\documentclass[12pt ]{article}
\setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true
cm}

\usepackage{hangul}
\usepackage{amscd,amsmath}
\usepackage{amsfonts}
\usepackage{amssymb,theorem}
\usepackage{longtable}
\usepackage{floatflt}
\usepackage{texdraw, epic, pstcol ,pstricks ,pst-3d, pst-poly,pst-grad,pst-node, pst-text}
\usepackage{floatflt}
\usepackage[arrow,matrix]{xy}
\usepackage{graphicx}

\newcommand{\Aff}{\mbox{\it Aff}}
\newcommand{\aff}{\mbox{\it aff}}


\newcommand{\alp}{\alpha}
\newcommand{\bet}{\beta}
\newcommand{\del}{\delta}
\newcommand{\gam}{\gamma}
\newcommand{\vep}{\varepsilon}
\newcommand{\eps}{\epsilon}
\newcommand{\lam}{\lambda}
\newcommand{\kap}{\kappa}
\newcommand{\sig}{\sigma}
\newcommand{\ome}{\omega}
\newcommand{\Gam}{\Gamma}
\newcommand{\Ome}{\Omega}
\newcommand{\Sig}{\Sigma}
\newcommand{\Del}{\Delta}
\newcommand{\Lam}{\Lambda}
\newcommand{\disi}{\displaystyle{\sum_{i=1}^n}}
\newcommand{\disj}{\displaystyle{\sum_{j=1}^m}}
\newcommand{\rb}{\mathbb{R}}
\newcommand{\zb}{\mathbb{Z}}
\newcommand{\bd}{\partial}
\newcommand{\key}[1]{\emph{\textbf{#1}}}
\newcommand{\rh}[1]{\widetilde{H}(#1)}
\newcommand{\hh}{\widetilde{H}}
\newcommand{\hhp}{\widetilde{H}_p}
\newcommand{\sn}{S^{n-1}}
\newcommand{\snn}{S^{n}}
\newcommand{\dn}{D^{n}}
\newcommand{\ttl}{T^{2}}
\newcommand{\is}{i_{1*}}
\newcommand{\sm}{S^n-s^r}

\newtheorem{thm}{����}
\newtheorem{cor}[thm]{��������}
\newtheorem{lem}[thm]{��������}
\newtheorem{prop}[thm]{����}
\newtheorem{cl}{Claim}


{\theorembodyfont{\rm}
\newtheorem{ex}{��}
\newtheorem{que}{����}
\newtheorem{notation}{Notation}[section]
\newtheorem{defn}{����}
\newtheorem{rem}{��}
\newtheorem{note}{Note}
}
\renewcommand{\thenote}{}
\renewcommand{\therem}{}

\newenvironment{pf}{{\bf ����}}{\hfill\framebox[2mm]{}}

\begin{document}

\parindent=0cm
\section*{I.3 Boundary identification of two manifolds $M_{1}, M_{2}$ with $f:\bd M_{1}
\overset{\cong}{\rightarrow} \bd M_{2}$}

{\bf Fact} Let $M$ be a topological (or $\mathcal{C}^{\infty}$)
manifold with $\bd M \neq \emptyset$. Then there exists an open
neighborhood $U$ of $\bd M$ such that $U
\underset{\underset{h}{\leftarrow}}{\cong}
\bd M \times [0,1)$ with $h(x,0) = x , \hspace{0.3em} x \in \bd M$.\\
$U$ is called a collar.\\

{\bf The idea of proof}\\
Use local collar and splice together using Partition of Unity.\\
(See Vick(homology theory) for topological case and
Milnor($h$-cobordism) for $\mathcal{C}^{\infty}$ case.)\\

Let $(X,A) = (M_{1}, \bd M_{1})$ (a collared pair), $Y=M_{2}$ and
let $f: \bd M_{1} \rightarrow \bd M_{2}$ be a homeomorphism. Then
we obtain a manifold $W=M_{1} \bigcup_{f} M_{2}$ identifying the
boundaries of $M_{1}$ and
$M_{2}$ via $f$.\\

{\bf Examples}\\
{\bf 1.(Sphere)}\\
Let $f:\sn \rightarrow \sn$ be a homeomorphism.\\ Then $W = \dn
\bigcup_{f} \dn \cong \snn$(homeomorphism). But, not
diffeomorphism, in general, for $\mathcal{C}^{\infty}$-category.\\

\begin{pf}
Let $\bar{f} : \dn \overset{\cong}{\rightarrow} \dn$ be an
extension of $f$.\\(Such extension always exists, e.g., a "radial extension".)\\

\[
\xymatrix @=1em @*[c] { %
\dn \coprod \dn \ar[d]^{p} \ar[r]^{\bar{f}\coprod id.} & \dn
\coprod \dn \ar[d]^{p} & & x,f(x) \ar[r] \ar[d] & f(x),f(x)\ar[d]\\
\dn \bigcup_{f} \dn \ar@{.>}[r] & \dn \bigcup_{id.} \dn = \snn  &
& x \sim f(x) & f(x) \sim f(x)  }
\]



������ �׸����� ������ map�� ������ �׸��� ���� $\dn \bigcup_{f}
\dn$�� $x \sim f(x) , x \in \bd D^{n}(= S^{n-1})$�� $\dn
\bigcup_{id.} \dn$�� $f(x) \sim f(x)$�� �������Ƿ� �� ���ǵȴ�.
���� �ڸ��� continuous, one-to-one �׸��� onto������ ��������
homeomorphism�� �ȴ�.

\end{pf}

Exactly same argument shows the following.\\

{\bf Theorem A}\\
Given $M_{i}$ with $\bd M_{i} \neq \emptyset$, if $f,g : \bd M_{1}
\overset{\cong}{\rightarrow} \bd M_{2}$ such that $g^{-1}\circ f :
\bd M_{1} \rightarrow \bd M_{1}$ can be extended to a
homeomorphism $M_{1} \rightarrow M_{1}$, then $M_{1} \bigcup_{f}
M_{2} \cong M_{1}
\bigcup_{g} M_{2}$.\\


\begin{pf}

\[
\xymatrix @=1em @*[c] { %
M_{1} \coprod M_{2} \ar[d]^{p} \ar[r]^{\overline{g^{-1}\circ f} \coprod
id.} & M_{1} \coprod M_{2} \ar[d] ^{p} & \hspace*{1.0em} &
x,f(x) \ar[d] \ar[r] & g^{-1}\circ f(x), f(x) \ar[d] \\
M_{1} \bigcup_{f} M_{2} \ar@{.>}[r]^{\phi} & M_{1} \bigcup_{g}
M_{2} & \hspace*{1.0em} & x \sim f(x) & f(x) \sim f(x) }
\]

������ �׸����� ���� $\phi$�� ������ �׸����� ������ ���� ����
������ �� ���ǵȴ�. ���� $f^{-1} \circ g = (g^{-1}\circ
f)^{-1}$�̹Ƿ� extension�� ������ ��ȿ�ϰ� ���� �������
$\overline{f^{-1} \circ g} = (\overline{g^{-1} \circ f})^{-1}$��
$\psi = \phi^{-1}$�� ���� �� �ִ�.\\

\end{pf}

{\bf 2.(Lens space)}\\
Let $M_{i} = D^{2} \times S^{1}$ with $\bd M_{i} = S^{1} \times
S^{1} = \ttl$. Consider $f : \ttl \overset{\cong}{\rightarrow}
\ttl$ and what
$M_{1} \bigcup_{f} M_{2}$ is.\\

{\bf Theorem B}\\
Let $M_{i}$ with $\bd M_{i} \neq \emptyset$ be manifolds and $f,g
: \bd M_{1} \overset{\cong}{\rightarrow} \bd M_{2}$ be
homeomorphisms isotopic to each other. Then $M_{1} \bigcup_{f}
M_{2} \cong M_{1}
\bigcup_{g} M_{2}$.\\

\begin{pf}
By Theorem A, it suffices to show the following statement : \\
{\bf claim} $f:\bd M \rightarrow \bd M$ isotopic to $id.$. Then
$f$ can be extended to a homeomorphism $\bar{f} : M \rightarrow
M$.\\


{\bf Proof of claim}\\

\psset{unit=0.7cm}
\begin{center}
\begin{pspicture}(-1.0,-2.0)(8.0,4.0)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]

\rput(0,0){
\pspolygon[fillstyle=solid,fillcolor=lightgray](0,0)(1,0)(1,4)(0,4)
\pscircle[fillstyle=solid,fillcolor=white](0.5,2.0){0.2}
\pspolygon[fillstyle=solid,fillcolor=gray](1,0)(2,0)(2,4)(1,4)
\psline[linecolor=red](0.7,2.0)(2,2) \psdots(2,2)
\rput(2.4,2){$x$} \psline{->}(2.3,0.5)(2.0,1.0)
\rput(2.5,0.3){$\small{\bd M}$} \rput(1.3,-0.8){$\small{\bd M
\times I}$} \psline(1,-0.1)(1.5,-0.3)(2,-0.1)
 }

\rput(5.0,0){
\pspolygon[fillstyle=solid,fillcolor=lightgray](0,0)(1,0)(1,4)(0,4)
\pscircle[fillstyle=solid,fillcolor=white](0.5,2.0){0.2}
\pspolygon[fillstyle=solid,fillcolor=gray](1,0)(2,0)(2,4)(1,4)
\psline[linecolor=red](0.7,2.0)(1.0,2.0)(2,3.5) \psdots(2,3.5)
\rput(2.5,3.5){$f(x)$} \psline{->}(2.3,0.5)(2.0,1.0)
\rput(2.5,0.3){$\small{\bd M}$} \rput(1.3,-0.8){$\small{\bd M
\times I}$}\psline(1,-0.1)(1.5,-0.3)(2,-0.1) }

\psline[linestyle=dashed]{->}(2.4,-0.5)(5.0,-0.5)
\rput(3.5,-1.0){$\exists \bar{f}$}

\end{pspicture}
\end{center}

such that $\bar{f}(x,t) = (f_{t}(x),t)$on a collar neighborhood $U \cong M \times [0,1)$ and define $\bar{f} = id.$
on the complement of $U$.

\end{pf}

{\bf Question : } What are homeomorphisms of $T^{2}$ up to
isotopy?\\
{\bf Answer : } Classical result $\Rightarrow Gl(2,\zb)$.\\
In fact, $Homeo(T^{2}) \overset{\phi}{\twoheadrightarrow}
Aut(H_{1}(T^{2}))(=Gl(2,\zb))$\\
\hspace*{7.5em} $f \mapsto f_{\ast}$\\
with $ker(\phi) = Homeo_{0}(T^{2}) =$ homeomorphism
homotopic(isotopic) to $id.$.\\

$\therefore \mathcal{H}(T^{2}) / \mathcal{H}_{0}(T^{2}) \cong
Gl(2,\zb)$.\\

Show $\phi$ is onto ; Given $g \in Gl(2,\zb)$, view $g : \rb^{2}
\to \rb^{2}$ preserving the integral lattice. $\Rightarrow \bar{g} : T^{2}
\to T^{2}$ is induced.\\

�Ϲ������� closed surface ���� "homotopy $\Rightarrow$ isotopy"��
�����ϸ� $ker(\phi)= Homeo_{0}(T^{2})$�� ���� exercise 2(3)��
�̿��Ͽ� $\tilde{f}$�� $id.$���̿� straight line homotopy�� �� ��
�ְ�,
�̰��� $T^{2}$�󿡼��� homotopy �� ���� �� �ִ�.\\

{\bf HW 7} Prove the followings.\\
\begin{prop}

\[
\xymatrix @=1em @*[c] { %
\widetilde{X} & \overset{\tilde{f}}{\rightarrow} & \widetilde{X'} &&
\forall \gam \in \Gamma = \textrm{deck group of $\widetilde{X}$},\\
\downarrow^{p} && \downarrow^{p} & \Leftrightarrow & \exists !
\gam' \in \Gamma' \textrm{such that}\\
X& \overset{\exists f}{\dashrightarrow} & X' && \widetilde{f} \circ \gam =
\gam' \circ \widetilde{f}}
\]

where $p$ is a regular covering.

\end{prop}

\begin{cor}
(1)

\[
\xymatrix @=1em @*[c] { %
\widetilde{X} \ar[d]^{p} \ar[r]^{\widetilde{f}}_{\cong} &
\widetilde{X'} \ar[d]^{p'} & \Leftrightarrow & C_{\widetilde{f}} :
\Gam \overset{\cong}{\rightarrow}
\Gamma'\\
X \ar@{.>}[r]^{\exists f}_{\cong} & X'&&}
\]

(2) ������ $X=X'$�� ��� �º� $\Leftrightarrow \widetilde{f} \in N_{G}(\Gamma),
$ when $G=Aut(\widetilde{X})$.(The type of the automorphisms depends on the category.)\\

(3) Universal covering case�� ��쿡��\\
$\widetilde{f} \in C_{G}(\Gamma)$ ( = Centralizer)
$\Leftrightarrow f_{\ast} = id. : \pi_{1}(X) \rightarrow
\pi_{1}(X)$ when $\widetilde{f}$ fixes a base point.\\

\end{cor}

Q. Suppose ${M_i}^3$ be a 3-manifold with $\bd M_i = T^2,
\hspace{1em} i= 1,2.$ \\ When is $M_1 \underset{f}{\cup} M_2 \cong
M_1 \underset{g}{\cup} M_2 ,$ where $f, g: T^2 \rightarrow T^2 $
are homeomorphism?\\

A. Theorem A $\Rightarrow$ If $g^{-1}\cdot f : T^2 \rightarrow
T^2$ can be extended to a homeomorphism: $M_1 \rightarrow M_1$,
then $M_1
\underset{f}{\cup} M_2 \cong M_1 \underset{g}{\cup} M_2 .$\\

\begin{prop} Suppose $M_1 = D^2 \times S^1 ,$ a solid torus. Then
$h : \bd M_1 = T^2 \rightarrow T^2 $ can be extended to $\bar{h} :
M_1 \rightarrow M_1$ iff $h_* : H_1 (T^2 ) = \mathbb{Z}\cdot m
\oplus \mathbb{Z}\cdot l \rightarrow H_1(T^2)$ is of the form
$\left(\begin
{array}{cc} \pm 1 & c\\
0 & \pm 1 \end{array}\right)$ , equivalently, $h$ sends meridian
to meridian.\end{prop}
\begin{pf} $(\Rightarrow )$
\[
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
H_1(T^2 ) = \mathbb{Z}\oplus\mathbb{Z} = <m,l> \ar[r]^<<<<<<{h_* :
\cong}\ar[d]^{i_*} &
\mathbb{Z}\ar[d]\oplus\mathbb{Z} & m \ar@{|->}[d]\ar@{|->}[r] & am+bl\ar@{|->}[d] \\
H_1(D^2 \times S^1 ) = \mathbb{Z} = <l> \ar[r]^<<<<<<{\bar{h_*}}&
\mathbb{Z} & 0 \ar@{|->}[r] & bl } \]
$\Rightarrow b=0$ and $a = \pm 1 (\because h_* : m \mapsto am :\cong )$\\
$\Rightarrow h_*= \left(\begin{array}{cc} \pm 1 & c\\
0 & d \end{array}\right) $ and $h_*$ is invertible.\\
$\Rightarrow h_*= \left(\begin{array}{cc} \pm 1 & c\\
0 & \pm 1 \end{array}\right)$(��ȣ���� �ƴ�)\\

($\Leftarrow)$ May assume $h$ is a "linear map" through isotopy in
the radial direction. And it is enough to check for $h =
\left(\begin{array}{cc}1&n\\0&1\end{array}\right) ,
\left(\begin{array}{cc}1&0\\0&-1\end{array}\right) , $ and
$\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$ , and each of
these is clearly a restriction of homeomorphism of a solid torus.
\end{pf}

\begin{cor} $M_1 = D^2 \times S^1 , \bd M_2 = T^2$ and let $h:\bd
M_1 \overset{\cong}{\rightarrow} \bd M_2 .$\\ Then the topological
type of $M_1 \underset{h}{\cup} M_2$ depends only on $|h_*(m)|$,
i.e., if $h_* (m) = \pm h'_* (m),$ then $M_1\underset{h}{\cup}M_2
\cong M_1\underset{h'}{\cup}M_2. $\end{cor}
\begin{pf} Suppose $h, h' : \bd M_1 \overset{\cong}{\rightarrow}
\bd M_2$ s.t. $h_* (m) = \pm h'_*(m).$ Then $(h^{-1}h')_* (m ) =
\pm m .$ Proposition and Theorem A $\Rightarrow
M_1\underset{h}{\cup}M_2 \cong M_1\underset{h'}{\cup}M_2.
$\end{pf}\\

\textbf{Lens Space}\\

$L(p,q) = D^2 \times S^1 \underset{h}{\cup} D^2\times S^1 ,
\hspace{0.5em} h(m) = qm+pl $, where $p$ and $q$ are relatively
prime.\footnote{Because $h_* $ is homeomorphism, determinant of
$h_*$ is $\pm 1$.
Then $qs -rp = \pm 1$, i.e., $p$ and $q$ are relatively prime.}\\

Note. $L(1,0 ) = S^3\\
\hspace*{2.7em} L(0,1) = S^2\times S^1 \\
\hspace*{2.7em} L(2,1) = \mathbb{R}P^2$\\

\textbf{���� 8.} (1) $L(p,q) \cong L(p,-q) \cong L(-p,q) \cong
L(-p,-q) \cong L(p,q+kp)\hspace{0.5em} \forall k\in \mathbb{Z}\\
\hspace*{3.7em}(2) \pi _1 (L(p,q)) = \mathbb{Z}/p$ (Use Van-Kampen Theorem.)\\

\textbf{Homology of $L(p,q)$}

\[
\xymatrix @M=1ex @C=2em @R=1em @*[c]{%
\cdots \ar[r]& H_q (S^1\times S^1 )\ar[r]^>>>>>{(i_*, h_* )} & H_q
(D^2\times S^1 )\oplus H_q(D^2\times S^1 ) \ar[r]& H_q(L)\ar[r] &
\cdots } \] $\Rightarrow H_q(L) = 0$ if $q\geq 4$.
\[
\xymatrix @M=1ex @C = 0.6em @R=1em @*[c]{%
0 \ar[r]& H_3(L)\ar[r]& H_2(S^1\times S^1)\ar[r]&\ar[r] 0
&H_2(L)\ar[r]& H_1(S^1\times S^1)\ar[r] &
\mathbb{Z}\oplus\mathbb{Z}\ar[r]&H_1(L)\ar[r]&\tilde{H_0}(S^1\times
S^1) }
\], where $h_* = \left(\begin{array}{cc} q&r\\ p&s
\end{array}\right)\Rightarrow $\\


$(i_*, h_* ) = \left(\begin{array}{cc} 0&1\\p&s\end{array}\right)
:H_1 (S^1\times S^1 ) = \mathbb{Z}\oplus \mathbb{Z} \rightarrow
H_1 (D^2\times S^1)\oplus H_1(D^2\times S^1) = \mathbb{Z}\oplus
\mathbb{Z}$\\

(1) $p \neq 0 \Rightarrow det \neq 0$ and $(i_* ,h_*)$ is
injective.$\Rightarrow$\[
\xymatrix @M=1ex @C = 1em @R=1em @*[c]{%
\ar[r] & H_2 (L)\ar[rr]\ar[dr] & & H_1 (S^1\times S^1)\ar[r] & \\
& & 0\ar[ur] & & }
\] $\Rightarrow H_2(L) = 0 $ and $H_1 (L) =
\mathbb{Z}\oplus
\mathbb{Z} /im (i_* , h_*) = \mathbb{Z}/p$\\

(2) $p=0 \Rightarrow h(m) = qm \overset{h
:isomorphism}{\Rightarrow} q = \pm 1$ (meridian$\mapsto$
meridian)\\

$H_2(L) = ker \left(\begin{array}{cc}0&1\\0&s \end{array}\right) =
\mathbb{Z}$ and $H_1 (L) = cok \left(\begin{array}{cc} 0&1 \\0
&s\end{array}\right) =\mathbb{Z}$\\

In this case, $L = S^2\times S^1$.\\

Remark. (1) $p \neq p' \Rightarrow L(p,q) \ncong L(p',q' )\\
\hspace*{4.1em}(2)$ Fact. $L(p,q)\cong L(p,q' ) \Leftrightarrow q'
=
\pm q^{\pm 1} (mod\hspace{0.3em} p)\\
\hspace*{8.3em} L(p,q) \simeq L(p,q') \Leftrightarrow q q' = \pm
m^2
(mod \hspace{0.3em}p)$ for some m\\
(Ref. M. Cohen. A course in simple homotopy theory)\\
\end{document}
