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\begin{document}
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\section*{Further examples}


    \subsection* {1. Projective space}
    \textbf{(a)} Define an equivalence relation $\thicksim$ on
    $\mathbb{R}^{n+1} \backslash \{0\}$ by
    $$x=(x_1,\ldots,x_{n+1}) \thicksim y=(y_1,\ldots,y_{n+1})$$
    if and only if $x=\lambda y$
    for some $\lambda \in \mathbb{R}\backslash \{0\}$.\\
    Now let $\mathbb{R}P^n :=\mathbb{R}^{n+1} / \thicksim$.
    Denote equivalence class of $x$ by
    $[x]=[x_1,x_2,\ldots,x_{n+1}]$. \\
    This is a manifold, in fact $C^\infty$-manifold. \\
    Consider
    $U_1 = \{[x] = [x_1,\ldots,x_{n+1}] : x_1 \neq 0\} $ and
    define $\phi_1 : U_1 \rightarrow \mathbb{R}^n$ by $$
    [x_1,\ldots,x_{n+1}] \mapsto \left(
    \frac{x_2}{x_1},\ldots, \frac{x_{n+1}}{x_1} \right).$$\\
    We claim that $\phi_1$ is a homeomorphism:\\
    Let $V_1=\{x=(x_1,\ldots,x_{n+1}) : x_1 \neq 0 \} \subset
    \mathbb{R}^{n+1} \backslash \{0\}$ , and define
    $\psi_1 : V_1 \rightarrow \mathbb{R}^n $ by
    $$ (x_1,\ldots,x_{n+1}) \mapsto  \left( \frac{x_2}{x_1},\cdots,\frac{x_{n+1}}{x_1} \right).$$
    Since $U_1$ is a quotient space of $V_1$ and $\phi_1$
    is induced by $\psi_1$, $\phi_1$ is a continuous map.
    Let $q : V_1 \rightarrow U_1$ be a quotient map and
    let $\sigma_1 : \mathbb{R}^n \rightarrow V_1$ be given by $(y_1,\ldots,y_n)
\mapsto (1,y_1,\ldots,y_n ) $.
    then
    $\phi_1 \circ(q\circ\sigma_1) = id_{\mathbb{R}^n}$ and
    $(q\circ\sigma_1)\circ\phi_1 = id_{\mathcal{U}_1}$.
    Hence $\phi_1$ is a homeomorphism.
    \\\\
    HW Show that $\mathbb{R}P^n$ is a $C^\infty$-manifold.\\\\

    \textbf{(b)}
    $\hspace{14em} i$

    $\hspace{14em} S^n\hookrightarrow \bR^{n+1}\backslash\{0\}$

    $\hspace{14em} \downarrow q\hspace{2em}\downarrow q$

    $\hspace{13em} S^n/\thicksim\rightarrow\bR\mathbf{P}^n$

    $\hspace{16em} \overline{i}$\\
    $\thicksim$ : antipodal identification,
    where antipodal map $A:S^n\rightarrow S^n$ is defined by $x\mapsto
    -x$. Equivalence relation on $S^n$ is given by $x\thicksim
    -x=Ax$.\\
    Then there exists a well-defined bijective continuous map $\overline{i}$,
    because $i$ is an embedding and $q$ is a quotient map. Since $S^n$
    is compact and $\bR\mathbf{P}^n$ is Hausdorff, $\overline{i}$ is a
    homeomorphism.\\\\
    (dim 1) $\bR\mathbf{P}^1=S^1/\thicksim=S^1$\\
    The quotient map may
    be given by $z\mapsto z^2$.\\\\
    (dim 2)
%    \parpic{\includegraphics[width=0.9\textwidth]{1.eps}}\\

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    $\bR\mathbf{P}^2$ with one point deleted is homeomorphic to open
    M\"{o}bius band.


    \subsection* {2. Product manifold}
    $M^m , N^n$ : manifolds $\Rightarrow$ $M\times N$ is an
    $(m+n)$-manifold.\\
    $\because \forall (p,q)\in M\times N$,  $p\in M$ has
    a coordinate neighborhood $(U,\phi)$  homeomorphic to an open subset of $\mathbb{R}^m$ and $q
    \in N$  has a coordinate neighborhood $(V,\psi)$ homeomorphic to an open subset of
    $\mathbb{R}^n$. It follows that $(p,q)$ has a coordinate neighborhood $(U \times
    V,\phi\times\psi)$
    homeomorphic to an open subset of $\mathbb{R}^m \times \mathbb{R}^n = \mathbb{R}^{m+n}$. \\

    \begin{ex}
    $\mathbb{R}^2=\mathbb{R}^1\times\mathbb{R}^1$, $T^2=S^1\times S^1$,  $T^3=S^1\times S^1\times
    S^1$, $\ldots$
    \end{ex}

    \subsection* {3. Other constructions}
    Connected sum, boundary identification, $\ldots$
    \begin{ex} $S^3$ can be obtained by identifying the boundaries
    of  two solid tori as follows:\\
    $D^2\times D^2=D^4$\\
    $\Rightarrow\partial(D^2\times D^2)=\partial(D^4)=S^3$\\
    $\Rightarrow(\partial D^2\times D^2)\bigcup(D^2\times\partial D^2)=(S^1\times
    D^2)\bigcup(D^2\times S^1)=S^3$\\
    \end{ex}
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    The picture shows the decomposition of $S^3$ as a union of two solid tori.\\\\
    HW Can you decompose $S^3$ as a union of two handle bodies of genus 2?


    \subsection* {4. Lie group}
    \begin{defn} A topological space $X$ is a \textbf{topological group} if

    $\hspace{2em} 1.$ $X$ is a group.

    $\hspace{2em} 2.$ $\mu : X \times X \rightarrow X$ given by
    $(x,y)\mapsto xy^{-1}$ is continuous.
    \end{defn}
    \begin{ex} Topological group
    \begin{enumerate}
    \item Any group $G$ with discrete topology.
    \item $\bR^n$ : additive group. \\$\because (x,y)\mapsto x-y$ is
    continuous.
    \item $S^1\subset \mathbb{C}$ is a multiplicative group.\\
    $\because S^1\times S^1\rightarrow S^1$ by $(z,w)\mapsto
    {{z}\over{w}}$ is continuous.
    \item General linear group
    $Gl(n,\mathbb{R})=\{A\in
    M_n(\mathbb{R}):detA\neq0\}\subset\mathbb{R}^{n^2}$\\
    $\because$ The map given by $(A,B)\mapsto AB^{-1}$ is
    continuous.
    \end{enumerate}
    \end{ex}
    \begin{defn} A \textbf{\textit{Lie group}} is a topological group $X$
    which is a smooth n-manifold such that $\mu$ is $C^{\infty}$
    \end{defn}

    Above examples are all \textit{Lie groups}.

    \subsection* {5. Manifold with boundary}
    \begin{defn} A Hausdorff space $M$ is an n-manifold with
    boundary if $\forall p\in M, \exists$ a coordinate chart $(U,
    \phi)$ of $p$ which is homeomorphic to either $\mathbb{R}^n$ or
    $\mathbb{H}^n$, where
    $\mathbb{H}^n=\{x=(x_1,\ldots,x_n)\in\mathbb{R}^n:x_n\geq0\}$,
    with $\phi (p)=0$.
    $\partial M=\{x\in M:x$ has a coordinate
    neighborhood homeomorphic to $\mathbb{H}^n\}$ is called a boundary of
    $M$.
    \end{defn}

    The notion of boundary point is well-defined by the following
    theorem.

    \begin{thm} \textbf{(Invariance of domain)} Let
    $U\subset\bR^n$ be an open set and $h:U\rightarrow\bR^n$
    be 1-1 and continuous map. The $h(
    U)$ is open in $\bR^n$.
    \end{thm}


    Invariance of domain implies that the image of an interior point by
    1-1 and continuous map is also an interior point and the image of
    a boundary point is also a boundary point.\\

    If $M$ is an $n$-manifold, then $\partial M$
    is an $(n-1)$-manifold.


\end{document}