\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}{\textrn{Ext}} \newcommand{\dlm}{\underset{\longrightarrow}{\lim}} \newcommand{\pon}{\underset{p=0}{\overset{n}{\oplus}}} \newcommand{\ak}{algebraic K\"{u}nneth formula} \newcommand{\Ak}{Algebraic K\"{u}nneth formula} \newcommand{\ku}{K\"{u}nneth formula} \newcommand{\ttx}{\textrm{Tor}} \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 정리의 증명}}{\hfill\framebox[2mm]{}} \newenvironment{pf2}{{\bf Proof}}{\hfill\framebox[2mm]{}} \newenvironment{pf3}{{\bf Sketch of proof}}{\hfill\framebox[2mm]{}} \newenvironment{lem1}{{\bf Lemma}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \section*{VII.2 \Ak \ and Universal Coefficient theorem for homology} {\bf {Algebraic K\"{u}nneth formula}}\\ \textbf{9. (Tensor product of chain complexes)}\\ $\cc, \cc'$ : chain complexes $\Rightarrow \cc\otimes\cc'$ :chain complex :\\ Define $(\cc\otimes\cc')_n := \pon (C_p\otimes C'_{n-p})$ and $\bd : (\cc\otimes\cc')_n\rightarrow (\cc\otimes\cc')_{n-1}$ by $\bd(c\otimes c')=\bd c\otimes c' +(-1)^p c\otimes \bd c'$ for $\forall c\otimes c'\in C_p\otimes C'_{n-p}\\ \Rightarrow \bd^2 =0 $: clear. ($\because (-1)^{p-1}\bd c\otimes \bd c' + (-1)^p\bd c\otimes \bd c' =0 $)\\ Want to compare $H(\cc\otimes \cc' )$ and $H(\cc)\otimes H(\cc')$:\\ We have a well-defined canonical homomorphism $i : H(\cc)\otimes H(\cc') \rightarrow H(\cc\otimes \cc')$\\ given by $H_p(\cc)\otimes H_{n-p}(\cc')\rightarrow H_n(\cc\otimes \cc').\\ \hspace*{5em}\{z\}\otimes \{z'\}\mapsto\{z\otimes z'\}$\\ Check this is well-defined : (1) $z,z'$: cycles $\Rightarrow z\otimes z'$ is a cycle. (clear)\\ \hspace*{12em}(2) $(z+\bd c)\otimes z'=z\otimes z' + (\bd c\otimes z') = z\otimes z' + \bd(c\otimes z')$\\ \textbf{10.} Assume $R$ : PID , $\cc$ : free\\ Show algebraic K\"{u}nneth :\\ Split chain complex $\cc$ into two short exact sequences as usual :\\ $(1) 0\rightarrow Z \overset{i}{\rightarrow}\cc \overset{\bd}{\rightarrow}\bar{B}\rightarrow 0\hspace{0.2em}(\bar{B}_p:=B_{p-1})\\ (2) 0\rightarrow B\overset{j}{\rightarrow} Z\rightarrow H(\cc)\rightarrow 0$\\ Start with tensoring (1) with $\cc'$ to compute $H(\cc\otimes \cc' ).$\\ Note $\bar{B}\subset \cc$ is free and (1) is a splitting s.e.s.\\ (1)$\Rightarrow 0\rightarrow Z\otimes \cc'\overset{i\otimes 1}{\rightarrow}\cc\otimes \cc'\overset{\bd\otimes 1}{\rightarrow}\bar{B}\otimes \cc'\rightarrow 0$ : s.e.s.\\ (i.e., $0\rightarrow Z_p\otimes C'_{n-p}\rightarrow C_p\otimes C'_{n-p}\rightarrow \bar{B}_p\otimes C'_{n-p}\rightarrow 0$ : s.e.s.) \\ Here we view $Z$ as a chain complex : $\rightarrow Z_p\overset{\bd=0}{\rightarrow}Z_{p-1}\overset{\bd=0}{\rightarrow}\cdots$ and similarly for $\bar{B}.$\\ Remark. $f : \cc\rightarrow \cc'$ and $g : \dd\rightarrow \dd'$ : chain map \\ $\Rightarrow f\otimes g :\cc\otimes\dd \rightarrow \cc'\otimes \dd'$ : chain map\\ \hspace*{4.5em}$c\otimes d\mapsto f(c)\otimes g(d)\\ \hspace*{4em}(\bd c\otimes d+(-1)^p c\otimes \bd d \mapsto f(\bd c)\otimes g(d) + (-1)^p f(c)\otimes g(\bd d)\\ \hspace*{14em}=\bd f(c)\otimes g(d)+(-1)^p f(c)\otimes\bd g(d))$\\ Now Snake lemma $\Longrightarrow \\ \cdots\rightarrow H_n(Z\otimes \cc')\rightarrow H_n(\cc\otimes \cc')\rightarrow H_n(\bar{B}\otimes \cc')\rightarrow H_{n-1}(Z\otimes \cc')\rightarrow \cdots$\\ Show $H(Z\otimes \cc')=Z\otimes H(\cc')$ and $H(\bar{B}\otimes \cc')=\bar{B}\otimes H(\cc')$:\\ $\rightarrow Z_p\overset{\bd=0}{\rightarrow}Z_{p-1}\overset{\bd=0}{\rightarrow}\cdots \\ \Rightarrow\rightarrow(Z\otimes \cc')_n\overset{\bd}{\rightarrow} (Z\otimes \cc')_{n-1}\rightarrow \cdots$ is given by\\ $\cdots \rightarrow Z_p\otimes C'_{n-p}\overset{"\bd"=(-1)^p 1\otimes\bd}{\longrightarrow}Z_p\otimes C'_{n-p-1}\rightarrow \cdots\\ \hspace*{5em}z\otimes c'\hspace{1em}\mapsto\hspace{1em} (-1)^pz\otimes \bd c'\\ = Z_p\otimes (\cdots \rightarrow C'_{n-p}\overset{\bd '=(-1)^p\bd }{\longrightarrow} C'_{n-p-1}\rightarrow \cdots )$\\ $= \left\{\begin{array}{l}Z_p\otimes (1)'_{n-p}\\ Z_p\otimes (2)'_{n-p}\end{array}\right\}$ (Since $Z_p$ is free, $Z_p\otimes $ preserves s.e.s.)\\ $\Rightarrow H_n(Z\otimes \cc') = \overset{n}{\underset{p=0}{\oplus}} H(Z_p\otimes \cc'_{n-p}) \cong \overset{n}{\underset{p=0}{\oplus}} Z_p\otimes H_{n-p}(\cc')= (Z\otimes H(\cc'))_n\\ \hspace*{3em}\{z\otimes z'\}\hspace{8em}\mapsto \hspace{8em}z\otimes \{z'\}$\\ Similarly, $H(\bar{B}\otimes \cc') =\bar{B}\otimes H(\cc')\\ \hspace*{4em} \{b\otimes z'\}\leftrightarrow b\otimes \{z'\}\\ \therefore \cdots \overset{\phi_n}{\rightarrow} (Z\otimes H(\cc'))_n\rightarrow H_n(\cc\otimes \cc')\rightarrow (\bar{B}\otimes H(\cc'))_n\overset{\phi_{n-1}}{\rightarrow}\cdots\\ \hspace*{4em}z\otimes\{z'\}\mapsto \{z\otimes z'\}\hspace{4em}(=(B\otimes H(\cc'))_{n-1})$\\ $\Rightarrow 0\rightarrow cok \phi_n\rightarrow H_n(\cc\otimes\cc')\rightarrow ker \phi_{n-1}\rightarrow 0$,\\ where $\phi_n : (B\otimes H(\cc'))_n\rightarrow (Z\otimes H(\cc'))_n$\\ Now, (2)$\Rightarrow 0\rightarrow B_p\overset{j}{\rightarrow}Z_p\rightarrow H_p(\cc)\rightarrow 0$ : s.e.s.\\ $\underset{ Z_p:free}{\overset{\otimes H_{n-p}(\cc')}{\Longrightarrow}}0\rightarrow Tor(H_p(\cc'),H_{n-p}(\cc'))\rightarrow B_p\otimes H_{n-p}(\cc')\rightarrow Z_p\otimes H_{n-p}(\cc')\rightarrow \hspace*{3em}H_p(\cc)\otimes H_{n-p}(\cc')\rightarrow 0$\\ $\therefore cok \phi_n = \pon H_p(\cc)\otimes H_{n-p}(\cc')$ and $ker \phi_{n-1} = \overset{n-1}{\underset{p=0}{\oplus}}Tor(H_p(\cc), H_{n-p-1}(\cc'))$\\ $\therefore 0\rightarrow \pon H_p(\cc)\otimes H_{n-p}(\cc')\rightarrow H_n(\cc\otimes \cc')\rightarrow \overset{n-1}{\underset{p=0}{\oplus}}Tor(H_p(\cc),H_{n-p-1}(\cc'))\rightarrow 0\\ \hspace*{5em}\{z\}\otimes\{z'\}\hspace{1em}\mapsto\hspace{1em}\{z\otimes z'\}$\\ or $0\rightarrow (H(\cc)\otimes H(\cc'))_n\rightarrow H_n(\cc\otimes \cc')\rightarrow (H(\cc)*H(\cc'))_{n-1}\rightarrow 0$.\\ Sequence splits if $\cc$ and $\cc'$ are free : Since $\bar{B}$ is free,\\ \[\xymatrix @C=2.8em @R=1em@*[l]{% 0 \ar[r]& Z\ar[r]^{i}\ar[d]^{p} & \ar@(ur,ul)[l]^{\exists \pi}\cc\ar[r]\ar[dl]^{\psi = p\cdot \pi} & \bar{B}\ar[r] & 0 & \Rightarrow & \cc\otimes\cc'\ar[r]^<<{\psi\otimes \psi'}&H(\cc)\otimes H(\cc')\\ &H(\cc)&&&&&c\otimes c'\ar@{|->}[r]&\{\pi(c)\}\otimes \{\pi'(c')\} }\],where again $H(\cc)$ is a trivial chain complex.\\ $\Rightarrow H(\cc\otimes \cc')\overset{(\psi\otimes\psi')*}{\longrightarrow}H(H(\cc)\otimes H(\cc'))= H(\cc)\otimes H(\cc')$\\ Naturality follows as before.\hspace{3em}$\square$\\ \textbf{11. Universal Coefficient Theorem for Homology}\\ In algebraic K\"{u}nneth, let $\cc'_n =\left\{\begin{array}{ll}G& ,n=0\\ 0&, otherwise\end{array}\right\}$, where $G $ is $R$(PID)-mod.\\ Then $H_q(\cc') =\left\{\begin{array}{ll}G& , q=0\\ 0&, q\neq 0\end{array}\right\}$ and $(\cc\otimes \cc')_n = C_n\otimes G$. Hence we have\\ \textbf{ U.C.T.} : If $\cc$ is free, then $0\rightarrow H_n(\cc)\otimes G\rightarrow H_n(\cc\otimes G)\rightarrow Tor(H_{n-1}(\cc),G)\rightarrow 0$ which splits (not canonical), where $H_n(\cc\otimes G)=H_n(\cc;G)$ homology with coefficient $G$.\\ Note. For a proof for splitting, see the proof of algebraic K\"{u}nneth : \[\xymatrix @C=2em @R=1em@*[l]{% 0 \ar[r]& Z'\ar[r]^{i}\ar[d]^{p} & \cc'\ar[r]\ar[dl]^{Want \hspace{0.2em}\psi = p\cdot \pi} & \bar{B'}\ar[r] & 0 \\ &H(\cc') }\] but $H_p(\cc')=Z'_p=0$ if $p\neq 0$ and for $p=0 , Z'_0=\cc'_0=H_0(\cc')=G$.\\ \textbf{Note 1.} In particular, if $\cc$ is a chain complex of free abelian group and $R,\hspace{0.2em}\forall$ commutative ring with 1 (so abelian group), then $0\rightarrow H_n(\cc)\otimes R\rightarrow H_n(\cc\otimes R)\rightarrow H_{n-1}(\cc)*R\rightarrow 0$, where $\otimes = \underset{\mathbb{Z}}{\otimes}$, i.e., abelian group tensor product and $* =\underset{\mathbb{Z}}{*}$,i.e., abelian group torsion product and hence if $\cc=S_n(X)$ (or $S_n(X,A)$), then $0\rightarrow H_n(X;\mathbb{Z})\otimes R\rightarrow H_n(X;R)\rightarrow Tor(H_{n-1}(X;\mathbb{Z}),R)\rightarrow 0$\\ \textbf{Note 2.} $A$: abelian group and $R$ : commutative ring with 1.\\ $\Rightarrow A\otimes R$ has a canonical $R$-module structure given by $r(a\otimes x)=a\otimes (rx)$\\ $R$-module $S_n(X;R)$ defined earlier is exactly $S_n(X)\otimes R$.\\ \textbf{숙제 28.} Compute $H(P^n;\mathbb{Z}/2)$ using $H(P^n;\mathbb{Z})$ and compare. (Use U.C.T.)\\ \textbf{숙제 29.} $R$ : PID $\Rightarrow \chi(X)=\chi(X;R)$ (Use U.C.T.)\\ \textbf{Note 3.} If $R$ is a field, then every $R$-module is free. Hence $H(\cc)\otimes H(\cc')\cong H(\cc\otimes\cc')$ (or $H_n(\cc\otimes\cc')\cong\pon H_p(\cc)\otimes H_{n-p}(\cc'))$ and $H_n(\cc)\otimes G\cong H_n(\cc\otimes G)\hspace{1em},\forall$ vector space $G$.\\ {\bf Note}\,\,(4) Let $R$ be a field with ch($R$)=0. $\Rightarrow H_{n}(X;\zb) \bigotimes R \cong H_{n}(X;R)$, since $R$ is a torsion free abelian group and hence $\ttx(H_{n-1}(X;\zb),R)=0$.\\ {\bf Proposition}\,\, Let $B$ be a torsion free abelian group. Then $B \ast A=0$. \begin{pf3} Note that any $R$-module is a direct limit of its finitely generated submodule. And finitely generated torsion free is free if $R$ is a P.I.D.(Structure theorem). Now tensor product commutes with direct limit(easy exercise). $\Rightarrow \ast$ commutes with direct limit. $\Rightarrow \ttx(C,A) = 0$ if $C$ is a torsion free $R$-module with $R$: P.I.D.\\ \end{pf3} {\bf 12. Eilenberg-Zilber Theorem} $S(X\times Y)$ and $S(X) \bigotimes S(Y)$ are naturally chain homotopy equivalent. Hence $H(X \times Y) \cong H(S(X) \bigotimes S(Y))$. \begin{pf2} \,Use Acyclic Model Theorem. Recall AMT. \vspace*{1.0em} \framebox{\parbox[b]{14cm}{ \textbf{Acyclic Model Theorem}\\ Let $F, F' :\mathcal{T}\rightarrow \mathcal{C}$ be functors and $\mathcal{M}\subset Ob(\mathcal{T}),$ where $\mathcal{C}$ is a category of chain complexes. Suppose \,\,(1) $F'$ is acyclic relative to $\mathcal{M}$, i.e., $F'(M)$ is acyclic $\forall M \in \mathcal{M}$. \,\,(2) $F$ is free relative to $\mathcal{M}$, i.e., $\forall p, \exists \textrm{indexed family} \{M_{\alp}\}_{\alp \in J_{p}} \textrm{and} \{i_{\alp}\}_{\alp \in J_{p}}, M_{\alp} \in \mathcal{M}, i_{\alp} \in F_{p}(M_{\alp})$ such that the indexed family $\{F(\sigma)i_{\alp}\}_{\alp \in J_{p}}, \sigma \in \textrm{hom}(M_{\alp},X)$ is a basis for $F_{p}(X)$. Then \,\,(1) $\exists $ a natural transformation $\tau :F\rightarrow F'$ which induces a given natural transformation $\tau_0:H_{0}(F)\rightarrow H_{0}(F')$. \,\,(2) Given two such natural transformations $\tau,\tau^{'}:F\rightarrow F'$ with $\tau_{0} = \tau_{0}^{'}, \tau\simeq\tau^{'}$}} \vspace*{1.0em} {\bf Eilenberg-Zilber Situation} Let $\mathcal{T}$ be the category of pairs $(X,Y)$ of topological spaces. Consider $F : \mathcal{T} \rightarrow \mathcal{C}, F(X,Y) = S(X \times Y)$ and $F^{'}: \mathcal{T} \rightarrow \mathcal{C}, F^{'}(X,Y)= S(X) \bigotimes S(Y)$. $\Rightarrow$ These are clearly functors : \[ \xymatrix @M=1ex @C=1em @R=1em @*[c] {% (X,Y) \ar[r] \ar[d]_{(f,g)} & S(X \times Y) \ar[d]^{F(f,g)=(f \times g)_{\sharp}} & \hspace*{1.5em} & S(X) \bigotimes S(Y) \ar[d]^{F^{'}(f,g) = f_{\sharp} \otimes g_{\sharp} = \overset{n}{\underset{p=0}{\bigoplus}} f_{\sharp p} \otimes g_{\sharp n-p}} \\ (X^{'}, Y^{'}) \ar[r] & S(X^{'} \times Y^{'}) & & S(X^{'}) \bigotimes S(Y^{'}) } \] Let $\mathcal{M} = \{(\triangle^{p},\triangle^{q}), \, p,q \geq 0\}$, where $\triangle^{p}$ is a standard $p$-simplex. (1) $F, F^{'}$ are both acyclic relative to $\mathcal{M}$ : That $F$ is acyclic relative to $\mathcal{M}$ is clear since $\triangle^{p} \times \triangle^{q}$ is contractible.\\ Consider $F^{'}.$ $H_{n}(S(X)\bigotimes S(Y)) \cong \overset{n}{\underset{p=0}{\bigoplus}} H_{p}(X) \bigotimes H_{n-p}(Y) \bigoplus \overset{n-1}{\underset{p=0}{\bigoplus}} \ttx(H_{p}(X),H_{n-p-1}(Y))$에서 $X=\triangle^{p}, Y=\triangle^{q}$라고 두자. 그러면 $\widetilde{H}(\triangle^{p}) = 0 = \widetilde{H}(\triangle^{q})$이고 $H_{0}(S(X)\bigotimes S(Y)) \cong H_{0}(X) \bigotimes H_{0}(Y) \cong R \bigotimes R = R$ 이며 $n >0$인 $n$에 대해서는 $H_{n}(S(X) \bigotimes S(Y)) = 0$이다.\\ (2) $F$ is free relative to $\mathcal{M}$ : For each $n$, choose $(\triangle^{n}, \triangle^{n}) \in \mathcal{M}$ and $d_{n} \in S_{n}(\triangle^{n} \times \triangle^{n})$, where $d_{n} : \triangle^{n} \rightarrow \triangle^{n} \times \triangle^{n}$ is the diagonal map $:t \mapsto (t,t)$. Now for each $(f,g) \in \textrm{hom}((\triangle^{n}, \triangle^{n}),(X,Y)), \{F(f,g)d_{n}\}$ form a basis for $S_{n}(X \times Y)$ since $ \xymatrix @=1em @R=2ex @*[c] { % & Y \\ \forall\sigma : \triangle^{n} \ar[ur]^{\sigma_{Y}} \ar[r] \ar[dr]_{\sigma_{X}} & X \times Y \ar[u]_{P_{Y}} \ar[d]^{P_{X}} \\ & X }$ can be written uniquely as $\sigma = (\sigma_{X}, \sigma_{Y}) = (\sigma_{X} \times \sigma_{Y})\circ d_{n}$.\\ (3) $F^{'}$ is free relative to $\mathcal{M}$ : For each $n$, choose $(\triangle^{p}, \triangle^{q}) \in \mathcal{M}$ with $p+q=n$ and $i_{p} \otimes i_{q} \in (S(\triangle^{p})\bigotimes S(\triangle^{q}))_{n} = \underset{i+j=n}{\bigoplus} S_{i}(\triangle^{p}) \bigotimes S_{j}(\triangle^{q})$, where $i_{p} = id. : \triangle^{p} \rightarrow \triangle^{p}$. For each $(\sigma, \tau) \in \textrm{hom}((\triangle^{p},\triangle^{q}), (X,Y)),$ $\{F^{'}(\sigma,\tau)(i_{p}\otimes i_{q})\}$ form a basis for $(S(X) \bigotimes S(Y))_{n} (\,\, \textrm{note that}\,\, F^{'}(\sigma,\tau)(i_{p}\otimes i_{q}) = \sigma_{\sharp} \otimes \tau_{\sharp}(i_{p}\otimes i_{q}) = \sigma_{\sharp}(i_{p}) \otimes \tau_{\sharp}(i_{q}) = \sigma \circ i_{p} \otimes \tau \circ i_{q} = \sigma \otimes \tau$) since $\{\sigma \otimes \tau | \sigma \in S_{p}(X), \tau \in S_{q}(Y)\}$ form a basis for $S_{p}(X) \bigotimes S_{q}(Y)$ and hence $\{\sigma \otimes \tau | \sigma \in S_{p}(X), \tau \in S_{q}(Y), p+q=n\}$ form a basis for $(S(X) \bigotimes S(Y))_{n}$.\\ {\bf Proof of Eilenberg-Zilber theorem} Let $\tau_{0} : H_{0}(F) (=H_{0}(X \times Y)) \leftrightarrow H_{0}(F^{'})(=H_{0}(X) \bigotimes H_{0}(Y))$ be a natural transformation(isomorphism) determined by path components, i.e., if $C, D$ are path-components of $X, Y$, respectively, then $C \times D$ is a path component of $X \times Y$ and $H_{0}(C \times D) \overset{\tau_{0}}{\leftrightarrow} H_{0}(C) \bigotimes H_{0}(D)$. AMT $\Rightarrow \tau_{0}$ gives rise to a natural chain homotopy equivalence. \end{pf2} {\bf 숙제 30} (29.14) Compute $H(\mathbb{P}^{2} \times S^{3}) \neq H(\mathbb{P}^{3} \times S^{2})$ and show $\pi_{*}(\mathbb{P}^{2} \times S^{3}) \cong \pi_{*}(\mathbb{P}^{3} \times S^{2})$.\\ {\bf 숙제 31} (29.15) Compare $S^{2} \times S^{4}$ and $\mathbb{C}P^{3}$. They both have same homology but $\pi_{4}$ are different. (If necessary, use the following fibration and the corresponding long exact sequence of homotopy groups : $S^{1} \to S^{7} \to \mathbb{C}P^{3} \Rightarrow \cdots \to \pi_{k}(S^{1}) \to \pi_{k}(S^{7}) \to \pi_{k}(\mathbb{C}P^{3}) \to \pi_{k-1}(S^{1}) \to \cdots$)\\ {\bf 숙제 32} (29.11.2) $\chi(X \times Y) = \chi(X)\chi(Y)$. (Use the Poincar\'{e} series of $X = f_{X}(t) := \bet_{0} + \bet_{1}t + \cdots + \bet_{n}t^{n}$ and notice that $\chi(X) = f_{X}(-1)$.)\\ {\bf 13. Homology cross product} {\scriptsize \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & (H(X) \bigotimes H(Y))_{n} \ar[r]^{i} \ar[dr] & H_{n}(X \times Y) \ar[r] \ar@2{-}[d] & (H(X)*H(Y))_{n-1} \ar[r] & 0 & & \{x\}\otimes\{y\} \ar@{|->}[dr] \ar@{|.>}[r] & \{x\}\times\{y\} \\ &&H_{n}(S(X) \bigotimes S(Y)) \ar[ur] &&&&&\{x\otimes y\} \ar@{|->}[u] } \]} {\bf Note} $ \phi : S(X)\bigotimes S(Y) \rightarrow S(Y) \bigotimes S(X)$ given by $\phi(x\otimes y) = (-1)^{pq} y \otimes x$ is a natural chain map. $\ulcorner \because \phi(\bd(x\otimes y)) = \phi(\bd x \otimes y + (-1)^{p} x \otimes \bd y) = (-1)^{(p-1)q} y \otimes \bd x + (-1)^{p}(-1)^{p(q-1)} \bd y \otimes x = \bd ( \phi(x \otimes y))\lrcorner$\\ $\Rightarrow \xymatrix @M=1ex @C=1em @R=1em @*[c]{% S(X \times Y) \ar[r]^{T_{\sharp}} \ar[d]^{\simeq}_{f_{1}} & S(Y \times X) \ar[d]^{\simeq}_{f_{2}} & , & T : X \times Y \rightarrow Y \times X & T(x,y) = (y,x) \\ S(X) \bigotimes S(Y) \ar[r]^{\simeq}_{\phi} & S(Y) \bigotimes S(X) &&&}$ All the maps a natural chain transformations. By the AMT, $T_{\sharp} \simeq "f_{2}^{-1}" \circ \phi \circ f_{1}.$ \\ $\Rightarrow T_{*}(\xi \times \eta) = (-1)^{pq}(\eta \times \xi), \xi \in H_{p}(X), \eta \in H_{q}(Y)$.\\ {\bf 14. Alexander-Whitney diagonal approximation} We can give a specific chain map( and hence a chain homotopy equivalence.)\\ $A : S(X \times Y) \rightarrow S(X) \bigotimes S(Y)$ given by $ \forall \ome \in S_{n}(X \times Y), \ome = (\sigma,\tau)$, $A(\ome) = \overset{n}{\underset{p=0}{\bigoplus}} \sigma\lam_{p} \otimes \tau \rho_{n-p} \in (S(X)\bigotimes S(Y))_{n}$.\\ (1) $A$ is natural : only to check the commutativity of the following diagram. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% S_{n}(X \times Y) \ar@{}[dr]|{\circlearrowright} \ar[r]^<<{A} \ar[d]_{(f \times g)_{\sharp}} & (S(X) \bigotimes S(Y))_{n} = \bigoplus S_{p}(X) \bigotimes S_{n-p}(Y) \ar[d]^{f_{\sharp} \otimes g_{\sharp} = \oplus f_{\sharp p} \otimes g_{\sharp n-p}} \\ S_{n}(X' \times Y') \ar[r]^<<{A} & (S(X') \bigotimes S(Y'))_{n} = \bigoplus S_{p}(X') \bigotimes S_{n-p}(Y') } \] Just note $f_{\sharp}(\sigma \lam_{p}) = (f_{\sharp}\sigma)\lam_{p} \Rightarrow !$.\\ (2) $A$ is a chain map, i.e., $\bd A = A \bd :$ (Proof is essentially same as the proof of derivation property of cup product.)\\ $\therefore$ Eilenberg-Zilber Theorem $\Rightarrow A_{*} : H_{*}(X \times Y) \overset{\cong}{\rightarrow} H_{*}(S(X) \bigotimes S(Y))$\\ {\bf 15. Relative K\"{u}nneth formula} K\"{u}nneth formula for $(X,A) \times (Y,B)$\\ {\bf Lemma \,\,(Relative Eilenberg-Zilber Theorem)} Suppose $\{X \times B, A \times Y\}$ : excisive couple in $ X \times Y$. Then $S(X \times Y) / S(X \times B \cup A \times Y) \simeq S(X)/S(A) \bigotimes S(Y)/S(B)$ naturally. Note that $S(X \times Y)/S(X \times B \cup A \times Y) := S((X,A) \times (Y,B)), S(X)/S(A)=S(X,A)\,\, \textrm{and}\,\, S(Y)/S(B)=S(Y,B)$. (Recall $\{A, B\}$ is excisive couple if $S(A) + S(B) \overset{i}{\underset{\simeq}{\hookrightarrow}} S(A \cup B)$.)\\ \begin{pf2} Excisive $\Rightarrow S(X \times B) + S(A \times Y) \overset{\simeq}{\hookrightarrow} S(X \times B \cup A \times Y)$ $\Rightarrow S(X \times Y)/(S(X \times B)+ S(A \times Y)) \overset{\simeq}{\hookrightarrow} S(X \times Y)/S(X \times B \cup A \times Y)$ naturally.\\ \hspace*{3.0em}$S(A) \bigotimes S(Y) \underset{\simeq}{\rightarrow} S(A \times Y)$ \hspace*{6.0em} $\cap$ \hspace*{4.0em} $\cap$ EZ $\Rightarrow S(X) \bigotimes S(Y) \underset{\simeq}{\rightarrow} S(X \times Y)$ naturally. \hspace*{6.0em} $\cup$ \hspace*{4.0em} $\cup$ \hspace*{3.0em}$S(X) \bigotimes S(B) \underset{\simeq}{\rightarrow} S(X \times B)$ $\Rightarrow$ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% S(X \times Y)/(S(X \times B)+S(A \times Y)) \ar[d]_{\simeq} & \ar[l]^<<<{\simeq} S(X) \bigotimes S(Y) / (S(X)\bigotimes S(B)+ S(A) \bigotimes S(Y)) \ar[d]^{\cong}_{(*)}\\ S(X \times Y)/S(X \times B \cup A \times Y) & S(X)/S(A) \bigotimes S(Y)/S(B) } \] $(*)$ follows from the general fact that$ (X \bigotimes Y)/(X \bigotimes B + A \bigotimes Y) \overset{\cong}{\rightarrow} X/A \bigotimes Y/B$. Indeed $\cong$ is induced from canonical map $ X \times Y \rightarrow X/A \bigotimes Y/B$(cf 1.(3)). \end{pf2} From the lemma, we have the following relative K\"{u}nneth formula. \vspace*{1.0em} \framebox{\parbox[b]{14cm}{ Let $R$ be a P.I.D. Then $0 \rightarrow (H(X,A)\bigotimes H(Y,B))_{n} \rightarrow H_{n}((X,A) \times (Y,B)) \rightarrow (H(X,A) * H(Y,B))_{n-1} \rightarrow 0$ , a natural short exact sequence which splits. By definition, $H_{n}((X,A)\times(Y,B)) = H_{n}(X \times Y, X \times B \cup A\times Y) $}} \vspace*{1.0em} \end{document}