\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{\ac}{\mathcal{A}} \newcommand{\dd}{\mathcal{D}} \newcommand{\ee}{\mathcal{E}} \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{\Tor}{\textrm{Tor}} \newcommand{\et}{\widetilde{\eps}} \newcommand{\bdt}{\widetilde{\bd}} \newcommand{\alt}{\widetilde{\alp}} \newcommand{\bett}{\widetilde{\bet}} \newcommand{\gat}{\widetilde{\gam}} \newcommand{\ten}{\otimes} \newcommand{\ak}{algebraic K\"{u}nneth formula} \newcommand{\Ak}{Algebraic K\"{u}nneth formula} \newcommand{\ku}{K\"{u}nneth formula} \newcommand{\xty}{X\times Y} \newcommand{\com}{\ar@{}[dr]|{\displaystyle{\circlearrowleft}}} \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*{VII. Products} \section*{VII.1 The \ku s and torsion product} Want a formula relating the homology of $X\times Y$ to the homology of $X$ and $Y$. \begin{thm}[The \ku] $R$: PID. There exists a natural s.e.s $$ 0 \to \bigoplus_{p=0}^{n} H_p(X)\otimes H_{n-p}(Y) \to H_n(\xty)\to \bigoplus_{p=0}^{n-1} \Tor( H_p(X) , H_{n-p-1}(Y)) \to 0$$ which splits (but not naturally). \end{thm} This result will be proved in 2 stages : \\ (1) The Eilenberg-Zilber theorem : {\it There exists a natural chain homotopy equivalence between $S(\xty)$ and $S(X)\otimes S(Y)$.} \\ (2) \Ak : {\it Let $\cc, \cc'$ be chain complexes over PID $R$. Suppose at least one of $\cc$ and $\cc'$ is a free chain complex. Then there exists a natural s.e.s $$ 0 \to \bigoplus_{p=0}^{n} H_p(\cc)\otimes H_{n-p}(\cc') \to H_n(\cc\otimes\cc')\to \bigoplus_{p=0}^{n-1} \Tor(H_p(\cc) ,H_{n-p-1}(\cc')) \to 0$$ which splits (but not naturally) if both $\cc$ and $\cc'$ are free.}\\ \underline{\bf Tensor product}\\ Work in the category of $R$-modules($R$: commutative ring with 1) and write $A\otimes B$ for $A\otimes_R B$. \\ 1. (1) Let $f: A\to B$, $f': A' \to B'$ be homomorphisms.\\ \hspace*{2em} $\Rightarrow \exists !$ homomorphism $f\otimes f' :$ \raisebox{-1.4ex}{\parbox{5cm}{$A\otimes A' \to B\otimes B'$\\ $ a\otimes a' \mapsto f(a)\otimes f'(a')$}}\\ (2) $A \overset{f}{\to} B \overset{g}{\to} C$, $A' \overset{f'}{\to} B' \overset{g'}{\to} C'$ homomorphisms \\ \hspace*{2em} $\Rightarrow (g\circ f)\otimes (g'\circ f')=(g\otimes g')\circ (f\otimes f') : A\otimes A'\to B\otimes B'\to C\otimes C'$\\ (3) $f : A\to B$, $f': A'\to B'$ : surjective\\ \hspace*{2em} $\Rightarrow f\otimes f'$ is surjective and $ker(f\otimes f')$ is generated by the elements of the form $a\ten a'$ with $ a\in ker f$ or $a' \in ker f'$.\\ \begin{pf} $f\ten f' : A\ten A' \to B\ten B'$. For all $b\ten b'\in B\ten B' $, there exists $a\in A $, $a'\in A'$ such that $f(a)=b$ and $ f'(a')=b'$. Hence, $f\ten f'$ is surjective. \\ Let $K = \langle a\ten a' | a\in ker f\ \textrm{or} \ a'\in ker f' \rangle$. Clearly $K\subset ker f\ten f' $. \\ Consider \[ \xymatrix @=1.5em @*[c] { % A\ten A' \ar[rr]^-{f\ten f'}\ar[rd]_-p &\ar@{}[d]|-{\displaystyle{\circlearrowleft}}& B\ten B' \ar[ld]^-{\displaystyle{\exists \varphi}}\\ & A\ten A'/K& }\] and define $\vph(b\ten b' )=(a\ten a')$ where $f(a)=b$ and $f'(a')=b'$. Then $\vph$ is well-defined since $$a\ten a' - a_0\ten a_0' = (a-a_0)\ten a' + a_0\ten (a'-a_0')\in K$$ Therefore $ker (f\ten f')\subset K$. \end{pf}\\ In particular, $A\overset{f}{\to} B\to 0 \Rightarrow A\ten G \overset{f\ten id}{\to} B\ten G \to 0$.\\ 2. Exactness of $\ten G$\\ (1) \xymatrix{A \ar[r]^f & B \ar[r]^g &C \ar[r] & 0} : exact \\ \hspace*{3ex}$\Rightarrow $ \xymatrix{A\ten G \ar[r]^{f\ten id} & B\ten G \ar[r]^{g\ten id } &C\ten G \ar[r] & 0} : exact \\ (2) \xymatrix{0 \ar[r]&A \ar[r]^f & B \ar[r]^g &C \ar[r] & 0} : split exact \\ \hspace*{3ex}$\Rightarrow $ \xymatrix{0 \ar[r]&A\ten G \ar[r]^{f\ten id} & B\ten G \ar[r]^{g\ten id } &C\ten G \ar[r] & 0} : split exact \\ \begin{pf} (1) 1.(3) $\Rightarrow g\ten id$ is onto and $ker(g\ten id)=\langle b\ten x | b\in ker g \rangle=im(f\ten id)$.\\ (2) There exists $p$ such that $p\circ f=id$. So $$(p\ten id )\circ(f\ten id)=(p\circ f)\ten id=id.$$ Therefore $f\ten id$ is 1-1 and the sequence is split exact. \end{pf}\\ \underline{Remark} \raisebox{-1.3em}{\parbox{4cm}{ \begin{center}$A\ten R \cong A \cong R\ten A$\\ $a\ten r \rightarrow ra \leftarrow r\ten a$\\ $a \ten 1 \leftarrow a\rightarrow 1\ten a$ \end{center}}} \\ e.g. $\zb \ten \zb/2=\zb/2$ \[ (\xymatrix{0\ar[r] &\zb\ar[r]^{\times 2}&\zb\ar[r]&\zb/2\ar[r]&0})\ten \zb/2 \] \[\Rightarrow \xymatrix @R=1ex {\zb/2 \ar[r]^{\times 2}_{\textrm{0-map}}&\zb/2 \ar[r]&\zb/2 \ar[r]&0 }\]\\ In general, \[ \xymatrix{0\ar[r] &\zb\ar[r]^{\times n}&\zb\ar[r]&\zb/n\ar[r]&0}\] \[\overset{\ten G}{\Rightarrow} \xymatrix @R=1ex {\zb\ten G \ar@{=}[d]\ar[r]&\zb\ten G \ar@{=}[d]\ar[r]&\zb/n\ten G \ar[r]&0\\ G\ar[r]^{\times n}&G }\] Therefore $\zb/n\ten G\cong G/nG$.\\ Exercise. $\zb/n\ten\zb/m=\zb/d$, $d=(n,m)$\\ Recall : $A\ten B\cong B\ten A$, $(A\ten B)\ten C\cong A\ten (B\ten C)$,\\ \hspace*{5ex} $(\oplus A_\alp)\ten B\cong \oplus(A_\alp\ten B)$, $A\ten (\oplus B_\alp)\cong \oplus (A\ten B_\alp)$\\ Use this to compute $\ten$ of finitely generated abelian groups or $R$-modules. ($R$: PID)\\ 3. If $G$ is free, then $\ten G$ preserves s.e.s.\\ ($G$ : free $\Rightarrow G=\oplus R $ and\\[2mm] \hspace*{5em} \xymatrix @C=2ex {0\ar[r]&A\ar[r]&B} : exact $\Rightarrow$ \xymatrix @R=1ex @C=2ex {0\ar[r] & A\ten R\ar[r]\ar@{=}[d] & B\ten R\ar@{=}[d]\\ & A&B} : exact \\ \hspace*{14em}$\Rightarrow$ \xymatrix @C=2ex {0\ar[r]&A\ten G \ar[r]&B\ten G} : exact) \\ 4. (Homology with coefficient $G$) \\ $\cc$ : a chain complex (over $R$) , $G$ : $R$-module.\\ $\Rightarrow \cc\ten G$ : $\cdots \to C_p\ten G \to C_{p-1}\ten G\to \cdots$ is a chain complex. \\ $\Rightarrow H(C; G):= $ homology of $\cc\ten G$ := homology of $\cc$ with coefficient $G$.\\ $f: \cc \to \cc'$, a chain map $\Rightarrow f_*: H(\cc;G)\to H(\cc'; G)$.\\ $\phi: G \to G'$, an $R$-module homomorphism $\Rightarrow \phi_*: H(\cc;G)\to H(\cc; G')$.\\ If $0\to \cc\to \dd\to\ee\to 0$ is a split s.e.s.(e.g. $\ee$ is free), by snake lemma there exists a natural l.e.s., $$ \cdots \to H_p (\cc;G)\to H_p(\dd;G)\to H_p(\ee;G)\overset{\bd}{\to} H_{p-1}(\cc;G)\to \cdots.$$\\ If $\cc$ is free and $0\to G'\to G\to G''\to 0$ is a s.e.s., by snake lemma there exists a natural l.e.s., $$ \cdots \to H_p (\cc;G')\to H_p(\cc;G)\to H_p(\cc;G'')\overset{\bet_*}{\to} H_{p-1}(\cc;G')\to \cdots.$$ In this case, $\beta_*$ is called Bockstein homomorphism.\\ \underline{\bf Torsion product}\\ 5. Let $$\cdots \to F_p\overset{\bd}{\to}\cdots\overset{\bd}{\to} F_1\overset{\bd}{\to}F_0\overset{\eps}{\to}A\to 0$$ be a free resolution of $A$. \\ Define $\Tor_p(A,G):= H_p(F;G)=H_p(F\ten G)$\\ \hspace*{10ex} = homology of $\cdots \to F_p\ten G {\to}\cdots \to F_1\ten G{\to} F_0\ten G{\to}0$.\\ This is well-defined by comparison theorem. ($\ext^p$의 경우와 마찬가지)\\ Note. (1) $\Tor_0(A,G)\cong A\ten G$ : \[ \xymatrix @=1.5em @*[c] { % F_1\ten G \ar[r]^{\bd\ten 1} & F_0\ten G \ar[r]^{\eps\ten 1} & A\ten G \ar[r] & 0 }\] is exact. So $\Tor_0(A,G)=F_0\ten G/ im(\bd\ten 1)=F_0\ten G/ ker(\eps\ten 1)=A\ten G$.\\ (2) If $p\geq 1$ and $F$ is free, $\Tor_p(F,G)=0$ and $\Tor_p(G,F)=0$.\\ 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 $$ \Tor_p(F,G)=0.$$ And since $\ten $free preserves resolution, $$\Tor_p(G,F)=0$$ (3) If $R$ is a PID, $\Tor_p(A,G)=0$ for $p\geq 2$.\\ A free resolution of $A$ is \[ \xymatrix @=1.5em @R=1ex @*[c] { % 0 \ar[r] & R\ar[r]\ar@{=}[d] & F \ar[r]^{\eps} \ar@{=}[d] & A \ar[r] & 0.\\ & ker\eps&free }\] So if $p\geq 2$, $C_p=0$ and $$ \Tor_p(A,G)=0.$$ In this case we simply denote $\Tor(A,G)$ or $A*G$ for $\Tor_1(A,G)$.\\ (4) $\Tor_p(A,G)$ is a covariant functor in both variables.\\ 6. If $$0\to A\to B\to C\to 0$$ is a short exact sequence, then there exists a natural long exact sequence,\\[2mm] \hspace*{1em} $ \begin{array}{l} \cdots \to \Tor_n(A,G)\to\Tor_n(B,G)\to\Tor_n(C,G) \to \Tor_{n-1}(A,G)\to \\ \cdots \to \Tor_1(C,G)\to A\ten G\to B\ten G\to C\ten G\to 0 \end{array} $\\ \begin{pf} \hspace{1ex} There exist free resolutions such that \[ \xymatrix @=1.5em @*[c] { % 0 \ar[r] &X\ar[r] \ar[d]&Y\ar[r]\ar[d]&Z\ar[r]\ar[d]&0\\ 0 \ar[r] &A\ar[r]&B\ar[r]&C\ar[r]&0 }\] And apply snake lemma. Naturality is same as before. \end{pf}\\ In particular, if $R$ is a PID, the l.e.s becomes $$ 0\to A*G\to B*G\to C*G\to A\ten G \to B\ten G\to C\ten G\to 0. $$ \bigskip 7.If $$0\to A\to B\to C\to 0$$ is a short exact sequence, then there exists a natural long exact sequence,\\[2mm] \hspace*{1em} $ \begin{array}{l} \cdots \to \Tor_n(G,A)\to\Tor_n(G,B)\to\Tor_n(G,C) \to \Tor_{n-1}(G,A)\to \\ \cdots \to \Tor_1(G,C)\to G\ten A\to G\ten B\to G\ten C\to 0 \end{array} $\\ \begin{pf} $F\to G\to 0 $ : free resolution of $G$. \\ $\Rightarrow 0\to F\ten A \to F\ten B\to F\ten C\to 0$ : s.e.s. of chain complex. \\ Apply the snake lemma. \end{pf}\\ \newpage 8. (1) $\Tor_n(A,B)\cong\Tor_n(B,A)$ for all $n$.\\ (2) $\Tor_n(\bigoplus A_\alp,B)\cong \bigoplus \Tor_n(A_\alp,B)$.\\ \hspace*{3ex}($\Tor_n(A, \bigoplus B_\alp)=\bigoplus\Tor_n(A,B_\alp)$)\\ (3) $ R/a* B=\Tor (R/a,B)\cong ker(B\overset{\times a}{\longrightarrow } B) $ and $R/a \ten B\cong B/aB $.\\ (4) $R$ : PID, $\Tor (R/a,R/b)=R/a*R/b\cong R/d$, $d=(a,b)$.\\ \begin{pf} (1) \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & F \ar[r] & B \ar[r] & 0 & \textrm{: a free resolution of }B} \] and let $K_p=ker\bd =im\bd\subset F_p$ so that \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & K_p \ar[r] & F_p \ar[r]^-\bd & K_{p-1} \ar[r] & 0 & \textrm{: s.e.s.}} \] By 7., \[ \xymatrix @M=1ex @C=1em @R=1ex @*[c]{% \cdots \ar[r] & \Tor_n(A,K_0) \ar[r] & \Tor_n(A,F_0) \ar[r] \ar@{=}[d] & \Tor_n(A,B) \ar[r] & \Tor_{n-1}(A,K_0) \ar[r] & \cdots \\&&0 } \] So, $\Tor_n(A,B)\cong\Tor_{n-1}(A,K_0)\cong \cdots \cong \Tor_1(A,K_{n-2})$.\\ Now consider \[ \xymatrix @M=1ex @C=1em @R=1ex @*[c]{% 0 \ar[r] & \Tor_1(A,K_{n-2}) \ar[r] & A\ten K_{n-1} \ar[r]^i & A\ten F_{n-1} \ar[r] & A\ten K_{n-2} \ar[r] &0 } \] Note that \[ \xymatrix @M=1ex @C=-1ex @R=2ex @*[c]{% &&&&0&&&&0\\ &&&A\ten K_n\ar[dr]\ar[ur]&&&&A\ten K_{n-2}\ar[ur]\\ \cdots \ar[rr] && A\ten F_{n+1} \ar[rr]^-{\bd_{n+1}}\ar[ur] && A\ten F_n \ar[dr]_-{\bd'} \ar[rr]^{\bd_n} && A\ten F_{n-1} \ar[rr]\ar[ur] && \cdots\\ &&&&&A\ten K_{n-1}\ar[ur]_i \ar[dr]\\ &&&&&&0} \] $ \begin{array}{rcl} \Tor_1(A, K_{n-2}) & = & ker i \cong ker \bd_n/ker \bd' = ker\bd_n/ im\bd_{n+1}\\ & = & H_n(A\ten F)=H_n(F\ten A)=\Tor_n(B,A) \end{array} $\\ (2) $\Tor_n(A,\bigoplus B_\alp):= H_n(F\ten (\bigoplus B_\alp))=\bigoplus H_n(F\ten B_\alp)=\bigoplus \Tor _n(A,B_\alp)$\\ The other follows from (1).\\ (3) Clear from the following. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & R \ar[r]^{\times a} & R \ar[r] & R/a \ar[r] & 0 & \textrm{free resolution of }R/a} \] $\Rightarrow$ \[ \xymatrix @M=1ex @C=1em @R=1ex @*[c]{% 0 \ar[r] & R/a*B \ar[r] & R\ten B \ar[r]\ar@{=}[d]& R\ten B \ar@{=}[d]\ar[r] & R/a\ten B \ar[r] & 0\\ &&B \ar[r]^{\times a} &B} \] (4) {\bf 숙제 27.} \\ $R/a*R/b = ker(R/b \overset{\times a}{\longrightarrow } R/b)= R/d$, $d=(a,b)$ \\ $R/a\ten R/b = coker(R/b \overset{\times a}{\longrightarrow } R/b)= R/d$. \end{pf} \end{document}