}
            The connected sum $\Sigma_1 \#
            \Sigma_2$ is a topological surface (no
            proof in this course).
        \end{lemma}

        \noindent
        If you want to learn more:
        \begin{itemize}
            \item Introduction to topological
                manifolds by Jack Lee.
            \item Introduction to smooth manifolds
                by Jack Lee.
        \end{itemize}
\end{enumerate}