\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} \usepackage{graphicx} \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{\vph}{\varphi} \newcommand{\Sig}{\Sigma} \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{\etx}{\textrm{Ext}} \newcommand{\cb}{\mathbb{C}} \newcommand{\sbb}{\mathbb{S}} \newcommand{\Ext}{\textrm{Ext}} \newcommand{\ext}{\textrm{Ext}} \newcommand{\Hom}{\textrm{Hom}} \newcommand{\et}{\widetilde{\eps}} \newcommand{\bdt}{\widetilde{\bd}} \newcommand{\alt}{\widetilde{\alp}} \newcommand{\bett}{\widetilde{\bet}} \newcommand{\gat}{\widetilde{\gam}} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{예} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{주} \newtheorem{note}{Note} } \renewcommand{\thenote}{} \renewcommand{\therem}{} \newenvironment{pf}{{\bf 증명}}{\hfill\framebox[2mm]{}\\} \newenvironment{pf1}{{\bf Proof}}{\hfill\framebox[2mm]{}} \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. \[ \xymatrix @M=1ex @C=.7em @R=1em @*[c] { % %뒤의 윗줄 & X \ar[rr] \ar[dl]_{\alt} \ar[dd] & & Y \ar[rr] \ar[dl]_\bett \ar[dd] \ar[dddl]^<<<<<}[d]^{\phi}_{\cong} & A \ar[r] \ar[d]^{=} & 0 \\ 0 \ar[r] & B \ar[r] & E^{'} \ar[r] & A \ar[r] & 0 \,\, . } \] 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. \[ \small{\xymatrix @M=1ex @C=1em @R=1em @*[c]{% &&0&&&0\ar[d]&&&0&&\\ 0 \ar[r] & H_{p-1}^{'} \ar[r] & Z_{p-1}^{'} \ar[u] \ar[rrr] &&& B_{p-1}^{'} \ar[r] \ar[d] & \etx_{p-1} \ar[r] & 0 & Z_{p+1}^{'} \ar[u] &&\\ &\cdots \ar[r] & C_{p-1}^{'} \ar[rrr]^{\del} \ar[u]&&& C_{p}^{'} \ar[rrr]^{\del} \ar[d]_{\widetilde{j}}&&& C_{p+1}^{'} \ar[r] \ar[u] & \cdots &\\ &&B_{p-2}^{'} \ar[u]& 0 \ar[r] & H_{p}^{'} \ar[r] & Z_{p}^{'} \ar[rrr]_{\widetilde{i}} \ar[d] &&& B_{p}^{'} \ar[u] \ar[r] & \etx_{p} \ar[r] & 0\\ &&0 \ar[u]&&&0&&&0\ar[u]&&}} \] %그림그리기 및 설명% \begin{center} \psset{unit=0.7cm} \begin{pspicture}(-0.5,-0.5)(10.5,9.5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] % S^2% \rput(1,1){\pscircle(0,0){0.6}} \rput(5,1){\pscircle(0,0){0.6} \pscircle(0,0){0.8} \psdot(0,0)} \rput(9,1){\psdot(0,0)} \rput(1,4.5){\pscircle(0,0){0.8}} \rput(5,4.5){\pscircle(0,0){0.8} \pscircle(0,0){0.6} \pscircle(0,0){0.4} \pscircle(0,0){0.2}} \rput(9,4.5){\pscircle(0,0){0.8} \psdot(0,0)} \rput(1,8){\pscircle(0,0){0.8}} \rput(5,8){\pscircle(0,0){0.8} \pscircle(0,0){0.4}} \rput(9,8){\pscircle(0,0){0.8} \psdot(0,0)} \psline(1,1.6)(5,1.6) \psline(1,0.4)(5,0.4) \psline(5,1.6)(9,1) \psline(5,0.4)(9,1) \psline(0.2,4.5)(0.2,8) \psline(1.8,4.5)(1.8,8) \psline(1,3.7)(5,4.3) \psline(1,5.3)(5,4.7) \psline(5,5.1)(9,4.5) \psline(5,3.9)(9,4.5) \psline[linestyle=dashed, linecolor=green](4.6,4.5)(5,1) \psline[linestyle=dashed,linecolor=green](5.4,4.5)(5,1) \psline[linestyle=dotted,linecolor=red](4.4,4.5)(4.4,1) \psline[linestyle=dotted,linecolor=red](5.6,4.5)(5.6,1) \psline(1,7.2)(5,7.6) \psline(1,8.8)(5,8.4) \psline(5,8.8)(9,8.8) \psline(5,7.2)(9,7.2) \psline(5,8.4)(9,8) \psline(5,7.6)(9,8) \psline[linestyle=dashed,linecolor=blue](4.2,8)(4.6,4.5) \psline[linestyle=dashed,linecolor=blue](5.8,8)(5.4,4.5) \psline[linestyle=dashed,linecolor=green](4.6,8)(4.8,4.5) \psline[linestyle=dashed,linecolor=green](5.4,8)(5.2,4.5) \rput(3,4.5){$\del$} \rput(7,4.5){$\del$} \rput(7,1){$\widetilde{i}$} \rput(4,2.75){$\widetilde{j}$} \rput(9,0.6){0} \rput(9,4.1){0} \rput(9,7.6){0} \psline{->}(9,1.5)(9,3.5) \rput(10,2.5){\textrm{1-1}} \end{pspicture} \end{center} $\widetilde{j}(\ker \del) = \ker \widetilde{i} \cong H_{p}^{'}$\\ 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 숙제 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 숙제 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)를 보이기 위해 $[X,S^{1}] \cong [X,S^{1}]_{*} \cong \htx(\pi_{1}X, \pi_{1}S^{1})$ 임을 증명하자. 여기서 $[X,S^{1}]_{*}$는 $X$와 $S^{1}$에 base point 가 정해져 있는 경우를 생각한 것이다. 그런 다음 $\htx(\pi_{1}X,\pi_{1}S^{1}) \cong [X,S^{1}]_{*}$을 보이기 위해서는 $X$에 있는 maximal tree 를 생각하고, 이 maximal tree 를 contract 시켜 base point로 잡은 후 $\pi_{1}X$와 $\pi_{1}S^{1}$사이의 주어진 map에 대해 대응하는 cellular map : $X \to S^{1}$을 잘 정의할 수 있다. ($\pi_{1}(X)$의 presentation을 이용하고, $\pi_{i}(S^{1})=0 , i \geq 2$이라는 사실을 이용하여) (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)를 보이기 위해서는 우선 $\pi_{1}(X)$를 abelize 하면 $H_{1}(X)$이 된다는 것에 주목하자. 따라서 우리는 왼쪽의 diagram을 commute시키는 map을 찾으면 된다.}} (3) (c)는 universal coefficient thoerem에 의해서 자명하다.\\ {\bf 숙제 19(Prove in detail)}\\ {\bf Fact} \, $[X,K(\pi,n)] \cong H^{n}(X;\pi)$\\ 여기서 $X \in K(\pi,n) \Leftrightarrow X$가 $\pi_{n}(X)=\pi$이고 $k \neq n$ 일때 $\pi_{k}(X)=0$인 space이다. 예를 들면, $S^{1} \in K(\zb,1)$이다. \end{document}