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\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}
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\newcommand{\vph}{\varphi}
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\newcommand{\Del}{\Delta}
\newcommand{\Lam}{\Lambda}
\newcommand{\rb}{\mathbb{R}}
\newcommand{\zb}{\mathbb{Z}}
\newcommand{\bd}{\partial}
\newcommand{\cc}{\mathcal{C}}
\newcommand{\dd}{\mathcal{D}}
\newcommand{\hh}{\widetilde{H}}
\newcommand{\tphi}{\widetilde{\phi}}
\newcommand{\hhp}{\widetilde{H}_p}
\newcommand{\hhq}{\widetilde{H}_q}
\newcommand{\hp}{H_p}
\newcommand{\hph}{H^{p}}
\newcommand{\cpc}{C^{p}(\cc;G)}
\newcommand{\snn}{S^{n+1}}
\newcommand{\htx}{\textrm{Hom}}
\newcommand{\im}{\textrm{im}}

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\newtheorem{que}{Áú¹®}
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\newtheorem{rem}{ÁÖ}
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\begin{document}
\parindent=0cm

\section*{IV.Cohomology of a chain complex}
{\bf 1.} Let $\cc = \{C_{p},\bd\}$ be a chain complex of abelian
groups
(or $R$-modules) and $G$ be an abelian group ( or an $R$-module).\\

$\hspace*{2.0em} \cpc = \htx(C_{p},G) : p$-dimensional cochain
group
of $\cc$.\\
\hspace*{4.0em}(or $\htx_{R}(C^{p},G) : p$-dimensional cochain
$R$-module of $\cc$.)\\

\hspace*{2.0em} coboundary operator $\del : C^{p} \to C^{p+1}$ is
the dual of $\bd :
C_{p+1} \to C_{p}$.\\
\hspace*{2.0em} $\Rightarrow \del^{2}=0$, since $\bd^{2}=0$.\\

$%
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\ar[r] & C_{p+2} \ar[r]^{\bd} & C_{p+1} \ar[r]^{\bd}
\ar[dr]_{\alp \circ \bd} & C_{p} \ar[r] \ar[d]^{\alp} & \cdots\\
&&&G& }$ \raisebox{-.5cm}{\hspace*{1.5em}\parbox{6cm}{$\del(\alp)
:= \alp \circ \bd$·Î Á¤ÀǵǸç, ¶ÇÇÑ $\del^{2}(\alp) := (\alp \circ
\bd)\circ \bd = \alp \circ \bd^{2} =0$ÀÌ ¼º¸³ÇÑ´Ù. µû¶ó¼­,}}

$\hspace*{2.0em}\cdots \rightarrow C_{p+1}
\overset{\bd}{\rightarrow} C_{p} \overset{\bd}{\rightarrow}
C_{p-1} \overset{\bd}{\rightarrow}
\cdots$ \\
$\hspace*{2.0em}\Rightarrow \cdots \leftarrow C^{p+1}
\overset{\del}{\leftarrow} C^{p} \overset{\del}{\leftarrow}
C^{p-1} \leftarrow \cdots :
\cc^{*} = \{C^{p},\del\}$ : cochain complex.\\


Homology of $\cc^{*}$ is the cohomology of $\cc$ :\\
\hspace*{2.0em} $Z^{p}(\cc;G):= \ker \, \del \subset C^{p},
B^{p}(\cc;G):= \im \, \del \subset Z^{p}(\cc;G)$\\
\hspace*{2.0em} $\hph(\cc;G):=Z^{p}(\cc;G)/B^{p}(\cc;G)$\\
\hspace*{4.0em} : cohomology of $\cc$ with coefficient $G$ in dim\,$p$\\

Simplicial cohomology if $\cc$ is the simplicial chain complex.\\
Singular cohomology if $\cc$ is the singular chain complex.\\


{\bf 2.} Cohomology of augmented chain complex,\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & C_{p} \ar[r] & \cdots \ar[r] & C_{1} \ar[r] &
C_{0} \ar[r]^{\eps} & \zb \ar[r] & 0 }
\]
\hspace*{2.0em} is called the reduced cohomology of $\cc$ and
denoted by $\widetilde{H}^{p}(\cc ;
G)$.\\

{\bf Note} $\left\{
\begin{array}{ll}
\widetilde{H}^{p}(\cc ; G) = H^{p}(\cc ; G) & \textrm{if}\,\, p > 0 \\
H^{0}(\cc ; G) = \widetilde{H}^{0}(\cc ; G) \bigoplus G &
\end{array}\right.$
(Exercise)\\


{\bf 3. Functorial property}\\
\hspace*{2.0em} A chain map $\phi : \cc \to \dd$ induces a chain
map $\widetilde{\phi} : \dd^{*} \to \cc^{*}$ between the
cochain complexes.\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & C_{p+1} \ar[r]^{\bd} \ar[d]_{\phi_{p+1}} & C_{p}
\ar[r] \ar[d]^{\phi_{p}} & \cdots & \Rightarrow & \cdots & \ar[l]
C^{p+1} & \ar[l]^{\del} C^{p} & \ar[l] \cdots \\
\cdots \ar[r] & D_{p+1} \ar[r]^{\bd} & D_{p} \ar[r] & \cdots &&
\cdots & \ar[l] D^{p+1} \ar[u]_{\widetilde{\phi_{p+1}}} &
\ar[l]^{\del} D^{p} \ar[u]^{\widetilde{\phi_{p}}} & \ar[l] \cdots
}
\]

$\phi \circ \bd = \bd \circ \phi \Rightarrow \widetilde{\phi \circ
\bd} = \widetilde{\bd \circ \phi} \Rightarrow \del \circ \tphi =
\tphi \circ \del$ÀÌ ¼º¸³.\\

Since $\tphi$ is a chain map, it induces a homomorphism
$\phi^{*} : \hph(\dd ; G) \to \hph(\cc ; G)$. Therefore,\\
\hspace*{2.0em} $f : X \to Y$\\
\hspace*{2.0em} $ \Rightarrow f_{\sharp} : S_{p}(X) \to S_{p}(Y)$
: singular chain map.\\
\hspace*{2.0em} $ \Rightarrow f^{\sharp} : S^{p}(Y) \to S^{p}(X)$
: singular cochain map.\\
\hspace*{2.0em} $ \Rightarrow f^{*} : \hph(Y;G) \to \hph(X;G)$\\
\hspace*{2.0em} ,where $\hph(X;G) = \hph(S(X) ; G)$\\
\hspace*{2.5em} : singular cohomology of $X$
with coefficient $G$.\\
Similarly for the simplicial case.\\


Now\\
$%
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
X \ar[r]^{f} & Y \ar[r]^{g} & Z \\
\triangle \ar[u]_{\sigma} \ar[ur]_{f \circ \sigma =
f_{\sharp}(\sigma)}&&}$\raisebox{-.5cm}{\hspace*{1.5em}\parbox{7cm}{$\hspace*{2.0em}\Rightarrow
(g \circ f)_{\sharp} = g_{\sharp} \circ f_{\sharp}$
\\
\hspace*{2.0em}$\Rightarrow (g \circ f)^{\sharp} = f^{\sharp} \circ g^{\sharp}$\\
\hspace*{2.0em}$\Rightarrow (g \circ f)^{*} = f^{*} \circ g^{*}$ in $\hph$\\
}}

And $id._{\sharp} = id. \Rightarrow id.^{*} = id.$\\
$\therefore \hph :\mathcal{T}op \to \mathcal{A}bel.\, groups$(or
$R-\mathcal{M}od$) : contravariant functor.\\

{\bf 4. Chain homotopy and equivalence}\\
Let $D : \phi \simeq \psi : \cc \to \cc'$ be a chain homotopy,
i.e., $\bd
D + D\bd = \phi - \psi$\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & C_{p+1} \ar[r]^{\bd} \ar[d] & C_{p} \ar[dl]^{D}
\ar[r]^{\bd} \ar[d]_{\phi}^{\psi} & C_{p-1} \ar[dl]^{D} \ar[r] &
\cdots & \Rightarrow & \cdots & \ar[l]
C^{p+1} & \ar[l]^{\del} C^{p} & \ar[l] C_{p-1} & \ar[l] \cdots \\
\cdots \ar[r] & C_{p+1}^{'} \ar[r]_{\bd} & C_{p}^{'} \ar[r]_{\bd}
&C_{p-1}^{'} \ar[r] &\cdots && \cdots & \ar[l] C^{'p+1} \ar[u]
\ar[ur]^{\widetilde{D}} & \ar[l]_{\del} C^{' p}
\ar[u]_{\widetilde{\phi}}^{\widetilde{\psi}}
\ar[ur]_{\widetilde{D}} & \ar[l] C_{' p-1} & \ar[l] \cdots }
\]

$\therefore \widetilde{D} \circ \widetilde{\bd} + \widetilde{\bd}
\circ \widetilde{D} = \widetilde{\phi} - \widetilde{\psi}$\\
$\Rightarrow \del \circ \widetilde{D} + \widetilde{D} \circ \del =
\widetilde{\phi} - \widetilde{\psi}$\\
$\Rightarrow \widetilde{D} : \widetilde{\phi} \simeq
\widetilde{\psi}$, cochain homotopy.\\
In this case, $\phi^{*} = \psi^{*}$.\\

$\phi : \cc \to \cc'$, a chain homotopy equivalence\\
\hspace*{2.0em} $\Rightarrow \phi_{*}$ and $\phi^{*}$ are
isomorphisms.\\

\begin{cor}
$f \simeq g : X \to Y \Rightarrow f_{\sharp} \simeq g_{\sharp} :
S(X)
\to S(Y).$\\
\hspace*{4.0em} $f^{*} = g^{*} : H^{*}(Y) \to H^{*}(X)$\\
Similarly for pairs, $f \simeq g : (X,A) \to (Y,B)$,\\
where $H^{p}(X,A;G):= H^{p}(S(X,A);G)$.\\
\end{cor}

{\bf 5. Long exact sequence for pairs.}\\
Recall\\
\begin{center}
$0 \rightarrow S(A) \rightarrow S(X) \rightarrow S(X)/S(A) =
S(X,A) \rightarrow 0$ : s.e.s.\\
$\overset{\textrm{snake}}{\Longrightarrow} \cdots \rightarrow
H_{p}(A) \rightarrow H_{p}(X) \rightarrow H_{p}(X,A)
\overset{\bd_{*}}{\rightarrow} H_{p-1}(A) \rightarrow \cdots$ :
l.e.s. of $(X,A)$.\\
\end{center}

More generally,\\
\begin{center}
$0 \rightarrow \cc \rightarrow \dd \rightarrow \mathcal{E} \rightarrow 0$ : s.e.s.\\
$\overset{\textrm{snake}}{\Longrightarrow} \cdots \rightarrow
H_{p}(\cc) \rightarrow H_{p}(\dd) \rightarrow H_{p}(\mathcal{E})
\overset{\bd_{*}}{\rightarrow} H_{p-1}(\cc) \rightarrow \cdots$ :
l.e.s.\\
\end{center}

If the dual sequence of a short exact sequence is short exact,
then we still obtain a long exact sequence by the snake lemma. But
in general,\\

\begin{center}
$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ : s.e.s.\\
$\nRightarrow 0 \leftarrow A^{*} \leftarrow B^{*} \leftarrow
C^{*} \leftarrow 0$ : s.e.s.\\
\end{center}
i.e., {\bf $\htx$ functor does not preserve short exact sequence!}\\

{\bf Exactness of $\htx$ functor}

\begin{thm}

(1) $B \overset{g}{\rightarrow} C \rightarrow 0$ : exact
$\Rightarrow \htx(B,G) \overset{\widetilde{g}}{\leftarrow}
\htx(C,G)
\leftarrow 0$ : exact.\\
(2) $A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C
\rightarrow 0 : $ exact\\ \hspace*{3.0em} $\Rightarrow \htx(A,G)
\overset{\widetilde{f}}{\leftarrow} \htx(B,G)
\overset{\widetilde{g}}{\leftarrow} \htx(C,G) \leftarrow 0$ :
exact.\\
(3) $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 :$
split exact.\\
$\Rightarrow 0 \leftarrow \htx(A,G)
\overset{\widetilde{f}}{\leftarrow} \htx(B,G)
\overset{\widetilde{g}}{\leftarrow} \htx(C,G) \leftarrow 0$ :
split exact.\\

\end{thm}

\begin{pf}
(1) $%
\xymatrix @M=1ex @C=2em @R=1em @*[c]{%
B \ar@{>>}[r]^{g} \ar[dr]_{\widetilde{g}(\alp)=\alp \circ g} & C \ar[d]^{\alp}\\
&G }$\raisebox{-.5cm}{\hspace*{1.5em}\parbox{8cm}{\hspace*{1.5em}
Show $\widetilde{g}$ is one to
one : $\widetilde{g}(\alp) = \alp \circ g = 0$\\
\hspace*{1.5em}$\Rightarrow \alp= 0 $\, since $g$ is onto.}}


(2) $g \circ f=0 \Rightarrow \widetilde{f} \circ \widetilde{g} =
0$.\\
$%
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
A \ar[r]^{f} \ar[dr]_{0} & B \ar[r]^{g} \ar[d]^{\bet} & C
\ar@{.>}[dl]^{\bar{\bet}}\\
&G&
}$\raisebox{-.5cm}{\parbox{9cm}{\hspace*{1.5em}$\widetilde{f}(\bet)
= \bet \circ f = 0$\\ \hspace*{1.5em} $\Rightarrow \ker \, \bet
\supset \im \, f = \ker \, g$ and
$C \cong B/ \ker\,g$\\
\hspace*{1.5em}$\Rightarrow \bet$ induces $\bar{\bet} : C \to G $
and \\ \hspace*{1.5em} $\widetilde{g}(\bar{\bet}) = \bar{\bet}
\circ g = \bet$}}


(3)Since short exact sequence splits, there exists $p : B \to A$
such that $p \circ f = id_{A}$.\\
\[
\xymatrix @M=1ex @C=1.5em @R=1em @*[c]{%
0 \ar[r] & A \ar[r]^{f} & B \ar@{.>}@(dl,ul)[l]^>>>>{p} \ar[r]^{g}
& C \ar[r] & 0}
\]

$\Rightarrow \widetilde{f} \circ \widetilde{p} = \widetilde{id.} =
id. \Rightarrow \widetilde{f}$ is onto and $\htx$-sequence splits.
\end{pf}

{\bf Remark}(1)
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
&0 \ar[r] & \zb \ar[r]^{\times 2}_{f} & \zb \ar[r] & \zb/2 \ar[r]
& 0& \textrm{exact}}
\]
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\Rightarrow & 0 & \ar@{.>}[l] \htx(\zb,\zb) &
\ar[l]_{\widetilde{f}} \htx(\zb,\zb) & \ar[l] \htx(\zb/2 , \zb) &
\ar[l] 0 & \textrm{exact(?)}}
\]

¸¦ »ìÆ캸¸é, ¿ì¼± $\htx(\zb , \zb) \cong \zb$ÀÌ°í µû¶ó¼­
$\widetilde{f}$´Â $\zb$ÀÇ 1À» 2·Î º¸³»´Â $\times 2$ÀÎ mapÀÓÀ» ¾Ë
¼ö ÀÖ´Ù. µû¶ó¼­ onto°¡ µÉ ¼ö ¾ø°í, ¹°·Ð exact°¡ ¾Æ´Ï´Ù.\\

(2) In general,
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
 0 \ar[r] & \zb \ar[r]^{\times n}_{f} & \zb \ar[r] & \zb/n \ar[r]
 & 0}
 \]

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
 \Rightarrow & \htx(\zb,G) & \ar[l]^{\widetilde{f}}_{\times n}
 \htx(\zb,G) & \ar[l] \htx(\zb/n,G) & \ar[l] 0}
\]

$%\because
\xymatrix @M=1ex @C=3em @R=1em @*[c]{%
\zb \ar[r]^{f} \ar[dr]_{\widetilde{f}(\alp)} & \zb \ar[d]^{\alp}\\
& G}$\raisebox{-.3cm}{\parbox{10cm}{\hspace*{2.0em}a homomorphism
$\alp : \zb \to G$ \\ \hspace*{2.0em} is determined  by $\alp(1)
\in
G$ and\\ \hspace*{2.0em} hence $\htx(\zb, G) \cong G$. \\
\hspace*{2.0em}$\widetilde{f}(\alp)(1) = \alp(f(1)) = \alp(n) = n
\alp(1) \Rightarrow \widetilde{f}(\alp) = n \alp$}}



$\Rightarrow \htx(\zb/n , G) \cong \ker\,(G \overset{\times
n}{\rightarrow} G)$\\
¸¸¾à $G=\zb/m$À̸é $\htx(\zb/n , G)$´Â ¾î¶»°Ô µÇ´Â°¡?(Exercise)
À̵é·ÎºÎÅÍ ¿ì¸®´Â ÁÖ¾îÁø finitely generated abelian group $A$¿¡
´ëÇؼ­ $\htx(A,G)$¸¦ °è»êÇÒ
¼ö ÀÖ´Ù.\\

{\bf Return to long exact sequence:}\\
¿ì¼± $S_{p}(X,A)$°¡ freeÀ̹ǷÎ\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & S_{p}(A) \ar[r] & S_{p}(X) \ar[r] & S_{p}(X,A) \ar[r] &
0}
\]
splits for each $p$, hence by the above argument,\\
\[
\xymatrix @M=1ex @C=1.5em @R=1em @*[c]{%
0 & \ar[l] S^{p}(A) & \ar[l] S^{p}(X) & \ar[l] S^{p}(X,A) & \ar[l]
0}
\]
is exact(split) for each $p$. Applying snake lemma, we obtain\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots & \ar[l] \hph(A;G) & \ar[l] \hph(X;G) & \ar[l] \hph(X,A;G)
& \ar[l]^{\del^{*}} H^{p-1}(A;G) & \ar[l] \cdots }
\]
Of course, it is also true for reduced cohomology.\\

{\bf Note.}\,In the above l.e.s., the connecting
homomorphism $\del^{*}$is given as follows.\\
\[
\xymatrix @M=1ex @C=0.5em @R=1em @*[c]{%
& \downarrow & \downarrow & \downarrow &&&& \uparrow & \uparrow &
\uparrow & \\
0 \ar[r] & C_{p+1} \ar[r] \ar[d]_{\bd} & D_{p+1} \ar[r]
\ar[d]_{\bd} & E_{p+1} \ar[r] \ar[d]_{\bd} & 0 & \Rightarrow & 0
&\ar[l] C^{p+1} & \ar[l] D^{p+1} & \ar[l] E^{p+1}& \ar[l] 0\\
0 \ar[r] & C_{p} \ar[r] & D_{p} \ar[r] &E_{p} \ar[r] & 0&& 0 &
\ar[l] C^{p} \ar[u]& \ar[l] D^{p} \ar[u] & \ar[l] E^{p} \ar[u]&
\ar[l] 0\\
& \downarrow & \downarrow & \downarrow &&&& \uparrow & \uparrow &
\uparrow &\\
&&&&&&&&&&\\
&&\heartsuit \ar[d] \ar[r] & \bullet \ar[d]
\ar@{.>}[dll]^>>>>>>>>>{\bd^{*}} &&&&
0& \ar[l] \spadesuit & \ar[l] \clubsuit&\\
&\clubsuit \ar[r] & \spadesuit \ar[r] &0 &&&&\bullet \ar[u]
\ar@{.>}[urr]_>>>>>>>>>{\del^{*}} & \ar[l] \heartsuit \ar[u]&&}
\]

Furthermore, long exact sequence is functorial.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \cc \ar[r] \ar[d] & \dd \ar[r] \ar[d] & \mathcal{E}
\ar[r] \ar[d] & 0 & \textrm{chain maps}\\
0 \ar[r] & \cc' \ar[r] & \dd' \ar[r] & \mathcal{E}' \ar[r]  & 0 &
}
\]
$\Rightarrow$
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 & \ar[l] \cc^{*} & \ar[l] \dd^{*} & \ar[l] \mathcal{E}^{*} &
\ar[l] 0 & \textrm{cochain maps}\\
0 & \ar[l] \cc^{'*} \ar[u] & \ar[l] \dd^{'*} \ar[u] & \ar[l]
\mathcal{E}^{'*} \ar[u] & \ar[l] 0 & }
\]
$\Rightarrow$ Functoriality of long exact sequence follows from
the earlier result. In particular, $f:(X,A) \to (Y,B) \Rightarrow
f_{*}(f^{*}$,resp.) induces a homomorphism for long exact sequence
of $(X,A)((Y,B)$,resp.) to the long exact sequence of
$(Y,B)((X,A)$,resp.).\\

Long exact sequence of triples : $A \subset B \subset X
\Rightarrow \exists$ a functorial long exact sequence,\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots & \ar[l] \hph(B,A) & \ar[l] \hph(X,A) & \ar[l] \hph(X,B) &
\ar[l]^{\del^{*}} H^{p-1}(B,A) & \ar[l] \cdots}
\]
¿Ö³ÄÇϸé, ¾Æ·¡ÀÇ short exact sequence¿¡¼­ $S(X)/S(B)$°¡
freeÀ̹ǷÎ, sequence°¡ splitsÇÏ°í À§¿¡¼­¿Í °°ÀÌ dualizeÇÏ°í snake
lemma¸¦ Àû¿ëÇÏ¸é µÇ±â ¶§¹®ÀÌ´Ù.\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & S(B)/S(A) \ar[r] & S(X)/S(A) \ar[r] & S(X)/S(B) \ar[r]
& 0}
\]

{\bf 6.(Excision)}\\

Let $\bar{U} \subset {\AA}$.\\
Then $i : (X-U,A-U) \hookrightarrow (X,A)$ induces an isomorphism
$i^{*} : H^{*}(X,A) \to
H^{*}(X-U,A-U)$.\\

\begin{pf}
{\bf (1st proof)}\\
Recall $i: S^{\mathcal{U}}(X) \hookrightarrow S(X)$ is a chain
homotopy equivalence, where $\mathcal{U} = \{X-U,A\}$, and hence
an isomorphism on cohomology.\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & S(A) \ar[r] \ar[d]^{=} & S^{\mathcal{U}}(X) \ar[r]
\ar[d]^{i} & S^{\mathcal{U}}(X)/S(A) \ar[r] \ar@{.>}[d]^{j} & 0\\
0 \ar[r] & S(A) \ar[r] & S(X) \ar[r] & S(X)/S(A) \ar[r] & 0 }
\]
Since $S^{\mathcal{U}}(X)/S(A)$ is free, we obtain the following
diagram.\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots & \ar[l] H^{p}(A) & \ar[l] H^{p}(S^{\mathcal{U}}(X)) &
\ar[l] H^{p}(S^{\mathcal{U}}(X)/S(A)) & \ar[l] \cdots\\
\cdots & \ar[l] H^{p}(A) \ar[u]^{=} & \ar[l] H^{p}(X)
\ar[u]_{\cong}^{i^{*}} & \ar[l] H^{p}(X,A) \ar[u]^{j^{*}} & \ar[l]
\cdots}
\]
By the 5-lemma, $j^{*}$ is an isomorphism.\\
Furthermore, $S^{\mathcal{U}}(X)/S(A) = \frac{S(X-U)+S(A)}{S(A)}
\cong \frac{S(X-U)}{S(X-U) \cap S(A)} = S(X-U,A-U)$ and this
completes
the proof.\\

\end{pf}

{\bf (2nd proof)Algebraic Mapping Cone}\\

(1)Construction\\

Let $f : \mathcal{C} \to \mathcal{D}$ be a chain map. Then mapping
cone $Cf=\mathcal{E}$ is defined by $E_{p}=D_{p} \bigoplus
C_{p-1}$ with $\bd(d,c)=(\bd d + f(c),-\bd c)$.\\

{\bf check} $\bd^{2}=0$ : \\

$\bd^{2} = \begin{pmatrix} \bd & f \\ 0 & -\bd \end{pmatrix}
\begin{pmatrix} \bd & f \\ 0 & -\bd \end{pmatrix} =
\begin{pmatrix} \bd^{2} & \bd f - f \bd \\ 0 & \bd^{2}
\end{pmatrix} = 0 $\\

Now
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & D_{p} \ar[r]^{i} & E_{p} \ar[r]^{p} & C_{p-1} \ar[r] &
0 }
\]
where $i(d) = (d,0)$ and $ p(d,c) = c$. And
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & D_{p+1} \ar[r] \ar[d]^{\bd} & D_{p+1} \bigoplus C_{p}
\ar[r] \ar[d]^{\bd = \scriptsize{\begin{pmatrix} \bd & f \\ 0 &
-\bd
\end{pmatrix}}} & C_{p} \ar[r] \ar[d]^{-\bd} & 0 &
\textrm{commutes} \\
0 \ar[r] & D_{p} \ar[r] & D_{p} \bigoplus C_{p-1} \ar[r] & C_{p-1}
\ar[r] & 0 & }
\]

$\Rightarrow$ \\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \mathcal{D} \ar[r]^{i} & \mathcal{E} \ar[r]^{p} &
\mathcal{C}^{'} \ar[r] & 0 & \quad \textrm{s.e.s. of chain
complexes.}}
\]
where $(C_{p}^{'}, \bd) = (C_{p-1},-\bd)$. Furthermore, by
applying snake lemma,\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\Rightarrow \cdots \ar[r] & H_{p}(\mathcal{D}) \ar[r] &
H_{p}(\mathcal{E}) \ar[r] & H_{p}(\mathcal{C}^{'}) \ar[r] &
H_{p-1}(\mathcal{D}) \ar[r] & \cdots }
\]

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\Rightarrow \cdots \ar[r] & H_{p}(\mathcal{D}) \ar[r] & H_{p}(Cf)
\ar[r] & H_{p-1}(\mathcal{C}) \ar[r]^{f_{*}} &
H_{p-1}(\mathcal{D}) \ar[r] & H_{p-1}(Cf) \ar[r] & \cdots }
\]

$\therefore f_{*} : H_{*}(\mathcal{C}) \to H_{*}(\mathcal{D})$ is
an isomorphism if and only if $H_{*}(Cf) = 0$.


Similarly for cohomology also, if $\mathcal{C}$ is free so that the above short exact sequence splits.\\

(2) Recall the following fact.\\
Let $\mathcal{C}$ be a free chain complex. Then
$H_{*}(\mathcal{C})=0$(i.e. $\mathcal{C}$ is acyclic) if and only
if $id. \simeq 0$(chain contractible).
It easily follows from the comparison theorem.\\

{\bf Review of comparison theorem}\\
\[\small{
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & X_{n} \ar[r]^{\bd} \ar@{.>}[d] & X_{n-1}
\ar[r]^{\bd} \ar@{.>}[d] & \cdots \ar[r] & X_{1} \ar[r]^{\bd}
\ar@{.>}[d]^{f_1} & X_{0} \ar[r]^{\eps} \ar@{.>}[d]^{f_0} & A
\ar[r] \ar[d]^{\gam} & 0 & \scriptsize{\textrm{augmented
free chain complex over $A$}} \\
\cdots \ar[r] & X_{n}^{'} \ar[r]^{\bd} & X_{n-1}^{'} \ar[r]^{\bd}
& \cdots \ar[r] & X_{1}^{'} \ar[r]^{\bd} & X_{0}^{'} \ar[r]^{\eps}
& A^{'} \ar[r] & 0 & \small{\textrm{resolution of $A^{'}$}} } }\]

$\Rightarrow$ 1.$\gam$ can be lifted to a chain map $f : X \to
X^{'}$.\\
\hspace*{1.0em} 2. Any two liftings are chain homotopic.\\


Let $f : \mathcal{C} \to \mathcal{D}$ be a chain map of free chain
complexes. Then the followings are equivalent.\\

1. $H_{*}(Cf) = 0$\\
2. $f$ is a chain homotopy equivalence.\\
3. $f_{*}$ is an isomorphism.\\

\begin{pf}
Clearly 3 implies 1 and 2 implies 3.\\
Remains to show 1 implies 2.\\

$H_{*}(\mathcal{E})=0 \Rightarrow \exists T : D_{p} \bigoplus
C_{p-1}(=E_{p}) \to D_{p+1} \bigoplus C_{p}(=E_{p+1})$ such that
$\bd T + T \bd = 1.$\\
Let $T_{p} = \begin{pmatrix} R_{p} & E_{p-1} \\ g_{p} & S_{p-1}
\end{pmatrix} \,\, , g_{p} : D_{p} \to C_{p}$.\\


$ 1= \bd T + T \bd = \begin{pmatrix} \bd & f \\ 0 & -\bd
\end{pmatrix} \begin{pmatrix} R & E \\ g & S
\end{pmatrix} + \begin{pmatrix} R & E \\ g & S
\end{pmatrix} \begin{pmatrix} \bd & f \\ 0 & -\bd
\end{pmatrix}\\ = \begin{pmatrix} \scriptsize{\bd R + fg} & \scriptsize{\bd E +fS} \\ \scriptsize{-\bd g} &
\scriptsize{-\bd S} \end{pmatrix} + \begin{pmatrix} \scriptsize{R
\bd} & \scriptsize{Rf+E(-\bd)} \\ \scriptsize{g \bd} &
\scriptsize{gf + S(-\bd)}
\end{pmatrix}$\\


$\Rightarrow \bd R + fg + R \bd = 1, \,\, -\bd g + g \bd =0, \,\,
-\bd S + gf - S\bd =1$\\


$\Rightarrow\left.\begin{array}{ll} \bd R + R \bd = 1- fg \\
\bd g = g \bd \end{array}\right\}
$\raisebox{-.1cm}{\parbox{10cm}{\hspace{1.0em}$\Rightarrow g$ is a
chain map
and $R$ is a chain\\ \hspace*{2.0em}homotopy : $1 \simeq fg$}} \\
$\hspace*{1.5em} \bd S + S \bd = gf -1 \Rightarrow S : 1 \simeq
gf$\\

$\Rightarrow$ chain map $g$ is a chain homotopy inverse of $f$ and
$f$ is a chain homotopy equivalence.\\

\end{pf}

{\bf 2nd proof of excision theorem}\\
\begin{pf}
Since $i_{*}$ is an isomorphism, $i$ is a chain homotopy
equivalence. Therefore, $i^{*}$ is an isomorphism.
\end{pf}\\

(3) {\bf Note} Let $\mathcal{C}$ and $\mathcal{D}$ be free chain
complexes , $R$ be a P.I.D. and $\gam_{p} : H_{p}(\mathcal{C}) \to
H_{p}(\mathcal{D}) , \forall \, p$. Then $\gam$ is induced by a
chain map (: $\mathcal{C} \to \mathcal{D}$).\\

\begin{cor}
$H_{*}(\mathcal{C}) \cong H_{*}(\mathcal{D}) \Rightarrow
H^{*}(\mathcal{C}) \cong H^{*}(\mathcal{D})$
\end{cor}

\begin{pf}
Let $\mathcal{C}$ and $\mathcal{C}^{'}$ be free chain complexes
and $\gam_{p} : H_{p}(\mathcal{C}) \to H_{p}(\mathcal{C}^{'})$ be
homomorphisms, $\forall \, p$.\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & B_{p} \ar[r]^{j} \ar@{.>}[d]^{\exists \alp} & Z_{p}
\ar[r] \ar@{.>}[d]^{\exists \bet} & H_{p} \ar[d]^{\gam_{p}} \ar[r]
& 0 &
\textrm{free}\\
0 \ar[r] & B_{p}^{'} \ar[r] & Z_{p}^{'} \ar[r] & H_{p}^{'} \ar[r]
& 0 & \textrm{acyclic}}
\]

By the comparison theorem, there exist $\alp, \bet$ such that the
above diagram commutes. We want $\phi$ such that the following diagram commutes.\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & Z_{p} \ar[r]^{i} \ar[d]^{\bet} & C_{p} \ar[r]_{\bd}
\ar@{.>}[d]^{\phi} & \ar@(ul,ur)[l]_{s} B_{p-1} \ar[r] \ar[d]^{\alp} & 0 \\
0 \ar[r] & Z_{p}^{'} \ar[r]_{i} & C_{p}^{'} \ar[r]^{\bd} &
\ar@(dl,dr)[l]^{s^{'}} B_{p-1}^{'} \ar[r] & 0}
\]

Since $B_{p-1}$ and $B_{p-1}^{'}$ are free, $C_{p} \cong Z_{p}
\bigoplus D_{p}$ and $C_{p}^{'} \cong Z_{p}^{'} \bigoplus
D_{p}^{'}$, where $D_{p} = s(B_{p-1})$ and $D_{p}^{'} =
s^{'}(B_{p-1}^{'})$.\\
Let $\phi = \begin{pmatrix} \bet & 0 \\ 0 & \alp \end{pmatrix}$.
Then the above diagram commutes. Hence,\\

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
C_{p} \ar[r] \ar[d]^{\phi} \ar@(ur,ul)[rrr]^{\bd} & B_{p-1}
\ar[r]^{j} \ar[d]^{\alp} &
Z_{p-1} \ar[r]^{i} \ar[d]^{\bet} & C_{p-1} \ar[d]^{\phi} \\
C_{p}^{'} \ar[r] \ar@(dr,dl)[rrr]_{\bd} & B_{p-1}^{'} \ar[r] &
Z_{p-1}^{'} \ar[r] & C_{p-1}^{'} }
\]
$\Rightarrow \phi$ is a chain map and $\phi|_{Z} = \bet$ certainly
induces $\gam$ in the first diagram.
\end{pf}\\

{\bf 7.} $H^{p}(\textrm{pt.} ; G) = \left\{\begin{array}{rr} G &
\textrm{if}\,\, p=0 \\ 0 & \textrm{if}\,\, p>0 \end{array}\right.$ \\

\begin{pf}
Recall\\
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & S_{3}(=\zb) \ar[r]^{0} & S_{2}(=\zb)
\ar[r]^{\cong} & S_{1}(=\zb) \ar[r]^{0} & S_{0}(=\zb) \ar[r] & 0 }
\]

$\Rightarrow$

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots & \ar[l]^{\cong} S^{3}(=G) & \ar[l]^{0} S^{2}(=G) &
\ar[l]^{\cong} S^{1}(=G) & \ar[l]^{0} S^{0}(=G) & \ar[l] 0 }
\]
\end{pf}

{\bf Remark}\\
$\left(\begin{array}{ll} \textrm{(contravariant) functoriality property} \\
\textrm{long exact sequence for pairs with the existence of} \,\, \del^{*} \\
\textrm{homotopy invariance}\\
\textrm{excision} \\
\textrm{dimension axiom 7}
\end{array} \right)$\\
$\Rightarrow$ Eilenberg-Steenrod axioms for (co)homology
theory and unique for finite CW-pairs (Reference : Vick)\\

{\bf 8.} Let $\{X_{\alp}\}$ be the family of path components of
$X$. Then $H^{p}(X) \cong \underset{\alp}{\prod} H^{p}(X_{\alp})$
for
any coefficient $G$.\\

\begin{pf}
$S_{p}(X) = \bigoplus S_{p}(X_{\alp}), Z_{p}(X) = \bigoplus
Z_{p}(X_{\alp}), B_{p}(X) = \bigoplus B_{p}(X_{\alp})$ and
$\htx(\bigoplus A_{\alp}, B) \cong
\underset{\alp}{\prod}\htx(A_{\alp},B)$
\end{pf}

{\bf 9.}(MV-sequence)\\
Same as homology case with reversed arrow of homs.\\
{\bf ¼÷Á¦ 16}\,\, Check!\\

{\bf 10.} $\widetilde{H}^{p}(S^{n};G) \cong
\left\{\begin{array}{ll} G &
\textrm{if}\,\, p=n \\ 0 & \textrm{otherwise} \end{array}\right.$\\
$H^{p}(D^{n},\bd D^{n} ; G) \cong \left\{\begin{array}{ll} G &
\textrm{if}\,\, p=n \\ 0 & \textrm{otherwise} \end{array}\right.$\\

Same MV-sequence for adjunction space, etc.\\

{\bf 11.} Let $X$ be a CW-complex with
$\mathcal{C}(X)=\{C_{p}(X),\bd\}$(cellular chain complex). Then
$H^{p}(\mathcal{C}(X);G) \cong H^{p}(X;G)$\\
\begin{pf}
See 6.(3) µû¸§Á¤¸® 1.($R$:P.I.D.)
\end{pf}

\end{document}