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% 23/01/2023 11AM

\begin{enumerate}[(i)]
    \setcounter{enumi}{5}
    \item[] 
        \begin{lemma}
            As a set, $\mathbb{RP}^2$ is naturally
            in bijection with the set of straight
            lines through $0$.
        \end{lemma}

        \begin{proof}
            Any straight line through $0 \in
            \RR^3$ meets $S^2$ in exactly a pair
            of antipodal points and each pair
            determines a straight line.
        \end{proof}

        \begin{lemma}
            $\mathbb{RP}^2$ is a topological
            surface with the \emph{quotient
            topology}.
        \end{lemma}

        \noindent
        Recall: Quotient topology: $q \colon X \to
        Y$, $V \subset Y$ is open if and only if
        $q^{-1}V \subset X$ is open in $X$ (so $q$
        is continuous).

        \begin{proof}
            First we check that $\mathbb{RP}^2$ is
            Hausdorff. If $[p] \neq [q] \in
            \mathbb{RP}^2$, then $\pm p, \pm q \in
            \S^2$ are distinct antipodal pairs.
            Take small open discs centred at $p$
            and $q$ and their antipodal image as
            in picture
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/7c9b17dc-5d64-4a5e-9c48-0f24af3029fb.png}
            \end{center}
            This gives open neighbourhood of $[p]$ and $[q]$
            in $\mathbb{RP}^2$. $q \colon S^2 \to
            \mathbb{RP}^2$, $q(B_\delta(p))$ is \emph{open} since
            \[ r^{-1}(q(B_\delta(p)) = B_\delta(p) \cup
            (-B_\delta(p)) \]
            $\mathbb{RP}^2$ is second countable:
            Let $\mathcal{U}_0$ be a countable
            base for $S^2$ and let
            \[ \ol{\mathcal{U}} = \{q(u) \colon U
            \in \mathcal{U}\} \]
            $q(u)$ is open:
            \[ q^{-1}(qU) = U \cup (-U) \]
            $\ol{\mathcal{U}}$ is clearly countable
            since $\mathcal{U}$ is Take $V \subset
            \mathbb{RP}^2$ open. By definition,
            $q^{-1} V$ is open in $S^2$ hence
            $q^{-1} V = \bigcup_\alpha U_\alpha$,
            $U_\alpha \in \mathcal{U}$.
            \[ V = q(q^{-1}V) = q \left(
            \bigcup_\alpha U_\alpha \right) =
            \bigcup_\alpha q(U_\alpha) \]
            Finally, let $p \in S^2$ and $[p] \in
            \mathbb{RP}^2$ its image. Let $\ol{D}$
            be a small closed disc neighbourhood
            of $p \in S^2$.
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/3d6634fa-6739-4221-80fa-41b90338b6a9.png}
            \end{center}
            $q |_{\ol{D}} \colon \ol{D} \to
            q(\ol{D}) \subset \mathbb{RP}^2$ is
            injective and continuous, from a
            compact space to a Hausdorff space.
            (Recall ``Topological inverse
            function theorem'').
            A continuous bijection from a compact
            space to a Hausdorff space is a
            \emph{homemorphism} (Analysis and
            Topology). So $q |_{\ol{D}} \colon
            \ol{D} \to q(\ol{D})$ is a
            homeomorphism. This induces a
            homemorphism
            \[ q|_D \colon D \to q(D) \subset
            \mathbb{RP}^2 \]
            ($D$ is open disc). So $[p] \in q(D)$
            has an open neighbourhood in
            $\mathbb{RP}^2$ homeomorphic to an
            open disc and we're done.
        \end{proof}
    \item Let $S^1 = \{z \in \ZZ \colon |z| =
        1\}$. The \emph{forms} $S^1 \times S^1$
        with the subspace topology of $\CC^2$
        (check: is the same as product topology).

        \begin{lemma}
            The forms is a topological surface.
        \end{lemma}

        \begin{proof}
            $\RR^2 \stackrel{e}{\to} S^1 \times
            S^1 \subset \CC \times \CC$, $(s, t)
            \mapsto (e^{2\pi is}, e^{2\pi i t})$.
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/e404dbe1-771a-47a7-bdbc-4e9057b4dbea.png}
            \end{center}
            Equivalence relation on $\RR^2$ given
            by translating by $\ZZ^2 = \ZZ \times
            \ZZ$. The map
            \[ [0, 1]^2 \injto \RR^2
            \stackrel{q}{\to} \RR^2 / \ZZ^2 \]
            is onto, so $\RR^2 / \ZZ^2$ is
            compact. Note that $\hat{e}$ is a
            continuous bijection. By Topological
            Inverse Function Theorem, $\hat{e}$ is
            a homeomorphism.
            \begin{note*}
                We already know that $S^1 \times
                S^1$ is compact, Hausdorff and
                second countable (closed and
                bounded in $\RR^4$).
            \end{note*}
            As for the
            case of $S^2 \to \mathbb{RP}^2$,
            pick $[p] \in q(p)$, $p \in \RR^2$
            and small closed disc $\ol{D}(p)
            \subset \RR^2$ such that $\forall
            (u, v) \neq (0, 0) \in \ZZ^2$,
            $\ol{D}(p) \cap (\ol{D}(p) + (u,
            v)) = \emptyset$.
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/1c078dd9-ed7b-4480-bf39-43d6725a867b.png}
            \end{center}
            Then $e|_{\ol{D}(p)}$ and
            $q|_{\ol{D}(p)}$ are injective.
            Now restricting to the open disc
            as before, we get an open disc
            neighbourhood of $[p] \in \RR^2 /
            \ZZ^2$ hence $S^1 \times S^1$ is a
            topological surface.
        \end{proof}
        \myskip
        Another viewpoint: $\RR^2 / \ZZ^2$ is
        also given by imposing on $[0, 1]^2$
        the equivalence relation $(x, 0) \sim
        (x, 1) ~\forall 0 \le x \le 1$, $(0,
        y) \sim (1, y) ~\forall 0 \le y \le
        1$.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/6c536ecc-c347-4c21-a270-75bbf13a1cd9.png}
        \end{center}
    \item Let $P$ be a planar Euclidean polygon
        (including interior). Assume the edges are
        \emph{oriented} and \emph{paired} and for
        simplicity assume Euclidean lengths of $e$
        and $\hat{e}$ are equal if $\{e,
        \hat{e}\}$ are paired.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/0398dd4c-d403-4f07-9264-34b52cae1f43.png}
        \end{center}
        Label by letters and describe orientation
        by a sign $\pm$ relative to the clock with
        orientation of $\RR^2$.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/f6f70541-be71-45d3-91d6-fff217c67382.png}
        \end{center}
        If $\{e, \hat{e}\}$ are parallel edges,
        there is a unique isometry from
        from $e$ to $\hat{e}$ respecting their
        orientation, say
        \[ f_{e\hat{e}} \colon e \to \hat{e} \]
        These maps generate an equivalence
        relation on $P$ where we identify $x \in
        \partial P$ with $f_{e\hat{e}}(x)$
        whenever $x \in e$.

        \begin{lemma}
            $P / \sim$ (with the quotient
            topology) is a topological surface.
        \end{lemma}
\end{enumerate}