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\begin{document}
\parindent=0cm

\section*{Ext and Universal coefficient theorem}
{\bf 1.} $A$ : $R$-module ($R$: commutative ring with 1)\\
Let
\[
\xymatrix @=2em @*[c] { %
\cdots \ar[r] & C_p\ar[r]^\bd  & \cdots \ar[r]^\bd  &
C_2\ar[r]^\bd  & C_1\ar[r]^\bd & C_0\ar[r]^\eps & A\ar[r]^\bd &0
}\]
be a free resolution of $A$.\\
Define $\Ext^p(A,G):= H^p(\cc;G)$, the homology of
\[
\xymatrix @=2em @*[c] { %
0 \ar[r] & C^0\ar[r] & C^1\ar[r]  & C^2\ar[r] & \cdots }\] where
$C^i=\Hom_R(C_i,G)$.\\
\underline{Check}: This is well-defined, i.e., independent of
choice of a free resolution:
\[
\xymatrix @=2em @*[c] { %
\cdots \ar[r]^\bd  & C_1\ar[r]^\bd \ar@/_1mm/[d]_{i_1} &
C_0\ar[r]^\eps \ar@/_1mm/[d]_{i_0} & A\ar[r] \ar@{=}[d]^{id} &0\\
\cdots \ar[r]^\bd  & C_1'\ar[r]^\bd  \ar@/_1mm/[u]_{j_1}&
C_0'\ar[r]^\eps \ar@/_1mm/[u]_{j_0} & A\ar[r] &0 }\]

$id : A\to A$ has a lifting and liftings are chain homotopic by
comparison theorem. Therefore
$$ j\circ i\simeq id, j\circ i \simeq id.$$
It follows that $i$ is a chain homotopy equivalence and induces a
cochain homotopy equivalence and hence an isomorphism
\[\xymatrix @=2em @*[c] { %
i^*: H^p(\cc';G) \ar[r]^\cong & H^p(\cc;G).}\]

\underline{Note}: Any such liftings $i$ are all chain homotopic
and hence induce a same homomorphism $i^*$ and this is induced
from $id : A\to A$ i.e., there exists a canonical isomorphism
between $H^p(\cc'; G)$ and $H^p(\cc; G)$.\\

{\bf Properties }\\
(1) $\Ext^0(A,G)=H^0(\cc;G)=\Hom(A,G)$
\[
\xymatrix @=1.5em @*[c] { %
0 \ar[r] & \Hom_R(A,G) \ar[r]^{\et} & \Hom_R(C_0,G)
\ar[r]^{\del=\bdt} & \Hom_R(C_1,G) \ar[r] & \cdots }\] is exact at
$\Hom_R(C_0,G)$. So $\ext^0(A,G)=ker\del=im\et\cong\Hom(A,G)$.\\

(2) $\ext^p(F,G)=0$, if $p\geq 1$ and $F$ is free.\\
A free resolution of $F$ is \[
\xymatrix @=1.5em @*[c] { %
\cdots \ar[r] & 0 \ar[r] & F \ar[r]^{id} & F \ar[r] & 0 }\] So if
$p\geq 1$, $C_p=0$ and  $$ \ext^p(F,G)=H^p(\cc;G)=0.$$

(3) Similarly $\ext^p(A,G)=0$ if $p\geq 2$ and $A$ is an abelian
group or a $R$-module with PID $R$.\\
A free resolution of $A$ is \[
\xymatrix @=1.5em @R=1ex @*[c] { %
\cdots \ar[r] & 0 \ar[r] & R\ar[r]\ar@{=}[d] & F \ar[r]^{\eps}
\ar@{=}[d] & A \ar[r] & 0.\\ && ker\eps&free }\] So if $p\geq 2$,
$C_p=0$ and  $$ \ext^p(F,G)=H^p(\cc;G)=0.$$
In this case we simply denote $\ext(A,G)$ for $\ext^1(A,G)$.\\

(4) $\ext^p(-,G)$ is a contravariant functor.\\
For $\gam: A\to A'$, by comparison theorem there exist liftings
$\gam_0$, $\gam_1$, $\cdots$ such that the following diagram
commutes.
\[
\xymatrix @=2em @*[c] { %
\cdots \ar[r]^\bd  & C_1\ar[r]^\bd \ar@{.>}[d]_{\gam_1} &
C_0\ar[r]^\eps \ar@{.>}[d]_{\gam_0} & A\ar[r] \ar[d]^{\gam} &0\\
\cdots \ar[r]^\bd  & C_1'\ar[r]^\bd  & C_0'\ar[r]^\eps
 & A'\ar[r]&0 }\]
Since such liftings are unique up to chain homotopy, $\gam_p^* :
\ext^p(A',G)=H^p(\cc'; G)\to H^p(\cc; G)=\ext^p(A,G)$ is
well-defined and
functorial property is obvious. \\

{\bf 2. (Hom-Ext sequence)}\\
(1) Given a short exact sequence
\[
\xymatrix @=1.5em @*[c] { %
0 \ar[r] &A\ar[r]^{i}&B\ar[r]^j&C\ar[r]&0, }\] want a long exact
sequence (functorial) associated to it.\\
idea : Find free resolutions so that the diagram,
\[
\xymatrix @=2em @*[c] { %
0 \ar[r] &X\ar[r] \ar[d]&Y\ar[r]\ar[d]&Z\ar[r]\ar[d]&0\\
0 \ar[r] &A\ar[r]&B\ar[r]&C\ar[r]&0
 }\]
commutes and apply snake lemma.\\

Start with $X$ and $Z$, and then find $Y$. An obvious candidate is
$Y=X\oplus Z$ with $\bd\oplus\bd$. Need to define $\eps$ at the
final stage.
\[
\xymatrix @=2em @*[c] { %
&\ar[d]&\ar[d]&\ar[d]&\\
0 \ar[r] &X_0\ar[r] \ar[d]^\eps &X_0\oplus Z_0
\ar[r]\ar@{.>}[d]^<<<\eps &Z_0\ar[r]\ar[d]^\eps
\ar[dll]_<<<<<<<<<<<<<<<<<<<<<<<<<<{f_0} \ar[dl]^<<<<<{k}
&0\\
0 \ar[r] &A\ar[d]\ar[r]^{i}&B\ar[d]\ar[r]^j&C\ar[d]\ar[r]&0\\
&0&0&0&
 }\]
Since $Z_0$ is free, there exist $k:Z_0\to B$ such that
$jk=\eps$.\\ For any $f_0:Z_0\to A$, let
$$\eps(x,z)=i\eps(x)+f_0(z)+k(z).$$
(Write $f_0$ for $if_0$.)\\
Then this is the most general form of $\eps$ such that the above diagram commutes.\\
We will try to find $f_0$ such that the middle vertical sequence
is exact.
$$\eps(\bd x, \bd z )=i\eps(\bd x) + f_0(\bd z)+ k(\bd z)$$
Since $jk(\bd z)=\eps (\bd z)=0$, $k(\bd z)\in i(A)$. So need
$f_0=-k$ on $\bd Z_i=ker \eps$.\\
But in general, it is not possible to find such $f_0$. And hence
we try to change boundary operator $\bd\oplus \bd $ in one higher
step.
\[
\xymatrix @=2em @*[c] { %
&\ar[d]&\ar[d]&\ar[d]&\\
0 \ar[r] &X_2\ar[r] \ar[d]^\bd &X_2\oplus Z_2
\ar[r]\ar@{.>}[d]^<<<\bd &Z_2\ar[r]\ar[d]^\bd
\ar[dll]_<<<<<<<<<<<<<<<<<<<<<<<<<<{f_2} %\ar[dl]^{k}
 &0\\
0 \ar[r] &X_1\ar[r] \ar[d]^\bd &X_1\oplus Z_1
\ar[r]\ar@{.>}[d]^<<<\bd &Z_1\ar[r]\ar[d]^\bd
\ar[dll]_<<<<<<<<<<<<<<<<<<<<<<<<<<{f_1} %\ar[dl]^{k}
&0\\
0 \ar[r] &X_0\ar[r] \ar[d]^\eps &X_0\oplus Z_0
\ar[r]\ar@{.>}[d]^<<<\eps &Z_0\ar[r]\ar[d]^\eps
\ar[dll]_<<<<<<<<<<<<<<<<<<<<<<<<<<{f_0} \ar[dl]^<<<<<{k}
&0\\
0 \ar[r] &A\ar[d]\ar[r]^{i}&B\ar[d]\ar[r]^j&C\ar[d]\ar[r]&0\\
&0&0&0&
 }\]
The most general form of $\bd$ such that the diagram commutes will
be $\begin{pmatrix}
  \bd & f_1 \\
  0 & \bd
\end{pmatrix}$ with $f_1:Z_1\to X_0$.
Now put $f_0=0$ and
$$
\begin{array}{rcl}
\eps(\bd(x,z))&=& \eps(\bd x + f_1z,\bd z)\\
&=&i\eps(\bd x +f_1 z) +k(\bd z)\\
&=&i\eps(f_1 z) +k(\bd z)\\
\end{array}
$$
So we need to find $f_1$ such that $i\eps(f_1 z) =-k(\bd z)$. But
this is possible because $Z_1$ is free and $jk(\bd z)=0$. \\

Similarly given $\begin{pmatrix}
  \bd & f_1 \\
  0 & \bd
\end{pmatrix}$ , choose $f_2: Z_2\to X_1$ \\
such that $0=\bd^2=\begin{pmatrix}
  \bd & f_2 \\
  0 & \bd
\end{pmatrix}
\begin{pmatrix}
  \bd & f_1 \\
  0 & \bd
\end{pmatrix}
=\begin{pmatrix}
  \bd^2 & \bd f_1+f_2\bd \\
  0 & \bd^2
\end{pmatrix}$.\\
This can be solved since $Z$ is free and $X$ is acyclic by the
argument of comparison theorem. Inductively find $f_3,f_4,\cdots
f_{p+1} : Z_{p+1}\to X_p$ to construct a chain complex $Y$. Since
$H(X)=H(Z)=0$, $H(Y)=0$ by snake lemma, and hence we have found
the desired resolutions.\\

{\bf Remark.} What we really obtained here is a mapping cone $Y_p
=X_p\oplus Z_p$ with respect to a chain map
\[
\xymatrix @=2em @*[c] { %
\cdots \ar[r]^{-\bd}  & Z_2\ar[r]^{-\bd} \ar[d]_{f_2} &
Z_1\ar[r]^{-\bd} \ar[d]_{f_1} &
Z_0\ar[r]^{-\eps} \ar[d]_{k} & C \ar[r] \ar@{=}[d]^{-id} &0\\
\cdots \ar[r]^\bd  & X_1\ar[r]^\bd  & X_0\ar[r]^{i\circ\eps}
 &B\ar[r]^j& C\ar[r]&0 }\]
(Check. Exercise)\\

(2) {\bf Theorem.} If $$0\to A\to B\to C\to 0$$ is a short exact
sequence, then there
exists a natural(functorial) long exact sequence,\\[2mm]
\hspace*{1em} $
\begin{array}{rcl}
  0 & \to & \Hom(C,G)\to\Hom(B,G)\to\Hom(A,G) \\
   & \to & \Ext^1(C,G)\to \Ext^1(B,G) \to\Ext^1(A,G)
\to\Ext^2(C,G)\to\cdots
\end{array}
$\\

\begin{pf1}
By (1) there exist free resolutions such that
\[
\xymatrix @=2em @*[c] { %
0 \ar[r] &X\ar[r] \ar[d]&Y\ar[r]\ar[d]&Z\ar[r]\ar[d]&0\\
0 \ar[r] &A\ar[r]&B\ar[r]&C\ar[r]&0
 }\]
Taking $\Hom(-,G)$, since $Z$ is free, we get a short exact
sequence
\[
\xymatrix @=2em @*[c] { %
0 &X^*\ar[l] &Y^*\ar[l]&Z^*\ar[l]&0.\ar[l]\\
 }\]
Applying snake lemma, we get a desired long exact sequence.\\

Functoriality follows from the following general consideration.\\
Suppose we have
\begin{center}
$
\xymatrix @=1.5em @*[c] { %
0 \ar[r] &A\ar[r] \ar[d]^\alp&B\ar[r]\ar[d]^\bet&C\ar[r]\ar[d]^\gam&0\\
0 \ar[r] &A'\ar[r]&B'\ar[r]&C'\ar[r]&0
 }$
\raisebox{-1.3em}{with resolutions \parbox{3cm}{$X,Y,Z$\\[1.2em]
$X',Y',Z'$}}
\end{center}

Since the snake lemma is functorial, it is enough to show that
there exist liftings such that the following diagram commutes.
\[
\xymatrix @=1.5em @*[c] { %
0 \ar[r] &X\ar[r] \ar[d]^\alt&Y\ar[r]\ar[d]^\bett&Z\ar[r]\ar[d]^\gat&0\\
0 \ar[r] &X'\ar[r]&Y'\ar[r]&Z'\ar[r]&0
 }\]


Consider the diagram.
\[
\xymatrix @M=1ex @C=.7em @R=1em @*[c] { %
%���� ����
& X \ar[rr] \ar[dl]_{\alt} \ar[dd]
 & & Y \ar[rr] \ar[dl]_\bett \ar[dd] \ar[dddl]^<<<<<<k
& &Z \ar[dl]^\gat \ar[dd] \ar[dddlllll]_<<<<<<<<<<<{\bar{k}}\\
%���� ����
X' \ar[rr] \ar[dd]_\bd & & Y' \ar[rr] \ar[dd] & & Z' \ar[dd] & \\
%���� �Ʒ���
& X_{-1} \ar[rr] \ar[dl]_\alp & & Y_{-1} \ar[rr] \ar[dl]^\bet& &
 Z_{-1} \ar[dl]^\gam\\
%���� �Ʒ���
X'_{-1} \ar[rr]_{i} & & Y'_{-1}  \ar[rr]_{j} & & Z'_{-1} & }
\]

Problem : Given two liftings $\alt,\gat$, find a lifting $\bett$
inductively such that top squares commute.\\
First, we can always find $\bett=\begin{pmatrix}
  \alt & f \\
  0 & \gat
\end{pmatrix}$ such that top square commutes. But this may not be
a lifting of $\bet$ and modify $\bett$ using $f$.\\
Consider $k=\bd \bett-\bet \bd$. Then
$$ jk=j(\bd \bett-\bet \bd)=0 ,\ \ ki=(\bd \bett-\bet \bd)i=0$$
So, $k$ induces $\bar{k}: Z\to X'_{-1}$ such that
$i\bar{k}j=k$.\footnote{$jk=0$ implies there exists $k': Y\to
X'_{-1}$ such that $ik'=k$. And since $ik'i=ki=0$ and $i$ is
injective, $k'i=0$. So, $k'$ induces $\bar{k}$. }


By induction hypothesis,
$$ \bd (\bd\bett-\bet\bd)=0$$
So $\bd \bar{k}=0$.\footnote{Since $\bd k=0$, $\bd i\bar{k}j =i\bd
\bar{k}j=0$. So $\bd \bar{k}=0$ because $i$ is injective and $j$
is surjective.}

Since $X'$ is acyclic and $Z$ is free, there exists $f:Z\to X'$
such that $\bar{k}=\bd f$. \\
Now $"\bett"=\bett-ifj$ is the desired lifting of $\bet$
%\footnote{$\bd "\bett"-\bet\bd=\bd\bett-\bd ifj-\bet\bd=k-i\bar{k}j=0$}
and note that $"\bett"$ still commutes the top squares. \\
Therefore we get commutative liftings of $\alp, \bet, \gam$ and
the functoriality follows from the functoriality of snake lemma.
\end{pf1}\\



{\bf 3. (Why is the name Ext?)}

Let $A$ be an abelian group or $R$-module with $R$: P.I.D. (so
that $\etx^{p} = 0$ if $p \geq 2$).

Show $\etx^{1}(A,B) \cong \etx(A,B) := \{ 0 \to B \to E \to A \to
0\}/\sim$,

where $\{0 \to B \to E \to A \to 0\} \sim \{0 \to B \to E^{'} \to
A \to 0\}$ if and only if

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & B \ar[r] \ar[d]^{=} & E \ar[r]
\ar@{.>}[d]^{\phi}_{\cong} & A
\ar[r] \ar[d]^{=} & 0 \\
0 \ar[r] & B \ar[r] & E^{'} \ar[r] & A \ar[r] & 0 \,\, . }
\]




The existence of
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & R \ar[r]^{i} & F \ar[r]^{\eps} & A \ar[r] & 0}
\]
implies the following long exact sequence
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \htx(A,B) \ar[r]^{\widetilde{\eps}} & \htx(F,B)
\ar[r]^{\widetilde{i}} & \htx(R,B) \ar[r] & \etx^{1}(A,B) \ar[r] &
0}
\]


Hence $\etx^{1}(A,B) \cong \htx(R,B)/\widetilde{i}(\htx(F,B))
(=\textrm{coker}\,\, \widetilde{i})$

Given an extension
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & B \ar[r] & E \ar[r] & A \ar[r] & 0 \,\, ,}
\]

Consider
\[
\xymatrix @M=1ex @C=2em @R=1em @*[c]{%
0 \ar[r] & R \ar[r] \ar[d]_{\exists \alp}
\ar@(dr,ur)[dd]^>>>>>>{\exists \alp^{'}} & F \ar[r]
\ar[d]_{\exists \bet} \ar@(dr,ur)[dd]^>>>>>{\exists \bet^{'}} & A
\ar[r]
\ar[d]_{=} & 0 & \textrm{free} \\
0 \ar[r] & B \ar[r] & E \ar[r] & A \ar[r] & 0 & \textrm{acyclic}\\
0 \ar[r] & B \ar[u]_{=} \ar[r] & E^{'} \ar[r]
\ar@{.>}[u]_{\cong}^{k} & A \ar[r] \ar[u]^{=}& 0 &}
\]

If $E^{'} \overset{k}{\sim} E$, then $k \circ \bet^{'} , \alp^{'}$
are liftings of $id.$. By the comparison theorem $\alp \simeq
\alp^{'}$,\,i.e., $\exists D : F \to B$ such that
$Di(=\widetilde{i}(D)) = \alp^{'} - \alp$.

\begin{center}
$\therefore [E] \overset{\Phi}{\longmapsto} [\alp] \in
\textrm{coker}\, \widetilde{i}, \,\, \forall E \in \etx(A,B)$
\end{center}

Show this correspondence is bijective :

Given $[\alp]$, want an extension $E$ such that
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & R \ar[r]^{i} \ar[d]^{\alp} & F \ar[r]^{\eps}
\ar@{.>}[d]^{\bet} & A \ar[r] \ar[d]^{=} & 0\\
0 \ar[r] & B \ar@{.>}[r]_{p} & {\Large E} \ar@{.>}[r]_{q} & A
\ar[r] & 0}
\]

Use "push-out" of $\alp$ and $i$ to get $E$ and $\bet$

Put $E = B \bigoplus F / N$ where $N=\{(-\alp(r),i(r))|r \in R\}$.
(this forces $\alp(r) = i(r)$ in $E$ ) and $\bet, p, q$ are
obvious maps.

Show if $\alp \sim \alp^{'}$, then $E \sim E^{'}$:

If $\alp \sim \alp^{'}$, then $\alp^{'} = \alp + Di$ where $D : F
\to B$. Let $E^{'} = B \bigoplus F / N^{'},
N^{'}=\{(-\alp^{'}(r),i(r)|r \in R\}$(of course, $-\alp^{'}(r) =
(-\alp - Di)(r)$). Then

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & N \ar[r] \ar[d]_{\cong}^{\gam|} & B \bigoplus F \ar[r]
\ar[d]_{\cong}^{\gam} & E \ar[r] \ar@{.>}[d]^{\bar{\gam}}
& 0 & \textrm{commutes} \\
0 \ar[r] & N^{'} \ar[r] & B \bigoplus F \ar[r] & E^{'} \ar[r] & 0
&}
\]

if we define $\gam = \begin{pmatrix} 1 & -D \\ 0 & 1 \end{pmatrix}
$ and $\gam$ induces $\bar{\gam} : E \to E^{'}$ and $\bar{\gam}$
is an isomorphism (by 5-lemma). Therefore $E \sim E^{'}$.

This construction is clearly the inverse of $\Phi$(check:
Exercise).\\




{\bf Note} The "push-out" has a universal property, i.e., when
with $h \circ i = k \circ \alp$,

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
R \ar[dd]^{\alp} \ar[rr]^{i}& & F \ar[dd] \ar[dr]^{h} &\\
 & &  & E\\
B \ar[urrr]^{k} \ar[rr] & & B \bigoplus F / N \ar@{.>}[ur]_{\quad
\quad \exists! \,\, \textrm{homo. s.t. the diagram commutes.}} &}
\]\\




{\bf Remark} Similar interpretation of $\etx^{p}(A,B)$

(See MacLane or Hilton and Stambach)\\




{\bf 4. Universal coefficient Theorem}

Let $\mathcal{C}$ be a free chain complex and $G$ an abelian group
or $R$-module with $R$, a P.I.D. Then there exists a natural short
exact sequence for all $p$
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \etx(H_{p-1}(\mathcal{C}),G) \ar[r] &
H^{p}(\mathcal{C};G) \ar[r] & \htx(H_{p}(\mathcal{C}),G) \ar[r] &
0}
\]
which splits (but not naturally).

\begin{pf}
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & C_{p} \ar[r]^{\bd} & C_{p-1} \ar[r]^{\bd} & \cdots
& : \mathcal{C} }
\]
Consider
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & B_{p} \ar[r]^{i} & Z_{p} \ar[r]^{p} & H_{p} \ar[r] & 0
& \textrm{and}\\
0 \ar[r] & Z_{p} \ar[r]_{j} & C_{p} \ar[r]_{\bd} & B_{p-1} \ar[r]
& 0 &}
\]
{\bf Notation} $A^{'} = \htx(A,G)$ and $\etx_{p} = \etx(H_{p},G)$

By applying $\htx$-functor, we obtain
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & H_{p}^{'} \ar[r] & Z_{p}^{'} \ar[r] & B_{p}^{'} \ar[r]
& \etx_{p} \ar[r] & 0 \\
0 \ar[r] & B_{p-1}^{'} \ar[r]^{\del} & C_{p}^{'} \ar[r]
\ar@(dl,dr)[l] & Z_{p}^{'} \ar[r] \ar@(dl,dr)[l] & 0&}
\]
Note that since $Z_{p}$ is free, $\etx(Z_{p},G)=0$. Now consider
the following diagram and figure.
\[
\small{\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
&&0&&&0\ar[d]&&&0&&\\
0 \ar[r] & H_{p-1}^{'} \ar[r] & Z_{p-1}^{'} \ar[u] \ar[rrr] &&&
B_{p-1}^{'} \ar[r] \ar[d] & \etx_{p-1} \ar[r] & 0  & Z_{p+1}^{'} \ar[u] &&\\
&\cdots \ar[r] & C_{p-1}^{'} \ar[rrr]^{\del} \ar[u]&&& C_{p}^{'}
\ar[rrr]^{\del} \ar[d]_{\widetilde{j}}&&& C_{p+1}^{'} \ar[r] \ar[u] & \cdots &\\
&&B_{p-2}^{'} \ar[u]& 0 \ar[r] & H_{p}^{'} \ar[r] & Z_{p}^{'}
\ar[rrr]_{\widetilde{i}} \ar[d] &&& B_{p}^{'} \ar[u] \ar[r] &
\etx_{p}
\ar[r] & 0\\
&&0 \ar[u]&&&0&&&0\ar[u]&&}}
\]


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$\widetilde{j}(\ker \del) = \ker \widetilde{i} \cong H_{p}^{'}$\\





Naturality follows from the naturality of $\htx-\etx$ sequence and
of the above construction in the proof.

\end{pf}

\begin{cor}
Let $(X,A)$ be a pair of spaces. Then there exists natural short
exact sequence which splits
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \etx(H_{p-1}(X,A),G) \ar[r] & H^{p}(X,A;G) \ar[r] &
\htx(H_{p}(X,A),G) \ar[r]& 0}
\]
(or as a direct sum)
\end{cor}

{\bf Note} If $H_{p-1}$ is free, then $H^{p}(\mathcal{C};G) \cong
\htx(H_{p}(\mathcal{C}),G)$.(e.g. $R$ is a field)\\




{\bf ���� 17} Compute $H^{*}(\textrm{closed surfaces}), H^{*}(\rb
P^{n})$ and $H^{*}(\mathbb{C} P^{n})$(See 5.)\\




{\bf 5. Computation of $\etx$}

Recall $\etx(\textrm{free},G) = 0$

\begin{thm}

(1) $\etx(\bigoplus A_{\alp},G) \cong
\underset{\alp}{\prod}\etx(A_{\alp},G)$

$\etx(A,\prod G_{\alp}) \cong \underset{\alp}{\prod}
\etx(A,G_{\alp})$

(2) $\etx(\zb/n,G) \cong G/nG$
\end{thm}

\begin{pf}
(1)
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & R_{\alp} \ar[r] & F_{\alp} \ar[r] & A_{\alp} \ar[r] & 0
& \textrm{free resolution}}
\]
$\Rightarrow$
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \bigoplus R_{\alp} \ar[r] & \bigoplus F_{\alp} \ar[r] &
\bigoplus A_{\alp} \ar[r] & 0 & \textrm{free resolution}}
\]
$\Rightarrow$
\[
\small{\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \htx(A_{\alp},G) \ar[r] & \htx(F_{\alp},G) \ar[r] &
\htx(R_{\alp},G) \ar[r] & \etx(A_{\alp},G) \ar[r] & 0 & \clubsuit \\
0 \ar[r] & \htx(\bigoplus A_{\alp},G) \ar[r] & \htx(\bigoplus
F_{\alp},G) \ar[r] & \htx(\bigoplus R_{\alp},G) \ar[r] &
\etx(\bigoplus A_{\alp},G) \ar[r] & 0 & \clubsuit\clubsuit }}
\]
Note that $\htx(\bigoplus A_{\alp},G) = \prod \htx(A_{\alp},G)$,
etc. Now apply $\underset{\alp}{\prod}$ to $\clubsuit$ and compare
with $\clubsuit\clubsuit$ to get the result.(5-lemma)

For 2nd isomorphism, similar argument using $\htx(A,\prod
G_{\alp})= \prod \htx(A,G_{\alp})$\\



(2)
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \zb \ar[r]^{\times n} & \zb \ar[r] & \zb/n \ar[r] & 0 &
\textrm{free resolution}}
\]
$\Rightarrow$
\[
\small{\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & \htx(\zb/n,G) \ar[r] & \htx(\zb,G)(=G) \ar[r]^{\times
n} & \htx(\zb,G)(=G) \ar[r] & \etx(\zb/n,G) \ar[r] & 0}}
\]
Hence $\etx(\zb/n,G) \cong \textrm{coker} (\times n) = G/nG$ and
$\htx(\zb/n,G) = \ker (\times n)$.

\end{pf}

{\bf ���� 18}

$\htx(\zb/n,\zb/m) = \etx(\zb/n,\zb,m) = \zb/d$, where
$d=(m,n)$.\\




{\bf 6.} Let $X$ be a CW-complex.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
[X,S^{1}] \ar[r]^{\cong} \ar[d]_{\textrm{(a)}}^{\cong} &
H^{1}(X,\zb) \ar[d]^{\cong}_{\textrm{(c)}}\\
\htx(\pi_{1}X,\pi_{1}S^{1}) \ar[r]_{\textrm{(b)}} &
\htx(H_{1}(X),\zb)}
\]
where $[X,Y]=\textrm{Maps}(X,Y)/\simeq = \{[f]:\textrm{homotopy
class}|f:X \to Y\}$.

(1) (a)�� ���̱� ���� $[X,S^{1}] \cong [X,S^{1}]_{*} \cong
\htx(\pi_{1}X, \pi_{1}S^{1})$ ���� ��������. ���⼭
$[X,S^{1}]_{*}$�� $X$�� $S^{1}$�� base point �� ������ �ִ� ��츦
������ ���̴�. �׷� ���� $\htx(\pi_{1}X,\pi_{1}S^{1}) \cong
[X,S^{1}]_{*}$�� ���̱� ���ؼ��� $X$�� �ִ� maximal tree ��
�����ϰ�, �� maximal tree �� contract ���� base point�� ���� ��
$\pi_{1}X$�� $\pi_{1}S^{1}$������ �־��� map�� ���� �����ϴ�
cellular map : $X \to S^{1}$�� �� ������ �� �ִ�. ($\pi_{1}(X)$��
presentation�� �̿��ϰ�, $\pi_{i}(S^{1})=0 , i \geq 2$�̶��
����� �̿��Ͽ�)

(2)
$
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\pi_{1}(X) \ar[dr] \ar[rr] &  & \zb \ar@(dr,dr)@{->}[dl] \\
& H_{1}(X) \ar@(ur,dl)@{.>}[ur]&}$\hspace*{3.0em} \raisebox{-.5cm}
{\parbox{6cm}{ (b)�� ���̱� ���ؼ��� �켱 $\pi_{1}(X)$�� abelize
�ϸ� $H_{1}(X)$�� �ȴٴ� �Ϳ� �ָ�����. ���� �츮�� ������
diagram�� commute��Ű�� map�� ã���� �ȴ�.}}



(3) (c)�� universal coefficient thoerem�� ���ؼ� �ڸ��ϴ�.\\




{\bf ���� 19(Prove in detail)}\\



{\bf Fact} \, $[X,K(\pi,n)] \cong H^{n}(X;\pi)$\\


���⼭ $X \in K(\pi,n) \Leftrightarrow X$�� $\pi_{n}(X)=\pi$�̰�
$k \neq n$ �϶�  $\pi_{k}(X)=0$�� space�̴�. ���� ���, $S^{1} \in
K(\zb,1)$�̴�.



\end{document}
