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\begin{document}
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\section*{III.2 $H_q (M^n , M-A ) ,\hspace{0.5em} q\geqslant n$}

idea :  Compare $H_n(M, M-A)$ with $\Gamma A.$\\

6. Suppose $U^{open} \subset M. $ Then $\bet_U \in H_n(M, M-U)$
can be viewed as a section as before and we have a homomorphism
$j_U : H_n(M, M-U) \rightarrow \Gamma U.\\ \hspace*{23em}
\bet_U\mapsto j_U(\bet_U ) : x\mapsto \bet_U |_x$ \\
In general, $\forall A\subset M, $ does $j_A : H_n(M,
M-A)\rightarrow \Gamma A$(defined by $"\bet" = j_A(\bet) :
x\mapsto \bet |_x )$ define a homomorphism, i.e., is $j_A(\bet)$
continuous section on $A$?\\

\begin{pf} Want $"\bet "$ is locally constant, i.e., $\forall a\in
A , \hspace{0.5em} \exists V$ and $\bet_V$ s.t. $\bet|_x = \bet
_V|_x\hspace{0.5em} ,\forall x\in A\cap V.$\\
Recall : Can represent $\bet = \{b\}$ with $\bd b \subset M-A .\\
|\bd b| :$ compact $\Rightarrow U = M- |\bd b|$ is open and choose
$V$, a coordinate ball neighborhood of $a$ with $V\subset U.$
\[
\xymatrix  @C=1em @*[c]{%
(M, M-U)\ar[r]\ar[d] & (M, M-V)\ar[d]^{\cong on H_n} &\Rightarrow&
\bet '= \{b\} \ar@{.>}[r]\ar[d] & \exists \bet_V \ar[d]^{\cong}
\\
(M, M-A)\ar[r] & (M, M-a) & & \bet ' |_A  = \bet \ar[r]& \bet |_a
\ar[u]
 } \]
And $\forall x\in A\cap V, \hspace{0.5em} \bet_V|_x = \bet'|_V|_x
= \bet'|_x = \bet'|_A|_x = \bet|_x.$\end{pf}\\

7. When is $j_A : H_n (M, M-A) \rightarrow \Gamma A$ an
isomorphism?\\
Know : true if $A = U$, a coordinate ball.\\
\[
\xymatrix  @C=3em @*[c]{%
 H_n (V, V-U )\ar[r]^{excision :\cong} \ar[d]^{\psi_* : \cong}
 & H_n(M, M-U)\ar[r]^{j_U} &\Gamma U \ar[d]^{\psi_* :\cong}\\
\mathbb{Z} = H_n (\mathbb{R}^n, \mathbb{R}^n\backslash D
)\ar[rr]^{j_D : \cong} &&\Gamma D = \mathbb{Z}
 } \]
 Also note $H_q(M, M-U )  \cong
 H_q(\mathbb{R}^n , \mathbb{R}^n\backslash D ) \cong
 \widetilde{H_{q-1}}(S^{n-1}) = 0 $ if $ q>n.$\\

Let $M = \mathbb{R}^n.$\\
If $A$ is a "nice " compact set, then $j_A$ is $\cong$.\\
e.g. $ A =D , [0,1]\times [0,1] , [0,1], $ point, $\cdots$ etc.\\
But note that if $A = M =\mathbb{R}^1, \hspace{0.5em}
H_1(\mathbb{R}^1, \mathbb{R}^1\backslash \mathbb{R}^1 ) = H_1
(\mathbb{R}^1 ) = 0$
but $\Gamma \mathbb{R}^1 = \mathbb{Z}$.\\

\textbf{Note.} $A$: closed. Then\[
\xymatrix  @C=1em @*[c]{%
 H_n(M,
M-A)\ar[rr]^{j_A} \ar[dr]^{j_A}& & \Gamma A
\\
  & \Gamma_c A\ar[ur]^{\subset} &
 } \]i.e., $j_A(\bet) \in \Gamma_c A \hspace{0.5em} ,\forall \bet$
where $\Gamma_c A$ consists of sections with compact support.\\

\begin{pf}Let $\bet = \{b\}$: relative cycle $\Rightarrow |b| :$ compact. \\Then $\forall x\in
(M-|b|)\cap A, \bet|_x = 0 $ since $|b| \subset M-x.$ Think of
this in chain level.
\[
\xymatrix  @C=1em @*[c]{%
0\ar[r] & S_n(M-A)\ar[d]\ar[r] & S_n(M) \ar[r]\ar[d]^{=}& S_n(M)/S_n(M-A)\ar[r]\ar[d] & 0\\
0 \ar[r]& S_n(M-x)\ar[r] & S_n(M)\ar[r] & S_n(M)/S_n(M-x)\ar[r] &
0
 } \]
 $\therefore \bet$ has a support $\subset |b|\cap A:$
 compact\end{pf}\\

 So the right statement is $j_A : H_n(M, M-A) \rightarrow\Gamma_c A .$
 \\Furthermore, $j_A$ is natural:\\
 \[
\xymatrix  @C=2em @*[c]{%
B\subset A\subset M \Rightarrow & H_n(M, M-A)\ar[r]^<<<{j_A}
\ar[d]^{\rho_B^A} &\Gamma A\ar[d]^{restriction}&\textrm{commute.}\\
&H_n(M, M-B) \ar[r]^<<<{j_B}&\Gamma B
 } \]

Exercise. \textbf{숙제 11.}
\[
\xymatrix  @C=2em @*[c]{%
f : (M, A)\ar[r]^{\cong} & (N, B)\Rightarrow &H_n(M,
M-A)\ar[r]^<<<{j_A}
\ar[d]^{f_* : \cong} & \Gamma A\ar[d]^{"f_*"}&\textrm{commute.}\\
& & H_n(N, N-B)\ar[r]^<<<{j_B} & \Gamma B&
 } \]\\

{\bf 8. (Theorem)} Let $M$ be an $n$-dimensional manifold and
 $A^{\textrm{closed}}\subset M$. Then\\
(1) $H_q(M,M-A)=0$ for $q>n$\\
(2) $H_n(M,M-A)\cong \Gam_cA$\\

\begin{pf}
\begin{lem} (MV) $A, B$ closed $\subset M.$ If the theorem is true
for $A, B$ and $A\cap B,$ then so is for $A\cup B$.\end{lem}
\begin{pf} $A, B$ : closed $\Rightarrow M-A, M-B$ open with
$(M-A)\cap(M-B) = M-A\cup B\\
\hspace*{20em}(M-A)\cup(M-B)=M-A\cap B$\\
relative MV:
\[\scriptsize{
\xymatrix  @C=1em @*[c]{%
\textrm{hypothesis}\ar[r]&0\ar[r]\ar[d]^{\cong}& H_n(M, M-A\cup
B)\ar[r]\ar[d]^{j}&H_n(M, M-A)\oplus H_n(M,
M-B)\ar[d]^{\cong}\ar[r] & H_n(M, M-A\cap B)\ar[d]^{\cong}
\ar[r] &0\\
\textrm{exact :}&0\ar[r]& \Gamma(A\cup B)\ar[r] &
\Gamma(A)\oplus\Gamma(B)\ar[r] &\Gamma(A\cap B)\ar[r]&0\\
&& s\ar@{|->}[r]&(s|_A, s|_B ) &&\\
&&&(a,b)\ar@{|->}[r]& a|_{A\cap B}-b|_{A\cap B}&
 } }\]
 (i) follows from relative MV-sequence.\\
 (ii) follows from 5-lemma.
 \end{pf}

\begin{lem}
If $M=\rb^n$ and $A$ is a compact subset of $\rb^n$, then the
theorem is true.
\end{lem}
\begin{pf}
Know : The theorem is true for a "nice" compact set $A\subset
\rb^n$, for instance
$A=$rectangle.\\
By lemma 1, the theorem is true if $A$ is a finite union of
rectangles by induction on number of rectangles.\end{pf} \\

이제 compact set $A$에 대하여 정리가 성립함을 보이자. \\
먼저 $j_A: H_n(M,M-A)\to \Gam A =\Gam_c A$가 onto임을 보인다. \\
(i) For $s\in \Gam A $, there exists an open set $U$ containing
$A$ such that $s$ can be extended to $\bar{s}$ on $U$.\\
pf) $s(A)$ is compact $\Rightarrow$ $s(A)$ lies in finitely many
sheets of $\rb^n_\mathcal{O} \cong \rb^n\times \zb$.\\
Let $A_i$=$s^{-1}(\rb^n\times i)$ for $i\in \zb$. $A_i$ is
compact.\\
So there exists an open set $U_i$ containing $A_i$ such that
$U_i$'s are pairwise disjoint and $s$ can be extended to $\bar{s}$
over $U=\bigcup U_i$.\\

(ii) Cover $A$ by finitely many rectangles in $U$ and let $A'$ be
the union of rectangles.\\
Then the following diagram commutes.
\[
\xymatrix @=2em @C=4em @*[c] { %
H_n(M,M-A') \ar[r]^<<<<<<<{j_{A'}}_<<<<<<<{\cong} \ar[d] & \Gam A'\ni\bar{s}|_{A'} \ar[d]^{\textrm{restriction}}\\
H_n(M,M-A) \ar[r]^<<<<<<<{j_{A}} & \Gam A \ni s}\]  임의의 $s\in
\Gam A$에 대하여 (i)에 의하여 $\bar{s}$가 존재하므로
$\bar{s}|_{A'}$은 $s$에 대응되는 $\Gam A'$의 원소이다. 또한
$j_{A'}$은 isomorphism이므로 $j_A$는 onto이다. \\

이제 $j_A$이 1-1이고 $H_q(M,M-A)=0$ for $q>n$임을 보이자. \\
Let $\alp\in H_q(M,M-A)$ and assume $j_A(\alp)=0$ if $q=n$.\\
Suppose $\alp=\{a\}$ with $\bd a\subset M-A$. Since $|\bd a|$ is
compact, $V=M-|\bd a| $ is open.
Let $\alp'=\{a\}\in H_q(M,M-V)$. \\

($q=n$) : Let $U\subset V $ be an open set containing $A$ each of
whose components intersects $A$. Let $A'$ be a finite union of
rectangles which covers $A$ and is contained in $U$.\\
Then $j_U(\alp'|_U)$ is a section on $U$ which has a zero on each
of its component.\\($\because$ Each component intersects $A$ and $j_A(\alp)=0$.)\\
By uniqueness of sections on connected sets, $j_U(\alp'|_U)=0$.
\[
\xymatrix @=1.5em @C=4em @*[c] { %
H_n(M,M-U) \ar[r]^<<<<<<<{j_{U}} \ar[d] & \Gam U \ar[d]&\alp'|_{U} \ar@{|->}[r] \ar@{|->}[d] &0\ar@{|->}[d]\\
H_n(M,M-A') \ar[r]^<<<<<<<{j_{A'}}_<<<<<<<{\cong} \ar[d] & \Gam A' \ar[d]&\alp'|_{A'} \ar@{|->}[r] \ar@{|->}[d]&0\ar@{|->}[d]\\
H_n(M,M-A) \ar[r]^<<<<<<<{j_{A}} & \Gam A &\alp \ar@{|->}[r] &0}\]


Therefore by the above diagram, $\alp'|_{A'}=0$ and hence
$\alp=0$.

($q>n$) : The theorem is true for $A'$. So $\alp'|_{A'}=0$ and
hence $\alp=0$.
\end{pf}\\


\begin{pf1}\\
{\bf Step 1.} $A^{\textrm{compact}}\subset M$:\\
$A$ is a finite union of compact sets each of which is contained
in a coordinate ball neighborhood($\approx \rb^n$) and apply lemma
1 and lemma 2.\\

{\bf Step 2.} $A\subset U^{\textrm{open}}\subset
\overline{U}^{\textrm{compact}}$ $\Rightarrow$ The theorem is true
for $A\subset U(=M)$\footnote{manifold의 open subset은
manifold이므로 $U=M$으로 하여도 무방하다.}\\
\hspace*{4em}where $A$ is a closed
subset of $U$:\\
$\bd U =\overline{U}-U$ is compact and $U-\overline{A}=U-A$.
\\
Consider $(M,M-\bd U, M-(\bd U\cup \overline{A} ))$.\\
($q>n$):
$$ H_{q+1}(M,M-\bd U) \to H_q(M-\bd U,M-(\bd U \cup
\overline{A}))\to  H_q(M,M-(\bd U \cup \overline{A})) \to \cdots
$$
By excision theorem, $H_q(M-\bd U,M-(\bd U \cup
\overline{A}))=H_q(U,U-A)$. \\By step 1, $H_{q+1}(M,M-\bd U)=0$
and $H_q(M,M-(\bd U \cup \overline{A}))=0$.\\ Therefore
$H_q(U,U-A)=0$.\\
($q=n$):
\[
\xymatrix @=1.5em @*[c] { %
0 \ar[r]&H_n(U,U-A) \ar[r] \ar[d]^{j_A} &H_n(M,M-(\bd U \cup
\overline{A})) \ar[r]\ar[d]^{\cong(\textrm{step 1})}
&H_n(M-\bd U) \ar[d]^{\cong(\textrm{step 1})}\\
0 \ar[r]&\Gam_cA \ar[r]_<<<<<<<<<<<<{\parbox{5em}{\scriptsize
extension by 0 outside support}}&\Gam(\bd U\cup \overline{A})
\ar[r]_<<<<<<<<<<{\textrm{restriction}} &\Gam(\bd U)
 }\]
So, $j_A$ is an isomorphism by 5 lemma.\\


{\bf Step 3.}(general case)\\
Show $j_A: H_n(M,M-A)\to \Gam_cA$ is onto:\\
$\forall s\in \Gam_cA$, let supp $s=K$. \\
Then K is compact and
$\exists U$ such that $K\subset U \subset
\overline{U}^\textrm{compact}$ \\
Let $A'=A\cap U$.
\[
\xymatrix @=1.5em @C=4em @*[c] { %
H_n(U,U-A') \ar[r]^<<<<<<<<{\cong\textrm{(step 2)}} \ar[d]^{i_*} &
\Gam_cA'\ni s|_{A'} \ar[d]^{\parbox{4.5em}{\scriptsize
extension by 0 outside {\it K}}} \\
H_n(M,M-A) \ar[r]^{j_A} & \Gam_cA\ni s
 }\]
So, $j_A$ is onto.\\

Show $j_A$ is 1-1 and $H_q(M,M-A)=0$ if $q>n$: \\
$\alp \in H_q(M,M-A) $ and assume $j_A(\alp)=0$ if $q=n$.\\
If $\alp=\{a\}$ with $\bd a \subset M-A$, $|a|$: compact
$\Rightarrow |a| \subset U \subset
\overline{U}^\textrm{compact}$.\\
Let $A'=A\cap U$. Apply the above diagram.\\
$q=n$ : Since $|a|\subset U$, $\alp'=\{a\}\in H_n(U,U-A')$. So in
the above diagram, $i_*(\alp')=\alp$. Since $j_A(\alp)=0$ and
$0|_{A'}=0$,  $j_{A'}(\alp')=0$. So $\alp'=0$ and hence
$\alp=0$.\\
$q>n$ : By step 2, $\alp'=\{a\}\in H_q(U,U-A')=0$. So $\alp=0$.
\end{pf1}\\


{\bf 9. Consequences of the theorem}\\
(1) Let $A$ be connected and closed but not compact. Then
$H_n(M,M-A)=0$. In particular if $M$ is connected but not compact then $H_n(M)=0$. \\
\begin{pf}
If $s\in \Gam_cA$, $\nu\circ s : A\to \zb^{\geq 0}$ is continuous
and $=0$ at some $a\notin $ supp $s$. So $\Gam_cA=0$
\end{pf}\\

(2) If $M$ is orientable along $A$, $A$ is compact and has $k$
components, then $H_n(M,M-A)\cong\zb^k$.\\
\begin{pf}
$\Gam_cA=\Gam A \cong \zb^k$ by 5. (2).
\end{pf}\\

(3) If $A\subset \rb^n$ is compact and has $k$ components, then
$\widetilde{H}_{n-1}(\rb^n-A)\cong\zb^k$.\\
\begin{pf}
From homology sequence of pair $(\rb^n, \rb^n-A)$, we get
$$ 0\to {H}_{n}(\rb^n, \rb^n-A) \to
\widetilde{H}_{n-1}(\rb^n-A)\to 0$$

So $\widetilde{H}_{n-1}(\rb^n-A)\cong {H}_{n}(\rb^n, \rb^n-A)
\cong\zb^k$ by (2).
\end{pf}\\


(4) If $M$ is connected and closed\footnote{A manifold $M$ is
closed if it is compact without boundary.}, then $H_n(M)=\left\{
\begin{array}{cl}
  \zb, & \textrm{if }M \ \textrm{ is orientable}  \\
  0, &\textrm{if }M \ \textrm{ is non-orientable}
\end{array}
\right. $\\
\begin{pf}
Clear from 5. (3) and (4).
\end{pf}\\

{\bf Remark.} For a PID $R$, $H_n(M)=\left\{
\begin{array}{cl}
  R, & \textrm{if }M \ \textrm{ is } R-\textrm{orientable}  \\
  0, &\textrm{if }M \ \textrm{ is not }R-\textrm{orientable}
\end{array}
\right. $\\

{\bf 10. Fundamental class of $M$ and degree}\\

A choice of generating section is an orientation and the
corresponding homology class $\zeta$\footnote{$[M]$으로 쓰기도
한다. } is called the \key{fundamental (orientation) class} of
$M$, i.e., $\zeta|_x\in H_n(M,M-x)$ is the
preferred orientation at $x$ for all $x\in M$.\\

Let $M^n, N^n$ be oriented closed connected manifolds. For $f:
M\to N$, if $f_*(\zeta_M)=k\cdot \zeta_N$, $k$ is called the
\key{degree} of
$f$. \\

{\bf 숙제 14} $\forall y\in N$, "regular value" i.e., $\forall
x\in f^{-1}(y),$ $f$ is a homeomorphism on a neighborhood $U_x$
($f|_{U_x}:U_x\overset{\cong}{\to} V_y)$. Then $$ deg f
=\sum_{x\in f^{-1}(y)}deg_x f$$ where $deg_xf$ is defined by
$f|_{U*}:H_n(U,U-x)\to H_n(V,V-y)$.\\
In particular, if $p:M\to N$ is a $k$-fold covering, $deg p =k$
(with respect to the induced orientation on M).\\

{\bf 숙제 15} (22.43) (22.49) (22.50)*
\end{document}
