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\begin{document}
\parindent=0cm

\section*{Homology of CW-complexes}
1. {\bf(Cellular chain complex)}\\
Let $X$ be a CW-complex and $X^{p}$, $p$-skeleton of $X$.\\
Consider $(X^{p}, X^{p-1}, X^{p-2})$.\\
\[
\xymatrix @M=0.3ex @C=0.7em @R=0.3em @*[c]{%
\textrm{l.e.s}: \cdots \ar[r] & H_{p}(X^{p-1},X^{p-2}) \ar[r] &
H_{p}(X^{p},X^{p-2}) \ar[r] & H_{p}(X^{p},X^{p-1})
\ar[r]^{\bd_{*}} & H_{p-1}(X^{p-1},X^{p-2}) \ar[r] & \cdots }
\]

Let $C_{p} := H_{p}(X^{p},X^{p-1})$ and $\bd : C_{p}(X) \to
C_{p-1}(X)$ be the connecting homomorphism $\bd_{*}$ in the above
l.e.s..\\
$\Rightarrow \{C_{p}(X), \bd \}$ is a chain complex and is called
a cellular chain complex.\\

\begin{pf}
Note
\[
\xymatrix @M=0.3ex @C=0.7em @R=1.0em @*[c]{%
\cdots \ar[r] & H_{p}(X^{p},X^{p-2}) \ar[r] & H_{p}(X^{p},X^{p-1})
\ar[r]^{\bd_{*}} & H_{p-1}(X^{p-1},X^{p-2}) \ar[r] & \cdots \\
\cdots \ar[r] & H_{p}(X^{p}) \ar[u] \ar[r] & H_{p}(X^{p},X^{p-1})
\ar[u]^{=} \ar[r]^{\bd_{*}'} & H_{p-1}(X^{p-1}) \ar[u]^{i_{*}}
\ar[r] & \cdots }
\]
(All homomorphisms are induced by inclusion maps)\\
$\Rightarrow \bd = \bd_{*} = i_{*} \circ \bd_{*}^{'}$


\[
\scriptsize{
\xymatrix @M=0.5ex @C=0.5em @R=1em @*[c]{%
&&& \ar[d] &&&\\
(X^{p+1}, X^{p}) : \cdots \ar[r] & H_{p+1}(X^{p+1}) \ar[r] &
H_{p+1}(X^{p+1},X^{p}) \ar[r]^{\bd_{*}'} \ar[dr]_{\bd_{*}} &
H_{p}(X^{p}) \ar[r] \ar[d]^{i_{*}} & \cdots && \\
&&& H_{p}(X^{p},X^{p-1}) \ar[d]^{\bd_{*}'} \ar[dr]^{\bd_{*}}
&& \downarrow &\\
(X^{p-1},X^{p-2}) : &&\cdots \ar[r] & H_{p-1}(X^{p-1}) \ar[d]
\ar[r]^{i_{*}} & H_{p-1}(X^{p-1},X^{p-2}) \ar[r] \ar[dr] &
H_{p-2}(X^{p-2}) \ar[r] \ar[d] & \cdots \\
&&& \vdots & \cdots \ar[r] & H_{p-2}(X^{p-2},X^{p-3}) \ar[r]
\ar[d] & \cdots\\
&&& (X^{p},X^{p-1}) && \vdots & \\
}}
\]


위의 diagram에서 $H_{p}(X^{p+1},X^{p}) \overset{\bd_{*}}{\to}
H_{p}(X^{p},X^{p-1}) \overset{\bd_{*}}{\to}
H_{p-1}(X^{p-1},X^{p-2})$ 즉, $C_{p+1}(X) \overset{\bd}{\to}
C_{p}(X) \overset{\bd}{\to} C_{p-1}(X)$를 보면, diagram
commutativity에 의해서 $\bd^{2} = \bd_{*}^{2} = 0$이 되는 것을 알
수 있다. 따라서 $\{C_{p}(X),\bd\}$는 chain complex가 된다.\\

\end{pf}


2. $X^{p} = \coprod D_{\alp}^{p} \cup_{f} X^{p-1}, f= \coprod
\varphi_{\alp}|_{\bd D_{\alp}}, \bar{f} = \coprod \varphi_{\alp}$.
Then by the theorem of homology of adjuction space, $H_{i}(\coprod
D_{\alp}, \coprod \bd D_{\alp}) \overset{\bar{f_{*}},
\cong}{\rightarrow} H_{i}(X^{p},X^{p-1})$.\\
Consequently,\\
$\left\{
\begin{array}{ll}
H_{i}(X^{p},X^{p-1}) = 0 & i \neq p\\
H_{p}(X^{p},X^{p-1}) \cong \bigoplus_{\alp}
H_{p}(D_{\alp}^{p}, \bd D_{\alp}^{p}) &\\
\end{array}\right.$

즉, $H_{p}(X^{p},X^{p-1})$는 free abelian group generated by
$\{\phi_{\alp_{*}}(\gam_{\alp})\}$,\\
\hspace*{2.0em} where $H_{p}(D_{\alp},\bd
D_{\alp}) = <\gam_{\alp}> \cong \mathbb{Z}$이다.\\

이것은 다음 diagram으로 부터 알 수 있다. 아래의 diagram에서 $\bd =
i_{*} \circ \bd_{*}$이다.

\[
\xymatrix @M=1ex @C=2em @R=1em @*[c]{%
\bigoplus H_{p}(D_{\alp}^{p},\bd D_{\alp})
\ar[d]^{\cong}_{\bar{f_{*}}} \ar[r]^{\bd_{*}} & \bigoplus
H_{p-1}(S_{\alp}^{p-1}) = \bigoplus <\sig_{\alp}> \ar[d]^{f_{*}} &&\\
H_{p}(X^{p},X^{p-1}) \ar[r]_{\bd_{*}} & H_{p-1}(X^{p-1})
\ar[r]_{i_{*}} & H_{p-1}(X^{p-1},X^{p-2}) \ar[r] & \cdots \\}
\]

여기에서 $\bd(\bar{f_{*}}(\gam_{\alp})) =
i_{*}f_{*}\bd_{*}(\gam_{\alp}) = i_{*}f_{*}(\sigma_{\alp})$


3.
\[
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
H_{p}(X) & \ar[l]^{(1), \cong}_{incl_{*}} H_{p}(X^{p+1})
\ar[r]^{(2), \cong}_{incl_{*}} \ar[d] ^{=} & H_{p}(X^{p+1},
X^{p-2})\\
& H_{p}(X^{p+1},X^{-1}(=\emptyset))&\\
}
\]

\begin{pf}
(1)
\[
\footnotesize{\xymatrix @M=1ex @C=1em @R=1em {%
(X^{n+1},X^{n}) \Rightarrow \cdots \ar[r] & H_{p+1}(X^{n+1},X^{n})
\ar[r] & H_{p}(X^{n}) \ar[r] & H_{p}(X^{n+1}) \ar[r] &
H_{p}(X^{n+1},X^{n}) \ar[r] & \cdots}}
\]
If $p, p+1 < n+1$, i.e., $p<n$, then $H_{p}(X^{n}) \cong H_{p}(X^{n+1}).$\\

\begin{center}
$\therefore H_{p}(X^{p+1}) \overset{\cong}{\to} H_{p}(X^{p+2})
\overset{\cong}{\to} H_{p}(X^{p+3}) \overset{\cong}{\to} \cdots$\\
\end{center}

Now consider $i_{*} : H_{p}(X^{p+1}) \to H_{p}(X)$.\\

{\bf $i_{*}$ is onto.}\\
$\forall a \in Z_{p}(X), |a|:$support of $a$, is compact, hence
$|a| \subset X^{N}$ for some $N$. And the chain follows from the following commutative diagram.\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
H_{p}(X^{p+1}) \ar[dr]_{\cong} & \longrightarrow & H_{p}(X) (\ni \alp = \{a\})\\
&H_{p}(X^{N}) (\ni \alp' = \{a\}) \ar[ur] & }
\]

{\bf $i_{*}$ is one-to-one.}\\
$i_{*}(\bet) = 0$. Then $i_{*}(\bet) =: \bet'=\{b'\} \Rightarrow
b'=\bd c, c \in S_{p+1}(X)$.\\
Furthermore, $|c|$ is compact, hence $|c| \subset X^{N}$ for some $N$. And\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
H_{p}(X^{p+1}) \ar[dr]_{\cong} & \rightarrow & H_{p}(X) &
& \bet \ar@{|->}[dr] & \overset{i_{*}}{\mapsto} & \bet'=0\\
& H_{p}(X^{N}) \ar[ur]&&&& \bet^{''} = \{\bd c\} = 0
\ar@{|->}[ur]& }
\]
$\Rightarrow \bet =0$\\

(2) $(X^{p+1},X^{i},X^{i-1})$로 부터 아래와 같은 l.e.s.를 얻는다.\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & H_{p}(X^{i},X^{i-1}) \ar[r] &
H_{p}(X^{p+1},X^{i-1}) \ar[r] & H_{p}(X^{p+1},X^{i}) \ar[r]
 & H_{p-1}(X^{i}, X^{i-1}) \ar[r] & \cdots }
\]
여기에서 만약 $i \neq p, p-1$인 경우 즉, $i <p-1$의 경우
$H_{p}(X^{i}, X^{i-1}) = H_{p-1}(X^{i}, X^{i-1}) = 0$이 되므로
$H_{p}(X^{p+1},X^{i-1}) \cong H_{p}(X^{p+1}, X^{i})$가 된다.
따라서 $H_{p}(X^{p+1}) = H_{p}(X^{p+1},X^{-1}) \cong
H_{p}(X^{p+1},X^{0}) \cong \cdots \cong H_{p}(X^{p+1},X^{p-2})$가
성립한다.
\end{pf}\\
4.
\begin{thm} $H_{p}(C(X)) \overset{\lam, \cong}{\rightarrow}
H_{p}(X), \forall p$.\\ Furthermore, $\lam$ is natural, where
$H_{p}(C(X))$ is homology of chain complex
$\{C_{p},\bd\} = \{H_{p}(X^{p},X^{p-1}),\bd_{*}\}$.\\
\end{thm}
\begin{pf}
Consider $(X^{p+1}, X^{p}, X^{p-1}, X^{p-2}).$\\
\[
\footnotesize{\xymatrix @M=1ex @C=0.5em @R=1em @*[c]{%
&& \downarrow & \downarrow&&\\
& 0 \ar[r] & H_{p+1}(X^{p+1},X^{p})
\ar[d] \ar[r]^{=} \ar[dr]_{\bd} & H_{p+1}(X^{p+1},X^{p}) \ar[d]^{\bd_{1}} &&\\
(X^{p}, X^{p-1}, X^{p-2}) : & 0 \ar[r] & H_{p}(X^{p},X^{p-2})
\ar[d]^{\hspace{4.0em} \Huge{\spadesuit}} \ar[r] &
H_{p}(X^{p},X^{p-1}) \ar[d] \ar[r]^{\bd_{0}} \ar[dr]_{\bd} &
H_{p-1}(X^{p-1},X^{p-2}) \ar[r] \ar[d]^{=} &
\cdots \\
(X^{p+1}, X^{p-1}, X^{p-2}) : & 0 \ar[r] & H_{p}(X^{p+1},X^{p-2})
\ar[d] \ar[r] & H_{p}(X^{p+1},X^{p-1}) \ar[d] \ar[r] &
H_{p-1}(X^{p-1},X^{p-2}) \ar[r] & \cdots \\
&&0&0&&\\
&&(X^{p+1},X^{p}, X^{p-2}) \hookrightarrow
&(X^{p+1},X^{p},X^{p-1})&&\\
}}
\]
\begin{center}
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\psline(0.9,0.2)(1.1,2.5) \psellipse(0.5,2.5)(0.15,0.07)
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\end{pspicture}
\end{center}

우선 위의 diagram으로부터 $H_{p}(C(X)) = ker \bd_{0} / im
\bd_{1}$임을 안다. 이제 위의 그림으로부터 $ker \bd_{0} / im
\bd_{1}$는 음영이 표시된 부분, 즉 $H_{p}(X^{p+1},X^{p-2})$인 것을
알고 따라서 3의 내용에 의해서 $H_{p}(X)$인 것이 증명된다.\\

{\bf Remark}\\
Let $f : X^{\textrm{CW}} \rightarrow Y^{\textrm{CW}}$ be a
cellular map, i.e., $f(X^{p}) \subset Y^{p}, \forall p$. Then, $f
: (X^{p}, X^{p-i}) \rightarrow (Y^{p},Y^{p-i})$ and induces
$f_{*}$ on various homology of pairs and hence induces a chain map
$f_{*} : C(X) \rightarrow C(Y)$.\\
Since everything is functorial and natural
\[
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
H_{p}(C(X)) \ar[d]^{f_{*}} \ar[r]^{\lam_{X}}_{\cong} & H_{p}(X)
\ar[d]^{f_{*}}\\
H_{p}(C(Y)) \ar[r]^{\cong}_{\lam_{Y}} & H_{p}(Y)}
\]
commutes, i.e., $\lam$ is natural.\end{pf}
\begin{cor}
Let $X$ be an n-dimensional CW-complex. Then $H_{q}(X) = 0, q>n$.
\end{cor}

\end{document}