
\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[all]{xy}

\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{\pa}{\partial}
\newcommand{\cb}{\mathbb{C}}
\newcommand{\cc}{\mathcal{C}}
\newcommand{\ccn}{\mathcal{C}^{\infty}}
\newcommand{\sbb}{\mathbb{S}}
\newcommand{\rb}{\mathbb{R}}
\newcommand{\zb}{\mathbb{Z}}
\newcommand{\bd}{\partial}
\newcommand{\key}[1]{\emph{\textbf{#1}}}
\newcommand{\td}[1]{\widetilde{#1}}
\newcommand{\ssx}[1][p]{\triangle^{#1}}
\newcommand{\sg}[2][X]{S_{#2}({#1})}
\newcommand{\sgi}[2][X]{S_{#2}({#1}\times I)}
\newcommand{\sgp}[1][X]{S_{p}({#1})}
\newcommand{\sgip}[1][X]{S_{p}({#1}\times I)}
\newcommand{\iosh}{i_{0\sharp}}
\newcommand{\ish}{i_{1\sharp}}
\newcommand{\ios}{i_{0*}}
\newcommand{\is}{i_{1*}}

\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{proof}{{\bf ����}}{\hfill\framebox[2mm]{}\\}
\newenvironment{proof1}{{\bf ��������}}{\hfill\framebox[2mm]{}}


\begin{document}
\parindent=0cm
\section*{VII.6 Relative Homology and Excision}

\begin{defn} $A \underset{i}{\hookrightarrow}X$, a subspace of a
topological space $X \Rightarrow
S_p(A)\underset{i_{\sharp}}{\hookrightarrow} S_p(X) $\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\Rightarrow & 0 \ar[r] & S_p(A) \ar[r]^i \ar[d]^{\bd} & S_p(X) \ar[r]^q \ar[d]^{\bd}
& S_p(X) / S_p(A) \ar[r] \ar@{.>}[d]^{\bar{\bd}} & 0\\
& 0 \ar[r] & S_{p-1}(A) \ar[r] & S_{p-1}(X) \ar[r] & S_{p-1}(X)/
S_{p-1}(A)\ar[r] & 0 }
\]
$\hspace{3em}$,where $\bar{\bd}$ is an induced boundary
homomorphism
s.t. ${\bar{\bd}} ^2 = \bar{{\bd}^2} = 0$.\\
$\Rightarrow \{ S_p (X,A) , "\bd" = \bar{\bd}\} = S(X,A) :$
relative chain complex or singular chain complex for a pair
$(X,A).\\
\Rightarrow H_p (S(X,A)) =: H_p (X,A) $: singular homology for a
pair $(X,A)$(or relative homology  X mod A).
\end{defn}

Remark 1. This is a general construction for $\forall$ chain
complexes $\mathcal{C}' \subset \mathcal{C}$, and hence can
talk about $H_p (K,A)$ for a simplicial pair $A<K$.\\

Remark 2. (geometric interpretation)\\
$S_p(X,A)$ can be viewed as a free group generated by p-simplices
not contained in $A$.\\

$\widetilde{Z_p}(X,A) := q^{-1}(ker \bar{\bd}) =
\bd^{-1}(S_{p-1}(A))=\{c\in S_p(X) | \bd c \in S_{p-1}(A)\}.$\\
$\widetilde{B_p}(X,A) := q^{-1}(im \bar{\bd}) = \bd S_{p+1}(X) +
S_p(A)\\ \hspace*{4em}
= \{ c\in S_p(X) : \exists a\in S_{p+1}(X) $ s.t. $ c \equiv \bd a $ mod $ S_p(A)\}.$\\

Then $H_p(X,A) \underset{\underset{q_*}{\leftarrow}}{\cong}
\widetilde{Z_p}(X,A) /\widetilde{B_p }(X,A)$. \\(by an isomorphism theorem, $M/p / N/p \cong M/N$)\\
\newpage
\textbf{1. Functorial Property}\\

$f : (X,A) \rightarrow (Y, B)$\\
\[
\xymatrix @=2em @*[c]{%
\Rightarrow & 0\ar[r] & S_p(A)\ar[r]\ar[d]^{f_{\sharp}|} &
S_p(X)\ar[r]\ar[d]^{f_{\sharp}} &
S_p(X)/S_p(A)\ar[r]\ar@{.>}[d]^{\bar{f_{\sharp}}="f_{\sharp}"} &
0&
 \textrm{and $"f_{\sharp}"$ is a chain map.}\\
            &0\ar[r] & S_p(B)\ar[r] & S_p(Y)\ar[r] & S_p(Y)/S_p(B)\ar[r] & 0&}
\]
$\Rightarrow f_* : H_p(X,A) \rightarrow H_p(Y,B).\\
\hspace*{5.5em} \{c\}\mapsto \{f\circ c\}$\\

(i) $id : (X,A)\circlearrowleft \Rightarrow id_* = id.$\\
(ii) $(X,A) \overset{f}{\rightarrow} (Y,B)
\overset{g}{\rightarrow} (Z,C) \Rightarrow (g\circ f)_* = g_*
\circ f_*$\\

Example.\\
(i) $\{X_{\alp}\} :$ path-components of $X$ and $A_{\alp} = A \cap
X_{\alp} \Rightarrow H_p(X,A) =
\underset{\alp}{\oplus}H_p(X_{\alp},A_{\alp})$\\
(ii) $X$ : path-connected and $A \neq \phi \Rightarrow H_0(X,A) =
0$.\\

\textbf{2. Homotopy Invariance}\\

$f \underset{F}{\simeq} g : (X,A)\rightarrow (Y,B)
\hspace{1em}(F(A\times I)
\subseteq B) \Rightarrow f_* = g_* .$\\

\begin{proof} Let $i_0 , i_1 :(X,A)\rightarrow(X\times I, A\times
I)$ be level maps as before.\\
Need a chain homotopy $D_{X,A} : S_p(X,A)\rightarrow
S_{p+1}(X\times I, A\times I)$\\ between $i_{0\sharp}$ and
$i_{1\sharp}$:
\[
\xymatrix @=2em @*[c]{%
0 \ar[r] & S_p(A)\ar[r]\ar[d]^{D_A = D_X |} & S_p(X)\ar[r]\ar[d]^{D_X}
& S_p(X,A)\ar[r]\ar@{.>}[d]^{\bar{D}_X = D_{X,A}} & 0\\
0 \ar[r] & S_{p+1}(A\times I)\ar[r] & S_{p+1}(X\times I)\ar[r] &
S_{p+1}(X\times I, A\times I)\ar[r] & 0}
\]
Check $\bar{\bd}\bar{D}\bar{\sigma} + \bar{D}\bar{\bd}\bar{\sigma}
= \bar{i_{0\sharp}}\bar{\sigma} - \bar{i_{1\sharp}}\bar{\sigma}$:
trivial.\\

$(X,A)\overset{i_0}{\underset{i_1}{\rightrightarrows}}(X\times I,
A\times I)\overset{F}{\rightarrow} (Y,B)$ s.t. $f= F\circ i_0$ and
$g = F\circ i_1$.\\
$\bar{D} : i_{0\sharp} \cong i_{1\sharp} \Rightarrow i_{0*} =
i_{1*} : H(X,A)\rightarrow H(X\times I, A\times I)\\
\therefore f_* = (F\circ i_0 )_* = F_* \circ i_{0*} = F_* \circ
i_{1*} =
(F\circ i_1 )_* = g_*$\end{proof}\\

\textbf{3.} $0\rightarrow S(A)\rightarrow S(X)\rightarrow
S(X,A)\rightarrow 0\\
\overset{\textrm{snake}}{\Rightarrow} \exists$ a functorial long
exact sequence
\[
\xymatrix @=2em @*[c]{%
\cdots \ar[r]& H_p(A) \ar[r]^{i_*} & H_p(X) \ar[r]^{q_*} &
H_p(X,A) \ar[r]^{\bd _*} & H_{p-1}(A)\ar[r] & \cdots}
\]
$\hspace*{20em}\{c\}\mapsto \{\bd c\}$\\

Remark. If $A \neq \phi $, define $\widetilde{H}(X,A) = H(X,A).$
Then 3 holds for reduced case also.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
 & \ar[d]& \ar[d]  &  \ar[d] & \\
0 \ar[r]& S_0(A)\ar[r]\ar[d]^{\epsilon} & S_0(X)\ar[r]\ar[d]^{\epsilon}
 & S_0(X,A)\ar[r]\ar[d] & 0\\
0 \ar[r]& \mathbb{Z}\ar[r]^{id}\ar[d] & \mathbb{Z} \ar[d]\ar[r]& 0\ar[r]\ar[d] & 0\\
  &  0  & 0   & 0 &}
\]

Example. $H_p(B^n, \bd B^n )$
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & \widetilde{H_p} (\bd B^n) \ar[r] &
\widetilde{H_p}(B^n) \ar[r] & \widetilde{H_p} (B^n , \bd B^n)
\ar[r]^{\bd} & \widetilde{H}_{p-1}(\bd B^{n-1}) \ar[r] & \cdots }
\]
$\widetilde{H_p}(B^n) =0 \hspace{0.5em},\forall
p\hspace{0.3em}\Rightarrow\hspace{0.3em} \bd : \cong
\hspace{0.3em}\Rightarrow \\H_p (B^n , \bd B^n)=\widetilde{ H}_p
(B^n , \bd B^n) \cong \widetilde{H}_{p-1}(S^{n-1}) =
\mathbb{Z}\hspace{0.8em} p=n\\
\hspace*{19em} = 0 \hspace{0.8em} p\neq n$\\

\textbf{4. Long Exact Sequence for Triple}\\

Let $A\subset B\subset X$. Then we have a functorial long exact
sequence \\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & H_p (B,A)\ar[r] & H_p(X,A) \ar[r] & H_p (X,B)
\ar[r]^{\bd} & H_{p-1}(B,A) \ar[r] & \cdots}
\]
\begin{proof}
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & S(B)/S(A) \ar[r] & S(X)/S(A) \ar[r] & S(X) /S(B) \ar[r]
& 0 & \textrm{: s.e.s.}}
\]
and apply snake lemma.\end{proof}

\textbf{5. Five Lemma}
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
A_1 \ar[r] \ar[d]^{f_1} & A_2 \ar[r] \ar[d]^{f_2} & A_3 \ar[r] \ar[d]^{f_3}
& A_4 \ar[r] \ar[d]^{f_4} & A_5 \ar[d]^{f_5 \hspace{2em} \textrm{all} \curvearrowright} & \textrm{:exact}\\
%    &     &     &     &     & \textrm{all} \curvearrowright \\
B_1 \ar[r] & B_2\ar[r] & B_3\ar[r] & B_4\ar[r] & B_5 &
\textrm{:exact}}
\]
$f_1 $: onto , $f_5 : 1-1$ and $f_2 , f_4 :\cong
\hspace{0.5em}\Rightarrow \hspace{0.5em}f_3 :\cong$.\\
(In particular, $f_i : \cong \hspace{0.5em} i = 1,2,4,5.
\hspace{0.5em} \Rightarrow \hspace{0.5em} f_3 : \cong$.)\\

\begin{proof} by diagram chasing. \end{proof}

Example. Show $ j : (B^n , S^{n-1}) \rightarrow (B^n, B^n
\setminus 0)$ induces an isomorphism\\ $j_* : H(B^n , S^{n-1})
\rightarrow H(B^n, B^n\setminus 0) $.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & H_p (S^{n-1}) \ar[r]\ar[d]^{j_* : \cong} &
H_p(B^n)\ar[r]\ar[d]^{j_* : = } & H_p(B^n ,
S^{n-1})\ar[r]\ar[d]^{\textrm{five lem.�� ����} j_* :\cong} &
H_{p-1}(S^{n-1}) \ar[r] \ar[d]^{j_* : =}& H_{p-1}(B^n )\ar[r]\ar[d]^{j_* : =}& \cdots\\
\cdots \ar[r] & H_p (B^n\setminus 0) \ar[r] & H_p(B^n)\ar[r] &
H_p(B^n , B^n\setminus 0) \ar[r] & H_{p-1}(B^n\setminus 0)\ar[r] &
H_{p-1}(B^n )\ar[r]& \cdots}
\]
\textbf{6. Split Short Exact Sequence}\\

Given short exact sequence $0\rightarrow A
\overset{i}{\rightarrow}
B \overset{p}{\rightarrow} C \rightarrow 0, $ TFAE :\\
(i) $\exists A \overset{j}{\leftarrow} B $ s.t. $j\circ i = id$.\\
(ii) $\exists B\overset{s}{\leftarrow} C$ s.t. $p\circ s
= id$.\\
In this case, $B = A\oplus C$ and the short exact sequence is said
to be split.\\

\begin{proof} (i)$\Rightarrow$(ii):\\
$0\rightarrow A \overset{j}{\underset{i}{\leftrightarrows}} B
\underset{p}{\rightarrow} C \rightarrow 0
\Rightarrow B = i(A) \oplus ker(j) : b = ij(b) +(b-ij(b))\\
(\because j(b) -jij(b) = 0 \Rightarrow b-ij(b) \in ker(j)$ and
$ji(a) = 0 \Rightarrow a=0\Rightarrow i(a) =0$.)\\

$p|_{ker(j)}: ker(j) \rightarrow C : \cong \hspace{0.3em}$ since
$ker (p) = i(A)
\Rightarrow \exists s = (p|)^{-1} $ s.t. $p\circ s = id.$\\

(ii)$\Rightarrow$(i) : exactly same.\end{proof}

Example. $A
\underset{\underset{r}{\leftarrow}}{\overset{i}{\hookrightarrow}}X
, $ a retract $\Rightarrow H_p (X) = H_p(A) \oplus H_p(X,A)$.
\newpage
\textbf{Excision}

\begin{thm} $(X,A) : $ a pair of spaces and let $U\subset X$ be
s.t. $\bar{U} \subset \AA.$\\
Then the inclusion $i :(X-U, A-U) \hookrightarrow(X,A)$ induces an
isomorphism $i_* : H_* (X-U, A-U) \overset{\cong}{\rightarrow}
H_*(X,A)$. (i.e., $U$ may be excised without altering relative
homology.)\end{thm}
\begin{proof} Let $\mathcal{U} = \{X-U , A\}$ so that
$\{(X-U)^{\circ}$ ,$ \AA\}$ covers X. (See the
footnote.\footnote{$(X-U)^{\circ}= (U^C)^{\circ} = \bar{U}^C$})\\
Then $i : S^{\mathcal{U}}(X)\hookrightarrow S(X)$ induces an
isomorphism on homology.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r]& S(A)\ar[r]\ar[d]^{i=i|} &
S^{\mathcal{U}}(X)\ar[r]\ar[d]^i
 & S^{\mathcal{U}}(X)/S(A)\ar[r]\ar@{.>}[d]^j & 0 \\
0 \ar[r]& S(A)\ar[r] & S(X)\ar[r] & S(X)/S(A)\ar[r] & 0}
\]where $j$ is induced as quotient map and becomes a chain map as before.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\Rightarrow & \cdots \ar[r]& H_p(A)\ar[r]\ar[d]^{i_* =id} &
H_p(S^{\mathcal{U}}(X))\ar[r]\ar[d]^{i_* \cong} & H_p
(S^{\mathcal{U}}(X)/S(A))\ar[r]\ar[d]^{j_* : \cong \textrm{by five
lem.}}
 & H_{p-1}(A)\ar[r]^{\bd_*}\ar[d]^= & \cdots\\
& \cdots\ar[r] & H_p(A)\ar[r] & H_p(X)\ar[r] & H_p(X,A)\ar[r] &
H_{p-1}(A)\ar[r]^{\bd_*} & \cdots}
\]
Now consider $S^{\mathcal{U}}(X)/S(A) : S_p ^{\mathcal{U}} =
S_p(X-U) + S_p(A)$ \\

$ \Rightarrow S_p ^{\mathcal{U}}/S_p(A) =
\frac{S_p(X-U)+S_p(A)}{S_p(A)}
\underset{\underset{i}{\leftarrow}}{\cong} S_p(X-U)/ (S_p(X-U)\cap
S_p(A)) \\ \hspace*{5em}= S_p(X-U)/S_p(A-U) = S_p(X-U, A-U)$\\

$\Rightarrow H_p(X-U, A-U )
\underset{\underset{i_*}{\rightarrow}}{\cong}
H_p(S^{\mathcal{U}}(X)/S(A))\underset{\underset{j_*}{\rightarrow}}{\cong}
H_p(X,A)$\end{proof}\\
Remark. Recall $X_1 , X_2 \subset X , X = X_1 \cup X_2.\\
\{X_1 , X_2\}$ : an excisive couple if $S(X_1) +S(X_2)
\hookrightarrow S(X_1 \cup X_2 = X)$ induces an isomorphism on
homology.\\
\newpage
\textbf{���� 24.} $\{X_1, X_2\}$ is an excisive couple.
$\Leftrightarrow (X_1 , X_1\cap X_2 )\hookrightarrow (X_1 \cup X_2
, X_2)$
is an excision map, i.e., induces an isomorphism on homology.\\
\begin{proof} Hint: Consider the following commutative diagram.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
S(X_1) /S(X_1\cap X_2) \ar[rr]^j \ar[dr]^i &&S(X_1 \cup X_2 )/S(X_2)\\
&S(X_1)+ S(X_2) / S(X_2)\ar[ur]^k &}
\]\end{proof}

Example.\\
\begin{floatingfigure}[l]{3cm}
\begin{pspicture}(2,1.5)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]
% S^2%
\rput(1,1){\psellipse(0,0)(.7,.2)
\pspolygon[fillstyle=solid,fillcolor=white,linestyle=none](-1,0)(1,0)(1,0.5)(-1,0.5)(-1,0)
\psellipse[linestyle=dashed](0,0)(.7,0.2) \pscircle(0,0){.7}%
\psellipse[linecolor=blue](0,-0.3)(.6,.12)
\pscurve[linecolor=blue](0.69,-0.3)(0.72,-0.5)(0.69,-0.7)\rput(1,-0.5){$V$}
}%
\end{pspicture}
\end{floatingfigure}

$X = S^n, E^n_+$: �Ϲݱ�, $E^n_- = A $: ���ݱ�, $U = \AA$�� �θ�
$U$�� excision theorem�� �̿��Ͽ� $X$���� �ٷ� excise�� �� ����.
($\because \bar{U}\nsubseteq \AA)$ �׷����� $V$�� �׸��� ����
��Ƽ� $V$�� excision theorem �� ���� excise�� ���� deformation
retract�� �̿��Ͽ� $U$�� "excise"�� �� �ִ�.\\

$H_p(S^n, E^n_-)\underset{\underset{i_*}{\leftarrow}}{\cong}
H_p(S^n -V, E^n- V)$\\

Note $j : (E_+^n , S^{n-1})\underset{\underset{\exists
r}{\leftarrow}}{\hookrightarrow}(S^n -V, E_-^n -V) $ where $r $ is
a deformation retraction\\(i.e., $rj = id, jr \simeq id)
\Rightarrow j_* :
\cong \Rightarrow H_p(S^n, E^n_-) \underset{\underset{i_*}{\leftarrow}}{\cong} H_p(E_n^+ , S^{n-1})$\\

$\therefore \{E_+^n , E_-^n\}: $ excisive couple and hence can
apply Mayer-Vietori sequence directly.\\

In general, $V\subset U \subset A$ and suppose that $V$ can be
excised. If $(X-U, A-U)$ is a deformation retract of $(X-V,A-V)$,
then $U$ can be excised by the exactly same argument as above.
Hence in this case, $\{X-U, A\}$ is an excisive couple.

\newpage
{\bf Equivalence of simplicial and singular homology}\\

Let $K$ be a simplicial complex. Recall that we have 3 homology
theories.\\
(1) $C(K) \rightsquigarrow H(K)$ ; simplicial homology\\
(2) $\bigtriangleup(K) \rightsquigarrow H^{\triangle}(K)$ ;
ordered simplicial homology\\
(3) $S(|K|) \rightsquigarrow H(|K|)$ ; singular homology\\

And we showed $\mu : \bigtriangleup(K) \to C(K)$ is a natural
chain equivalence,\\
$\hspace*{7.0em} (v_{0}, \cdots v_{p}) \mapsto [v_{0}, \cdots,
v_{p}]$\\
so that $\mu_{*} : H^{\triangle}(K) \to H(K)$ is a natural
isomorphism.\\

���� (2)�� (3)�� ���غ���.\\
Let $\theta : \bigtriangleup(K) \to S(|K|) \\ \hspace*{1.0em}
(v_{0}, \cdots v_{p}) \mapsto l$\\
\hspace*{1.0em} where $l : \triangle^{p} \to |K|$ is an affine map
determined by $l(e_{i}) = v_{i} , i=0,1, \cdots , p$.\\

$\Rightarrow \theta$ is a natural(augmentation-preserving) chain
map so that $\theta_{*} : H^{\triangle}(K) \to H(|K|)$ is a
natural transformation.\\

If $K_{0} < K$, a subcomplex, then
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \bigtriangleup(K_{0}) \ar[r]^{i} \ar[d]^{\theta |} &
\bigtriangleup(K) \ar[r] \ar[d]^{\theta} &
\bigtriangleup(K)/\bigtriangleup(K_{0}) \ar[r]
\ar@{.>}[d]^{\bar{\theta}="\theta"} & 0\\
0 \ar[r] & S(|K_{0}|) \ar[r] & S(|K|) \ar[r] & S(|K|)/S(|K_{0}|)
\ar[r] & 0 }
\]

$\Rightarrow \theta_{*} : H^{\triangle}(K,K_{0}) \to
H(|K|,|K_{0}|)$ is naturally well-defined.\\


\framebox{\hspace*{1em}\parbox[b]{14cm}{$\theta_{*} :
\widetilde{H^{\triangle}}(K) \to \widetilde{H}(|K|)$ is a natural
isomorphism. (Hence so is $\theta_{*} : H^{\triangle}(K) \to
H(|K|)$ and $H_{*} : H^{\triangle}(K,K_{0}) \to H(|K|,|K_{0}|)$ if
$K_{0} < K$ by 5-lemma.)}}\\



{\bf step 1} Assume that $K$ is finite.\\
Prove by induction on $n$(the number of simplices in $K$).\\

$n$=1 \hspace{2.0em} $K=\{v\} \Rightarrow
\widetilde{H^{\triangle}}(K)
= \widetilde{H}(|K|) = 0$.\\

$n > 1$\\
Let $\sigma$ be a simplex of $K$ of maximal dimension so that
$K_{1} := K - \{\sigma\}$ is a subcomplex of $K$.\\
$\underline{\sigma} := \sigma$ as a subcomplex of $K$\\
$bd \underline{\sigma}$ := subcomplex of $\underline{\sigma}$
consisting of proper faces of $\sigma$\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\widetilde{H_{p}^{\triangle}}(K) \ar[r]^{\theta_{*}}
\ar[d]^{i_{*}} &
\widetilde{H_{p}}(|K|) \ar[d]^{i_{*}}\\
H_{p}^{\triangle}(K,\underline{\sigma}) \ar[r]^{\theta_{*}} &
H_{p}(|K|,\sigma)\\
H_{p}^{\triangle}(K_{1}, bd \underline{\sigma})
\ar[r]^{\theta_{*}} \ar[u]^{j_{*}} & H_{p}(|K_{1}|,bd \sigma)
\ar[u]^{j_{*}} }
\]


���� diagram���� $j_{*}$�� excision theorem�� ���ؼ�
isomorphism�̰�, ���� �Ʒ��� $\theta_{*}$�� induction hypothesis��
���� isomorphism�̴�. ���������� $i_{*}$�� ���� $(|K|,\sigma), (K,
\underline{\sigma})$ pair�� ���� long exact sequence�� ���ؼ�
isomorphism�̴�.($ \because \sigma$�� ���� homology�� �׻� 0
�̹Ƿ�) ����,

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\widetilde{H_{p}^{\triangle}}(K) \cong \widetilde{H_{p}}(|K|)}
\]

{\bf step2} Assume that $K$ is infinite.\\
(1) $\theta_{*}$ is onto.\\
Given $\{z\} \in \widetilde{H_{p}}(|K|), |z|:=$ support of $z$.
Then $|z|$ is compact, and hence is contained in $|L|$, a finite
subcomplex of $K$.

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\widetilde{H_{p}^{\triangle}}(L) \ar[r]^{\theta_{*}}_{\cong}
\ar[d]^{i_{*}} & \widetilde{H_{p}}(|L|) \ar[d]^{i_{*}}  \ni & \{z\} \ar[d] \\
\widetilde{H_{p}^{\triangle}}(K) \ar[r]^{\theta_{*}} &
\widetilde{H_{p}}(|K|)  \ni & \{z\}}
\]

step 1 ���� ���� $L$�� ���� $\theta_{*}$�� $\cong$�̰� diagram��
commute�ϹǷ� $K$�� ���� $\theta_{*}$�� onto�̴�.\\


(2) $\theta_{*}$ is 1-1.\\
Suppose $ \{z\} \in \widetilde{H_{p}^{\triangle}}(K)$ and
$\theta_{*}\{z\} = \{\theta(z)\} = 0.$ Then $\exists c \in
\widetilde{S_{p+1}}(|K|)$ such that $\bd c = \theta(z)$. $|c|$ is
compact, hence contained in $L$, a finite subcomplex of $K$. By
above diagram, $\theta_{*}$ is 1-1.\\


{\bf Note} Given a continuous map $f : |K| \to |L|$, let $g : K'
\to L$ be a simplicial approximation of $f$. �׷���

\[
\xymatrix @M=1ex @C=1ex @R=3ex @*[c] { %
% ����
& \ar[dl]_{\mu_{*}(\cong)} H^{\triangle}(K') \ar[rr]^{g_{*}}
\ar[dd]^<<<<{\theta_{*}(\cong)}
& & \ar[dl]^{\mu_{*}(\cong)} H^{\triangle}(L) \ar[dd]^{\theta_{*}(\cong)}\\
% ��� ��
H(K') \ar[rr]^{g_{*}} \ar[dr]_{\eta_{*}=\theta_{*} \circ
\mu_{*}^{-1}} & &
H(L) \ar[dr]_{\eta_{*}}& \\
% �Ʒ���
& H(|K'|) \ar[rr]_{g_{*}=f_{*}} && H(|L|) }
\]

���� �׸����� ��� diagram�� commute�ϰ�, $\mu_{*} , \theta_{*}$��
natural isomorphism�̹Ƿ� $\eta_{*}$�� natural isomorphism��
�ȴ�.\\

\newpage

{\bf Relative Mayer-Vietoris sequence}

\psset{unit=2cm}
\begin{center}
\begin{pspicture}(4,2)%
%\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green]
\rput(2,1){ \psellipse(0,0)(1.5,0.8) \psellipse(-0.6,0)(0.8,0.5)
\psellipse(0.6,0)(0.8,0.5) \psellipse(-0.3,0)(0.5,0.2)
\psellipse(0.3,0)(0.5,0.2) \rput(0,0.9){$X$}
\rput(-0.6,0.6){$X_{1}$} \rput(0.6,0.6){$X_{2}$}
\rput(-0.3,0.3){$A_{1}$} \rput(0.3,0.3){$A_{2}$} }
\end{pspicture}
\end{center}

���� ���� ��� �Ʒ��� ���� sequence���� ������ �� �ִ�.


\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
& 0 \ar[d] & 0 \ar[d] & 0 \ar[d] && \\
0 \ar[r] & S(A_{1} \cap A_{2}) \ar[r]^{\phi|} \ar[d]^{i} &
S(A_{1}) \bigoplus S(A_{2}) \ar[r]^{\psi|} \ar[d]^{i_{1}\oplus
i_{2}} & S(A_{1}) + S(A_{2}) \ar[r] \ar[d]^{j} & 0 &
\textrm{s.e.s}\\
0 \ar[r] & S(X_{1} \cap X_{2}) \ar[r]^{\phi} \ar[d] & S(X_{1})
\bigoplus S(X_{2}) \ar[r]^{\psi} \ar[d] & S(X_{1}) + S(X_{2})
\ar[r] \ar[d] & 0 & \textrm{s.e.s}\\
0 \ar[r] & \frac{S(X_{1} \cap X_{2})}{S(A_{1} \cap A_{2})}
\ar@{.>}[r]^{\bar{\phi}} \ar[d] & \frac{S_(X_{1})}{S(A_{1})}
\bigoplus \frac{S(X_{2})}{S(A_{2})} \ar@{.>}[r]^{\bar{\psi}}
\ar[d] & \frac{S(X_{1}) + S(X_{2})}{S(A_{1}) + S(A_{2})} \ar[r]
\ar[d] & 0
& \textrm{$(\star)$}\\
& 0 & 0 & 0 & & }
\]

(1) $\bar{\phi}$ and $\bar{\psi}$ are chain maps. ($\because$
chain map�� quotient�� chain map)\\
(2) $(\star)$ is exact by 9-lemma.\\

Suppose $\{X_{1},X_{2}\}$ is an excisive couple of $X_{1} \cup
X_{2}$ and $\{A_{1},A_{2}\}$ is an excisive couple of $A_{1} \cup
A_{2}$. Then $(\star)$ induces an long exact sequence

\[
\xymatrix @=1em @*[c]{%
(\ast) \cdots \ar[r] & H_{p}(X_{1} \cap X_{2}, A_{1}\cap A_{2})
\ar[r] & H_{p}(X_{1},A_{1}) \bigoplus H_{p}(X_{2},A_{2}) \ar[r] &
H_{p}(X_{1} \cup X_{2}, A_{1}\cup A_{2}) \ar[r] & \cdots }
\]

���⼭ $H_{p}(X_{1} \cup X_{2}, A_{1} \cup A_{2})$�� �Ǵ� ������ \\
$S(X_{1}) + S(X_{2}) \overset{i}{\hookrightarrow} S(X_{1} \cup
X_{2})$ �� $S(A_{1}) + S(A_{2}) \overset{i}{\hookrightarrow}
S(A_{1} \cup A_{2})$�� homology���� isomorphism�� induce�ϰ�,
5-lemma �� ���ؼ� $\frac{S(X_{1})+S(X_{2})}{S(A_{1})+S(A_{2})}
\hookrightarrow \frac{S(X_{1} \cup X_{2})}{S(A_{1} \cup A_{2})}$
�� homology���� isomorphism�� induce�ϱ� �����̴�.

$(\ast)$�� {\bf Relative Mayer-Vietoris sequence} ��� �θ���.\\


{\bf Special cases}\\
(1) $X=X_{1} \cup X_{2} \Rightarrow H_{p}(X_{1} \cup X_{2} , A_{1}
\cup A_{2}) = H_{p}(X, A_{1} \cup A_{2})$ \\
(2) Furthermore, if $X_{1} = X_{2} = X$, then
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & H_{p}(X, A_{1} \cap A_{2}) \ar[r] & H_{p}(X,
A_{1}) \bigoplus H_{p}(X,A_{2}) \ar[r] & H_{p}(X, A_{1} \cup
A_{2}) \ar[r] & \cdots } \]

(3) If $X = X_{1} \cup X_{2}$ and $A_{1} = A_{2} = A$, then
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & H_{p}(X_{1} \cap X_{2}, A) \ar[r] & H_{p}(X_{1},
A) \bigoplus H_{p}(X_{2},A) \ar[r] & H_{p}(X,A) \ar[r] & \cdots}
\]

\end{document}
