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\begin{document}
\parindent=0cm
\section*{VIII. Application of Homology}
1. The Jordan-Brouwer Separation theorem.
\begin{thm}
$e^r$= a closed $r$-cell (i.e., a homeomorph of $D^r$) in $S^n$
$\Rightarrow \rh{S^n-e^r} =0$.
\end{thm}
\begin{pf}
use induction on $r$.\\
$r=0$ : $e^0$�� �� point�̹Ƿ� $S^n-e^0$�� contractible�Ͽ�
�翬�ϴ�.\\
$r>0$ : $e^r$�� $r$-cell�̹Ƿ�, $\phi :I^r\to e^r\subset S^n$��
homeomorphism�� �����Ѵ�. $z$�� $S^n-e^r$�� ������ reduced
$p$-cycle�̶� ����.
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induction�� ������ ���Ͽ� $\rh{S^n-e_t^{r-1}}=0$�̹Ƿ�, $\bd
c_t=z$�� �Ǵ� $c_t\in S_{p+1}(S^n-e_t^{r-1})$�� �����Ѵ�. �׷���,
$|c_t|$= support of $c_t$�� compact�̹Ƿ�
$d(e_t^{r-1},|c_t|)=\eps_t>0$�̴�. ����, �׸��� ����
$\phi(U_t\times I^{r-1})\cap |c_t|=\phi$�� �Ǵ� $t$�� open
neighborhood $U_t$�� �����Ѵ�. \\
������ $t\in I$�� ���Ͽ� $U_t$�� �����ϹǷ� $\{U_t\}$�� compact
set $I$�� covering�� �ǰ� �̿� ���� Lebesgue number
$\frac{1}{m}$�� ����. \\
$I_j=[\frac{j}{m},\frac{j+1}{m}]$�� �θ�, $I_j$�� � $U_t$��
���ϹǷ� ������ $I_j$�� ���Ͽ� $\bd c_j=z$�� �Ǵ� $c_j\in
S^n-\phi(I_j\times I^{r-1})$�� �����Ѵ�. ���� ���� claim���κ���
������ ������. \\

\underline{Claim}  Let $J_1=[a,t]$, $J_2=[t,b]$ for $a,b\in I$ and
$e_i^r=\phi(J_i\times I^{r-1})$. \\ If $c_i\in S_{p+1}(S^n-e_i^r)$
is such that $\bd c_i=z$ for $i=1,2$, then there exists $c\in
S_{p+1}(S^n-e_1^r \cup e_2^r)$ such that $\bd c=z$.\\[2mm]
\underline{proof of Claim}\\
$X_i=S^n-e_i^r$�̶� �θ�, $X_1\cup X_2=S^n-e_t^{r-1}$, $X_1\cap
X_2=S^n-e_1^r\cup e_2^r$�̹Ƿ� ������ ���� MV-sequence�� ��´�.

\[
\xymatrix @=1em @*[c] { %
\cdots \ar[r] & \hh_{p+1}(X_1\cup X_2) \ar[r] & \hh_{p}(X_1\cap
X_2) \ar[r]^<<<i & \hh_p(X_1)\oplus\hh_p(X_2) \ar[r] &\cdots }
\]
induction�� ������ ���Ͽ� $\hh_{p+1}(X_1\cup X_2)=0$�̹Ƿ� $i$��
injection�̴�. claim�� �������κ��� $\{z\}$�� $\hh_p(X_1)$,
$\hh_p(X_2)$���� ���� 0�̰�, $i$�� ������ ��ǥ�� ����
inclusion���κ��� induce�� map�̹Ƿ� $\hh_{p}(X_1\cap X_2)$����
$\{z\}= 0$�̴�.
\end{pf}
\begin{cor}
$e^r\subset \rb^n\Rightarrow \hh(\rb^n-e^r)\cong\hh(S^{n-1})$.
\end{cor}
\begin{pf} Exercise. \end{pf}
\begin{thm}
Let $s^r\subset S^n$ be a homeomorph of $S^r$. Then\\
(1) $r\leq n$ \\
(2) $r=n\Rightarrow s^n=S^n$ \\
(3) $r<n \Rightarrow \hh_p(\sm)=\{\begin{array}{ll} \zb (or\ R)&,
p=n-r-1\\ 0&, otherwise \end{array}$
\end{thm}
\begin{pf}
������ ���Ͽ� homeomorphism $\phi :S^r\to s^r$�� �����Ѵ�. $S^r$��
(������ �����ϴ�) �Ϲݱ��� $E^r_+$, ���ݱ��� $E^r_-$�� �ΰ�,
$e^r_+=\phi(E^r_+)$, $e^r_-=\phi(E^r_-)$�� ����. $X_+=S^n-e^r_+$,
$X_-=S^n-e^r_-$�� �ϸ�, $X_+\cup
X_-=S^n-s^{r-1}$, $X_+\cap X_-=S^n-s^{r}$�̴�.\\
���� $\sm=\phi$�� ��쿡�� $S^n=s^r$�̹Ƿ� $n=r$���� �� �� �ִ�.\\
$\sm\neq\phi$��� ������ MV-sequence�� ��´�.
\[
\xymatrix @=1em @*[c] { %
\cdots \ar[r] & \hh_{p+1}(X_1)\oplus\hh_{p+1}(X_2) \ar[r] &
\hh_{p+1}(X_1\cup X_2) \ar[r] & \hh_{p}(X_1\cap X_2) \ar[r]
&\cdots }
\]
���� 1�� ���Ͽ� $\hh(X_1)=\hh(X_2)=0$�̹Ƿ�,
$$ \hh_{p+1}(S^n-s^{r-1})\cong \hh_p(\sm)$$
�� ��´�. �� ���� �ͳ������� �����ϸ�,
$$  \hh_p(\sm)\cong\hh_{p+r}(S^n-s^0)$$
���� �� �� �ִ�. �׷��� $s^0$�� 2 points�̹Ƿ� $S^n-s^0\simeq
S^{n-1}$�̰�, ����
$$\hh_p(\sm) =\{\begin{array}{ll} \zb (or\ R)&,
p+r=n-1\\ 0&, otherwise \end{array}$$ �̴�. �� ��, ���� $r\geq
n$�̶�� $p=n-r-1<0$�� ���Ͽ� $\hh_p(\sm)=\zb$�� �ǹǷ� ����̴�.
���� (1)(2)(3)�� ��� �����Ǿ���.
\end{pf}

\newpage
\begin{cor}
$s^r\subset\rb^n\Rightarrow r<n$ and $
\hh_p(\rb^n-s^r)=\{\begin{array}{ll} \hh_p(\sm) &, p\neq n-1\\
 \hh_p(\sm)\oplus \zb &, p=n-1\end{array}$
 \end{cor}
\begin{pf} Exercise. \end{pf}\\
��,\\
(1) $r\neq 0 \Rightarrow \hh_p(\rb^n-s^r)=\{\begin{array}{ll} \zb &, p=n-r-1,\ n-1\\
 0 &, otherwise\end{array}$,\\
(2) $r= 0 \Rightarrow \hh_p(\rb^n-s^r)=\{\begin{array}{ll} \zb\oplus\zb &, p=n-1\\
 0 &, otherwise\end{array}$.\\
�̷κ��� ������ ���������� ��´�.
\begin{cor}
$s^r\subset S^n$ disconnects iff $r=n-1$\\ and $s^r\subset \rb^n$
disconnects iff $r=n-1$.
\end{cor}
\begin{pf}
$r=n-1$�̸� $\hh_0(\sm)=\hh_0(\rb^n-s^r)=\zb$�̹Ƿ� �ڸ��ϴ�.
\end{pf}
\begin{thm}[Jordan-Brouwer Separation theorem]
\hfill\\ For all $s^{n-1}\subset S^n$, $S^n-s^{n-1}$ consists of 2
connected components both having $s^{n-1}$ as boundary.
\end{thm}

\begin{pf}
$\hh_0(S^n-s^{n-1})=\zb$�̹Ƿ� $S^n-s^{n-1}$�� 2����
path-component�� ���´�. �̸� $O_1, O_2$��� �ΰ� ������
boundary�� $s^{n-1}$���� ���̸� ������ ������.\\
$O_1, O_2$�� ��� open set�̹Ƿ�, $x\in bdy(O_i)$��� $x \notin
O_i$�� �Ǿ� $x\in s^{n-1}$�̴�.\\
������ $s^{n-1}\subset bdy(O_i)$�� ����� ���ؼ� ������ $x\in
s^{n-1}$�� ���Ͽ� $x$�� ������ open neighborhood $U$�� $U\cap
O_i\neq\emptyset$�� �������� ���̸� �ȴ�.\\
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���� $y_1\in O_1$, $y_2\in O_2$�� ������ ��´�. homeomorphism
$\phi$�� ���� $s^{n-1}=\phi(S^{n-1})$�� �ΰ� $x\in A\subset U\cap
s^{n-1}$�� �����ϵ��� $y=\phi^{-1}(x)\in S^{n-1}$�� closed ball
neighborhood $B$�� $A=\phi(B)$�� ������, $e^{n-1}:=
s^{n-1}-\phi(\overset{\circ}{B})$�� closed $(n-1)$-cell�� �ȴ�.
���� ���� 1�� ���Ͽ� $\hh_0(S^n-e^{n-1})=0$�̹Ƿ�,
$S^n-e^{n-1}$�� path
connected�̴�. \\
$y_1$, $y_2$�� �մ� $S^n-e^{n-1}$�� path�� $\sig$�� ����.
$\emptyset\neq\sig(I)\cap s^{n-1}=\sig(I)\cap A$�� closed
set�̹Ƿ� $\sig^{-1}(\sig(I)\cap A)$�� closed set�̰�, ����
complement $W$�� open interval���� countable union�̴�. ����,
$I_1=[0,x_1)$�� $I_2(x_2,1]$�� $W$�� ó���� ������ open
interval�̶� ����.

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����, $\sig(I_1)\subset S^n-s^{n-1}$�̰� $O_1$�� path
component�̹Ƿ� $\sig(I_1)\subset O_1$�̰�, ����������
$\sig(I_2)\subset O_2$�̴�. �׷��� $\sig(x_i)\in A\subset
U$�̹Ƿ�, $U\cap O_i\neq\emptyset$�̴�.
\end{pf}


\begin{cor}
For all $s^{n-1}\subset \rb^n$, $n\geq 2 \Rightarrow
\rb^n-s^{n-1}$ has 2 connected components both having $s^{n-1}$ as
boundary.
\end{cor}
\begin{pf}
$S^n=\rb\cup\{\infty\}$�̹Ƿ� ���� �����κ��� �ڸ��ϴ�.
\end{pf}\\
�� ��, $\infty$�� �����ϴ� component�� unbounded component,
�׷����ʴ� component�� bounded component�� �θ���.\\


{\bf Remark}\\
If $n=2$, the above corollary is the Jordan Curve Theorem. \\

\underline{\bf Schoenflies Theorem}\\
If $n=2$, $\overline{O}_1$ and $\overline{O}_2$ are closed disks.
In fact, any embedding $h:S^1(\subset \rb^2) \to \rb^2$ can be
extended to a homeomorphism $\bar{h}:\rb^2\to \rb^2$.\\

If $n=3$, this is not true as Alexander horned sphere shows.

\begin{figure}[h]
   \centerline{\includegraphics*[scale=0.3,clip=true]{alexander.eps}}
\end{figure}
\begin{center}
�׸� : Alexander horned sphere \footnote{It is homeomorphic with
the ball $D^3$, and its boundary is therefore a sphere. It is
therefore an example of a wild embedding in $\rb^3$. The outer
complement of the solid is not simply connected, and its
fundamental group is not finitely generated. Furthermore, the set
of nonlocally flat ("bad") points of Alexander's horned sphere is
a Cantor set. \\
From http://mathworld.wolfram.com/AlexandersHornedSphere.html}
\end{center}
\newpage
\begin{thm}[Invariance of domain]
\hfill\\
Let $U\subset \rb^n$(resp. $S^n$) be an open set. If $f: U\to
\rb^n$(resp. $S^n$) is one-to-one continuous, then $f(U)$ is open
in $\rb^n$(resp. $S^n$) and hence $f$ is an embedding. (since $f$
is an open map.)
\end{thm}
\begin{pf}
$S^n$�� ���ؼ��� ���̸� ����ϴ�.\\
������ $y=f(x)\in f(U)$�� ���Ͽ� $x\in B_\eps\subset
\overline{B_\eps}\subset U$�� open ball $B_\eps$�� ����.
$S_\eps=\bd B_\eps$, $s^{n-1}=f(S_\eps)$�� �θ�, ���� ������
���Ͽ� $S^n-s^{n-1}$�� 2���� connected component $O_1$�� $O_2$��
���´�. �� �� $y$�� �����ϴ� component�� $O_1$�̶� �ΰ�
$f(B_\eps)=O_1$���� ���̸� ������ ������.\\
$f(B_\eps)\subset O_1$ : $f(B_\eps)$�� connected�̹Ƿ� �ڸ��ϴ�.\\
$f(B_\eps)=O_1$ : $f(\overline{B}_\eps)$�� $n$-cell�̹Ƿ�
$S^n-f(\overline{B}_\eps)$�� connected�̴�. �׷���,
$$S^n-f(\overline{B}_\eps)=S^n-f(B_\eps)-f(S_\eps)=(O_1-f(B_\eps))\sqcup
O_2=(O_1-f(\overline{B}_\eps))\sqcup O_2$$ �̰� �º�����
connected�ؾ� �ϹǷ�, $O_1-f(\overline{B}_\eps)=\phi$�� �Ǿ�
$f(B_\eps)=O_1$�� �ȴ�.
\end{pf}

\end{document}
