\magnification=2000 \overfullrule=0pt \parindent=0pt \nopagenumbers \input amstex %\voffset=-.6in %\hoffset=-.5in %\hsize = 7.5 true in %\vsize=10.4 true in \voffset=1.8 true in \hoffset=-.6 true in \hsize = 10.2 true in \vsize=8 true in \input colordvi \def\cltr{\Red} % Red VERY-Approx PANTONE RED \def\cltb{\Blue} % Blue Approximate PANTONE BLUE-072 \def\cltg{\PineGreen} % ForestGreen Approximate PANTONE 349 \def\cltp{\DarkOrchid} % DarkOrchid No PANTONE match \def\clto{\Orange} % Orange Approximate PANTONE ORANGE-021 \def\cltpk{\CarnationPink} % CarnationPink Approximate PANTONE 218 \def\clts{\Salmon} % Salmon Approximate PANTONE 183 \def\cltbb{\TealBlue} % TealBlue Approximate PANTONE 3145 \def\cltrp{\RoyalPurple} % RoyalPurple Approximate PANTONE 267 \def\cltp{\Purple} % Purple Approximate PANTONE PURPLE \def\cgy{\GreenYellow} % GreenYellow Approximate PANTONE 388 \def\cyy{\Yellow} % Yellow Approximate PANTONE YELLOW \def\cgo{\Goldenrod} % Goldenrod Approximate PANTONE 109 \def\cda{\Dandelion} % Dandelion Approximate PANTONE 123 \def\capr{\Apricot} % Apricot Approximate PANTONE 1565 \def\cpe{\Peach} % Peach Approximate PANTONE 164 \def\cme{\Melon} % Melon Approximate PANTONE 177 \def\cyo{\YellowOrange} % YellowOrange Approximate PANTONE 130 \def\coo{\Orange} % Orange Approximate PANTONE ORANGE-021 \def\cbo{\BurntOrange} % BurntOrange Approximate PANTONE 388 \def\cbs{\Bittersweet} % Bittersweet Approximate PANTONE 167 %\def\creo{\RedOrange} % RedOrange Approximate PANTONE 179 \def\cma{\Mahogany} % Mahogany Approximate PANTONE 484 \def\cmr{\Maroon} % Maroon Approximate PANTONE 201 \def\cbr{\BrickRed} % BrickRed Approximate PANTONE 1805 \def\crr{\Red} % Red VERY-Approx PANTONE RED \def\cor{\OrangeRed} % OrangeRed No PANTONE match \def\paru{\RubineRed} % RubineRed Approximate PANTONE RUBINE-RED \def\cwi{\WildStrawberry} % WildStrawberry Approximate PANTONE 206 \def\csa{\Salmon} % Salmon Approximate PANTONE 183 \def\ccp{\CarnationPink} % CarnationPink Approximate PANTONE 218 \def\cmag{\Magenta} % Magenta Approximate PANTONE PROCESS-MAGENTA \def\cvr{\VioletRed} % VioletRed Approximate PANTONE 219 \def\parh{\Rhodamine} % Rhodamine Approximate PANTONE RHODAMINE-RED \def\cmu{\Mulberry} % Mulberry Approximate PANTONE 241 \def\parv{\RedViolet} % RedViolet Approximate PANTONE 234 \def\cfu{\Fuchsia} % Fuchsia Approximate PANTONE 248 \def\cla{\Lavender} % Lavender Approximate PANTONE 223 \def\cth{\Thistle} % Thistle Approximate PANTONE 245 \def\corc{\Orchid} % Orchid Approximate PANTONE 252 \def\cdo{\DarkOrchid} % DarkOrchid No PANTONE match \def\cpu{\Purple} % Purple Approximate PANTONE PURPLE \def\cpl{\Plum} % Plum VERY-Approx PANTONE 518 \def\cvi{\Violet} % Violet Approximate PANTONE VIOLET \def\clrp{\RoyalPurple} % RoyalPurple Approximate PANTONE 267 \def\cbv{\BlueViolet} % BlueViolet Approximate PANTONE 2755 \def\cpe{\Periwinkle} % Periwinkle Approximate PANTONE 2715 \def\ccb{\CadetBlue} % CadetBlue Approximate PANTONE (534+535)/2 \def\cco{\CornflowerBlue} % CornflowerBlue Approximate PANTONE 292 \def\cmb{\MidnightBlue} % MidnightBlue Approximate PANTONE 302 \def\cnb{\NavyBlue} % NavyBlue Approximate PANTONE 293 \def\crb{\RoyalBlue} % RoyalBlue No PANTONE match %\def\cbb{\Blue} % Blue Approximate PANTONE BLUE-072 \def\cce{\Cerulean} % Cerulean Approximate PANTONE 3005 \def\ccy{\Cyan} % Cyan Approximate PANTONE PROCESS-CYAN \def\cpb{\ProcessBlue} % ProcessBlue Approximate PANTONE PROCESS-BLUE \def\csb{\SkyBlue} % SkyBlue Approximate PANTONE 2985 \def\ctu{\Turquoise} % Turquoise Approximate PANTONE (312+313)/2 \def\ctb{\TealBlue} % TealBlue Approximate PANTONE 3145 \def\caq{\Aquamarine} % Aquamarine Approximate PANTONE 3135 \def\cbg{\BlueGreen} % BlueGreen Approximate PANTONE 320 \def\cem{\Emerald} % Emerald No PANTONE match %\def\cjg{\JungleGreen} % JungleGreen Approximate PANTONE 328 \def\csg{\SeaGreen} % SeaGreen Approximate PANTONE 3268 \def\cgg{\Green} % Green VERY-Approx PANTONE GREEN \def\cfg{\ForestGreen} % ForestGreen Approximate PANTONE 349 \def\cpg{\PineGreen} % PineGreen Approximate PANTONE 323 \def\clg{\LimeGreen} % LimeGreen No PANTONE match \def\cyg{\YellowGreen} % YellowGreen Approximate PANTONE 375 \def\cspg{\SpringGreen} % SpringGreen Approximate PANTONE 381 \def\cog{\OliveGreen} % OliveGreen Approximate PANTONE 582 \def\pars{\RawSienna} % RawSienna Approximate PANTONE 154 \def\cse{\Sepia} % Sepia Approximate PANTONE 161 \def\cbr{\Brown} % Brown Approximate PANTONE 1615 \def\cta{\Tan} % Tan No PANTONE match \def\cgr{\Gray} % Gray Approximate PANTONE COOL-GRAY-8 \def\cbl{\Black} % Black Approximate PANTONE PROCESS-BLACK \def\cwh{\White} % White No PANTONE match \loadmsbm \input epsf \def\ctln{\centerline} \def\u{\underbar} \def\ssk{\smallskip} \def\msk{\medskip} \def\bsk{\bigskip} \def\hsk{\hskip.1in} \def\hhsk{\hskip.2in} \def\dsl{\displaystyle} \def\hskp{\hskip1.5in} \def\lra{$\Leftrightarrow$ } \def\ra{\rightarrow} \def\mpto{\logmapsto} \def\pu{\pi_1} \def\mpu{$\pi_1$} \def\sig{\Sigma} \def\msig{$\Sigma$} \def\ep{\epsilon} \def\sset{\subseteq} \def\del{\partial} \def\inv{^{-1}} \def\wtl{\widetilde} %\def\lra{\Leftrightarrow} \def\del{\partial} \def\delp{\partial^\prime} \def\delpp{\partial^{\prime\prime}} \def\sgn{{\roman{sgn}}} \def\wtih{\widetilde{H}} \def\bbz{{\Bbb Z}} \def\bbr{{\Bbb R}} \def\bbq{{\Bbb Q}} \def\bbc{{\Bbb C}} \def\hdsk{\hskip.7in} \def\hdskb{\hskip.9in} \def\hdskc{\hskip1.1in} \def\hdskd{\hskip1.3in} \def\Hom{\text{Hom}} \def\Ext{\text{Ext}} \def\larr{\leftarrow} {\bf Cohomology:} There is ``dual'' theory to the homology theory, called ``cohomology theory'', which is based on the following observation: if we have a chain complex \ssk \ctln{$\cdots \ra C_n\buildrel{\del_n}\over\ra C_{n-1} \ra \cdots \ra C_0\ra 0$} \ssk with, we assume for the moment, finitely-generated free abelian chain groups, then, thinking in terms of $\bbz$-vector spaces, the boundary maps are represented by integer matrices $d_n:C_n\ra C_{n-1}$. The {\it transposes} of these matrices give rise to linear maps $d_n^{\text{T}}:C_{n-1}\ra C_n$, which form a chain complex (since $ d_{n}^{\text{T}}d_{n-1}^{\text{T}} = (d_{n-1}d_n)^{\text{T}} = 0^{\text{T}} = 0$) {\it running the opposite direction} \ssk \ctln{$0 \ra C_0\ra C_{1} \ra \cdots \ra C_{n-1}\buildrel{\delta_n}\over\ra C_n\ra \cdots$} \ssk whose homology groups we can then compute in the usual way. This is the basic idea behind cohomology. The transpose is not ``really'' a map from $C_{n-1}$ to $C_n$, though, it is more properly thought of as a map between the dual vector spaces $C_{n-1}^*,C_n^*$. The formal way to construct the theory is to introduce the ``Hom'' functor: for a (fixed) abelian group $G$ and an abelian group $A$, \ssk \ctln{$\Hom(A,G) = \{f: A\ra G : f\text{ is a homomomorphism}\}$ .} \ssk For $\bbr$-vector spaces $V$, for example, the usual dual vector space is $V^*=\Hom(V,\bbr)$. The basic point is that Hom ``turns arrows around''; given a homomorphism $\varphi:A\ra B$, there is an induced homomorphism $\varphi^*:\Hom(B,G)\ra\Hom(A,G)$ given by $\varphi^*(f)=f\circ\varphi$ . This satisfies $(\varphi\circ\psi)^*=\psi^*\circ\varphi^*$ and $I^*=I$, as a quick computation establishes. Just as important for us is that $0^*=0$. \vfill \eject Given a chain complex \hskip.2in $\cdots \ra C_n\buildrel{\del_n}\over\ra C_{n-1} \ra \cdots \ra C_0\ra 0$ \hskip.2in (with no assumptions on the chain groups) and a ``coefficient'' group $G$, we can dualize the complex to obtain \ssk \ctln{(*) \hskip.2in $0 \ra \Hom(C_0,G)\ra \Hom(C_{1},G) \ra \cdots \ra \Hom(C_{n-1},G)\buildrel{\del_n^*}\over\ra \Hom(C_n,G)\ra \cdots$} \ssk We will usually write $\del_n^* = \delta_n$, calling them the {\it coboundary} maps. These coboundary maps are defined by $\del_n^*(f)(x) = f(\del_n x)$ for $f\in \Hom(C_{n-1},G)$ and $x\in C_n$ (since $\del_n^*(f)\in \Hom(C_n,G)$). Just as in the finitely generated case, $\delta_{n+1}\circ\delta_n=0$, since for any $f\in\Hom(C_{n-1},G)$, and $x\in C_{n+1}$, \ssk \ctln{$\delta_{n+1}\circ\delta_n(f)(x)=\delta_n(f)(\del_{n+1}x) = f(\del_n\del_{n+1}x) \ f(0)=0$,} \ssk so $\delta_{n+1}\circ\delta_n(f) = 0$. Consequently, (*) is a chain complex (although technically, with its indices rising it is called a {\it cochain complex}), with the attendant homology groups (which we call {\it cohomology groups}!). We could treat this, really, as homology, by renumbering so that $C_{-n}^\prime = \Hom(C_n,G)$; then the coboundary map decreases index, although our homology groups then live in negative dimensions! But cohomology is inherently arrow-reversing, so it seems better to just live with index-increasing maps? \msk In particular, starting with a space (or $\Delta$-complex or CW-complex) $X$, dualizing the standard singular (or simplicial or cellular) chain complex $C_n(X)$ we obtain the singular (or...) cochain complex $(C^n(X;G),\delta_{n+1})$, whose elements are {\it cochains}, and whose homology groups are the singular (or...) cohomology groups of $X$, with coefficients in $G$. As with homology, it is not immediately clear that the simplicial and cellular cohomology groups are topological invariants, but the singular homology groups are; the only input is $X$, from which we build $C_n(X)$ and $C^n(X)=\Hom(C_n(X),G)$. \vfill \eject %%% So what do the singular cohomology groups \u{measure}? We shall see that much of the edifice that we have built around singular homology goes through, with small changes made necessary by the reversal of arrows. One way to see a large part of this is by the fact that the \cltr{homology groups of a chain complex \u{determine} the associated cohomology groups.} To see how this might be formulated, we first note that there is a homomorphism $h:H^n({\Cal C};G)\ra \Hom(H_n({\Cal C}),G)$, for any chain complex ${\Cal C}$, defined as follows: given $[f]\in H^n({\Cal C};G)$ and $[z]\in H_n({\Cal C})$, we have $f\in \Hom(C_n,G)$ (with $\delta f = 0$; it is a {\it cocycle}), and $z\in Z_n\subseteq C_n$, so the element $f(z)\in G$ makes sense. So why not just try $h([f])([z]) = f(z)$ ? We can show that this is well-defined; if $[z]=[z^\prime]$, then $[z]-[z^\prime] = [z-z^\prime]=0$, so $z-z^\prime=\del w$ for some $w\in C_{n+1}$, and then $f(z)-f(z^\prime) = f(z-z^\prime)=f(\del w) = (\delta f)(w)=0$, since $\delta f = 0$. OTOH, if $[f]=[f^\prime]$, then $f-f^\prime = \delta g$ for some $g\in \Hom(C_{n-1},G)$, and then $f(z)-f^\prime(z) = \delta g(z) = g(\del z) = g(0)=0$, since $z$ is an $n$-cycle. So $h$ is well-defined. And since $h([f]-[f^\prime])([z]) = (f-f^\prime)(z) = f(z)-f^\prime(z) = h([f])([z]) - h([f^\prime])([z])$, we have $h([f]-[f^\prime]) = h([f])-h([f^\prime])$, so $h$ is a homomorphism. \ssk Even more, though, \cltg{if the chain groups $C_n$ are free abelian,} then $h$ is onto. To see this, note that any $\varphi:H_n{\Cal C}=Z_n/B_n\ra G$ gives rise to a homom $\varphi_1:Z_n\ra G$, by $\varphi_1(z)=\varphi([z])$. But since $C_n$ is free abelian, $Z_n=\ker \del_n$ is a direct summand of $C_n$; $B_{n-1}=\text{im}\ \del_n\subseteq C_{n-1}$ is a subgroup of a free abelian group, so is free abelian, and a basis for $B_{n-1}$, pulled back to a collection of elements $\{v_i\}$ of $C_n$, together with a basis for $Z_n$, gives a basis for $C_n$. [Showing that the two bases span complementary subspaces, and together span $C_n$, is straightforward.] The point to this is that our homom $\varphi_1$ can be extended to a homom $\varphi_2:C_n\ra G$ by declaring that $\varphi_2(v_i)=0$ for all $i$ and that $\varphi_2=\varphi_1$ on $Z_n$. \u{Then} $\delta(\varphi_2)=0$, since $\delta(\varphi_2)(x)=\varphi_2(\del x) = \varphi_1(\del x) = \varphi([\del x]) = \varphi(0)=0$ for all $x$. [$\varphi_2(\del x) = \varphi_1(\del x)$ since $\del x\in B_n\subseteq Z_n$ .] \vfill \eject So $\varphi_2$ is a cocycle, and $h([\varphi_2])([z]) = \varphi_2(z) = \varphi_1(z) =\varphi([z])$, so $h([\varphi_2]) = \varphi$, as desired. So we have the beginnings of a short exact sequence; \ssk \ctln{$0\ra \ker h\ra H^n({\Cal C};G)\buildrel{h}\over\ra \Hom(H_n({\Cal C}),G)\ra 0$} \ssk This sequence splits; our construction above actually describes a homom $k:\varphi\mapsto \varphi_2$, since $\varphi\mapsto \varphi_1$ is a homom, and $\varphi_2$ is essentially $\varphi_1$ extended by $0$ to a subspace complementary to $Z_n$ (in matrix terms, we pad the matrix for $\varphi_1$ with columns of $0$'s). Since $h(\varphi_2) = \varphi$, $k$ is a right inverse to $h$. This then implies, by the Splitting Lemma (?), that \ssk \ctln{$H^n({\Cal C};G) \cong \Hom(H_n({\Cal C}),G)\oplus \ker h$} \ssk and so to show that cohomology depends only on the homology groups of ${\Cal C}$, it remains to show that $\ker h$ can be computed from the homology groups. \ssk \ctln{$\ker h = \{ [f] \text{ } : \text{ } f:C_n\ra G \text{ and } f(z)=0 \text{ for all } z\in C_n \text{ with } \del z=0\}$} %%% \ctln{$ = \{ f:C_n\ra G \text{ } \text{ } : \text{ } f(z) = 0 \text{ for all } z\in Z_n\}/\{\delta g \text{ } : \text{ } g:C_{n-1}\ra G\}$} \msk There is another map $j:\Hom(B_{n-1},G)\ra H^n({\Cal C};G)$ given by $j(\varphi) = [\psi]$, where $\psi:C_n\ra G$ is defined by $\psi(x)=\varphi(\del x)$ [note that $\delta\psi(x) = \psi(\del x) = \varphi(\del^2 x) = 0$ for all $x$, so $\psi$ is a cocycle], and again, this map is a homomorphism. Further, $\text{im}\ j = \ker h$, since $h(j(\varphi))([z]) = h([\psi])([z]) = \psi(z) = \varphi(\del z) = \varphi(0)=0$ for all $[z]$, giving one containment, and given $\psi$ with $h(\psi)=0$, we \u{define} $\varphi: B_{n-1}\ra G$ by $\varphi(\del x) = \psi(x)$; if $\del x = \del y$, then $\psi(x-y)=0$ since $\del(x-y)=0$, so $\psi(x)=\psi(y)$, and so $\varphi$ is well-defined. Yet again, $\varphi$ is a homom. And certainly $j(\varphi) = \psi$ (by pretending that the equation $\varphi(\del x) = \psi(x)$ defines $\psi$ !). \vfill \eject \ctln{$j(\varphi) = [\psi:x\mapsto \varphi(\del x)]\in H^n({\Cal C},G)\}$} \ssk Therefore, by one of the isomorphism theorems, $\ker h \cong \Hom(B_{n-1},G)/\ker j$. But $\ker j$ consists of those maps $\varphi:B_{n-1}\ra G$ for which $x\mapsto \varphi(\del x)$ is a coboundary $(\delta\psi)(x) = \psi(\del x)$ for some $\psi:C_{n-1}\ra G$. On the face of it, it looks like $\varphi$ itself could stand in for $\psi$, but the point is that \cltr{$\varphi$ and $\psi$ have different domains}. $\psi$ has domain $C_{n-1}$, while $\varphi$ has domain $B_{n-1}$. But this means that $\ker j$ is the image of the map $\Hom(C_{n-1},G)\ra \Hom(B_{n-1},G)$ dual to the inclusion map $B_{n-1}\hookrightarrow C_{n-1}$ . Note that we can put a third term in the middle of these two; \ctln{$\Hom(C_{n-1},G)\ra \Hom(Z_{n-1},G)\ra \Hom(B_{n-1},G)$} since $B_{n-1}\hookrightarrow Z_{n-1}\hookrightarrow C_{n-1}$. But the map $\Hom(C_{n-1},G)\ra \Hom(Z_{n-1},G)$ is surjective, since $Z_{n-1}$ is a direct summand of $C_{n-1}$ (as before, we extend a map $\varphi$ from $z_{n-1}$ by zero of a complementary subspace to build a map from $C_{n-1}$ whose image is $\varphi$). So $\ker j$ is also the image of the dual to the inclusion $i:B_{n-1}\hookrightarrow Z_{n-1}$. \msk The reason for tinkering with things in this way is that $B_{n-1}$ and $Z_{n-1}$ fit into a short exact sequence \ssk \ctln{(**) $0\ra B_{n-1} \ra Z_{n-1} \ra H_{n-1}({\Cal C}) \ra 0$} \ssk with dual sequence \ssk \ctln{(***) $0\leftarrow \Hom(B_{n-1},G) \leftarrow \Hom(Z_{n-1},G) \leftarrow \Hom(H_{n-1}({\Cal C}),G) \leftarrow 0$} \ssk This sequence is not exact, but it is a cochain complex, and so has its own homology groups. Note that the group that we are after is the homology of this cochain complex at the spot $\Hom(B_{n-1},G)$. Since by hypothesis $C_{n-1}$ is free abelian, so are $B_{n-1}$ and $Z_{n-1}$; (**) is then an example of a {\it free resolution} of the abelian group $H_{n-1}({\Cal C})$. \vfill \eject Since I am getting tired of doing homological algebra and not topology, we will finish our proof that $\ker h$ (which we now know to be the (co)homology group mentioned above) depends only on the homology of ${\Cal C}$ by appealing to: \msk \cltr{The homology groups of the cochain complex dual to a free resolution of a group $H$ depend only on the group $H$, and not on the particular free resolution chosen.} \msk The particular homology group that we are interested in is known in the literature a $\Ext(H,G)$. We will not be interested in knowing why it is called that, but only in the fact that, as a consequence we have the \ssk {\bf Universal coefficients Theorem:} For any chain complex ${\Cal C}$, $H^n({\Cal C};G) \cong \Hom(H_n({\Cal C}),G)\oplus \Ext(H_{n-1}({\Cal C}),G)$ . \ssk together with some observations on how to calculate Ext, based on its indifference to the resolution used to compute it. From the exact sequence \ssk $0\ra 0 \ra \bbz \ra \bbz \ra 0$, the dual $0\larr 0 \larr G \larr G \larr 0$ gives $\Ext(\bbz,G) = 0$ ; from \ssk $0\ra \bbz \buildrel{\times n}\over\ra \bbz \ra \bbz_n\ra 0$, the dual \ssk $0\larr G \buildrel{\times n}\over\larr G \ra \Hom(\bbz_n,G)\ra 0$ gives $\Ext(\bbz_n,G) = G/nG$ ; and the fact that \ssk $\Ext(H_1\oplus H_2,G) \cong \Ext(H_1,G)\oplus \Ext(H_2,G)$ (by taking the direct sum of two resolutions, and using the fact that $\Hom(-,G)$ respects direct products, \u{and} that homology respects direct products) \ssk suffice to compute $\Ext(H,G)$ for any finitely-generated abelian group $H$, which will usually suffice for our purposes. \msk Applying all of this homological algebra (there is no topology underlying any of the above work, except as motivation) to our chain complexes from a space $X$, we find that \ssk $H^n(X;G) \cong \Hom(H_n(X),G)\oplus \Ext(H_{n-1}(X),G)$ . \ssk Since the groups on the right are the same whether we use singular, simplicial, or cellular chain complexes to build them, the same is true for the left. So the singular, simplicial, and cellular cohomology groups are all isomorphic (when any two of them are defined)! If we were to chase through the computations above, we could recover the fact that the isomorphisms can be induced by the inclusion maps of the various chain groups. \vfill \end