\classheader{2012-10-15} There will be an in-class midterm the week after next. Until further notice, $H_i(X)$ will denote the singular homology groups; also could be simplicial, really just pretend I'm doing both at the same time. We've defined simplicial and singular homology, and proved that singular homology is a homotopy functor. However, we can't compute singular homology. Thus, we want to show that $H_i\cong H_i^S$. \subsection*{Relative Homology} Suppose that $X$ is a topological space and $A\subset X$ a subspace, with $i:A\hookrightarrow X$ the inclusion. This obviously induces an injective chain map $i_\#:C_n(A)\hookrightarrow C_n(X)$; we'll identify $C_n(A)$ with its image under $i_\#$. There is an induced map on homology $i_*:H_n(A)\to H_n(X)$. Note that $i_*$ is often not injective, for example with $X=\D^2$ and $A=\S^1$, we get $H_1(\S^1)\to H_1(\D^2)$ is a map from $\Z$ to 0. \begin{definition} The relative chain group is defined to be \[C_n(X,A):=C_n(X)/C_n(A).\] \end{definition} Because $\partial_n$ on $C_n(X)$ preserves $C_n(A)$, we get an induced map $\partial_n':C_n(X,A)\to C_{n-1}(X,A)$. Then \[\cal{C}'(X,A)=\{C_n(X,A),\partial_n'\}\] is a chain complex, and we define the relative homology group to be $H_i(X,A)=H_i(\cal{C}'(X,A))$. A cycle $\sigma\in Z_n(X,A)$ is just a chain $\sigma\in C_n(X)$ such that $\partial\sigma\in C_{n-1}(A)$. \begin{center} \begin{tikzpicture}[scale=3] \begin{scope} \node (center) at (45:1) {}; \draw[thick] (center) circle (0.261); \clip (0,0) rectangle (1.,1.1); \draw[very thick] (0,0) circle (1); \fill[gray!40] (0,0) circle (1); \end{scope} \begin{scope}[every node/.style={fill,circle,inner sep=0pt,outer sep=1pt,minimum size=0.15cm}] \node (a) at (30:1) {}; \node (b) at (60:1) {}; \end{scope} \node (A) at (45:0.6) {$A$}; \node (s) at (45:1.35) {$\sigma$}; \end{tikzpicture} \end{center} \begin{definition} Suppose that $A\subset Y$. Then $A$ is a deformation retract of $Y$, denoted $Y\sesquigarrow A$, if there is a homotopy $F:Y\times [0,1]\to Y$ such that $F_0=\id_Y$, $F_1(Y)\subseteq A$, and for all $t\in [0,1]$, we have $F_t|_A=\id_A$. \end{definition} \begin{definition} A pair $(X,A)$ of spaces is reasonable if there is a neighborhood $Y\supset A$ such that $Y\sesquigarrow A$. This is the case for all $\Delta$-complexes. This rules out crazy things like the topologist's sine curve. \end{definition} \begin{center} \begin{tikzpicture}[scale=1] \fill[gray!50] (0,0) circle (2); \draw[very thick] (0,0) circle (2); \fill[gray!25,rotate=30] (0,0) ellipse (1.3 and 1); %\fill[gray!25] (0,0) circle (1.3); \draw[rotate=30] (0,0) ellipse (1.3 and 1); %\draw[thick] (0,0) circle (1.3); \begin{scope} \tikzset{>=latex} \clip[rotate=30] (0,0) ellipse (1.3 and 1); \draw[->] (0:1.5) -- (0:0.6); \draw[->] (45:1.5) -- (45:0.6); \draw[->] (90:1.5) -- (90:0.6); \draw[->] (135:1.5) -- (135:0.6); \draw[->] (180:1.5) -- (180:0.6); \draw[->] (225:1.5) -- (225:0.6); \draw[->] (270:1.5) -- (270:0.6); \draw[->] (315:1.5) -- (315:0.6); \end{scope} \fill[white] (0,0) circle (0.6); \draw[thick] (0,0) circle (0.6); \node (X) at (60:1.7) {$X$}; \node (Y) at (120:1.2) {$Y$}; \node (A) at (0,0) {$A$}; \end{tikzpicture} \end{center} \begin{theorem}[Homology of quotient spaces] Let $(X,A)$ be a reasonable pair. Then the quotient $P:(X,A)\to(X/A,A/A)$ induces an isomorphism for all $i\geq 0$, \[P_*: H_i(X,A)\to H_i(X/A,A/A)\cong \widetilde{H}_i(X/A).\] \end{theorem} \subsection*{Fundamental Theorem of Homological Algebra} Let \[\xymatrix{ \cdots \ar[r] & A_{n+1} \ar[r]^{\phi_{n+1}} & A_n\ar[r]^{\phi_n} & A_{n-1}\ar[r] & \cdots}\] be a sequence of abelian groups, or chain complexes, or $R$-modules, $\ldots$ with $\phi_n$ appropriate morphisms. The sequence is exact at $A_n$ if $\im(\phi_{n+1})=\ker(\phi_n)$. \begin{example} A short exact sequence (SES) is a sequence \[0\to A\xrightarrow{\;\phi\;} B\xrightarrow{\;\psi\;} C\to 0\] which is exact at $A$, $B$, and $C$. Being exact at $A$ is equivalent to $\phi$ being injective, being exact at $C$ is equivalent to $\psi$ being onto. Key example: \[0\to \bigoplus C_n(A)\to\bigoplus C_n(X)\to\bigoplus C_n(X,A)\to 0\] This is a SES of chain complexes. \end{example} \begin{theorem}[Fundamental Theorem of Homological Algebra] Let $\cal{A}$, $\cal{B}$, and $\cal{C}$ be chain complexes, and suppose that \[0\to\cal{A}\to\cal{B}\to\cal{C}\to0\] is a SES of chain complexes. Then there exists a connecting homomorphism $\delta:H_n(\cal{C})\to H_{n-1}(\cal{A})$ and a ``long exact sequence'' of abelian groups \[\xymatrix{\cdots\ar[r] & H_n(\cal{A})\ar[r]^{\phi_*} & H_n(\cal{B})\ar[r]^{\psi_*} & H_n(\cal{C})\ar[r]^{\delta} & H_{n-1}(\cal{A}) \ar[r]^{\phi_*} & H_{n-1}(\cal{B})\ar[r] & \cdots}\] \end{theorem} \begin{proof}[Start of proof] We need to define $\delta:H_n(\cal{C})\to H_{n-1}(\cal{A})$. The natural candidate is \[\xymatrix@!C=0.6in{Z_n(\cal{C}) \ar[d] \ar[r]& Z_{n-1}(\cal{A}) \ar[r] & H_{n-1}(\cal{A})\\ H_n(\cal{C})=Z_n(\cal{C})/B_n(\cal{C}) \ar@{-->}[rru]}\] \end{proof} A short exact sequence of chain complexes just means a diagram like this: \[\xymatrix@M=0.13in{ & \vdots\ar[d] & \vdots\ar[d] & \vdots\ar[d] & \\ 0 \ar[r] & A_{n+1}\ar[r]^{\phi_{n+1}} \ar[d]^{\partial_{n+1}}& B_{n+1}\ar[r]^{\psi_{n+1}}\ar[d]^{\partial_{n+1}'} & C_{n+1}\ar[r] \ar[d]^{\partial_{n+1}''} & 0\\ 0 \ar[r] & A_{n}\ar[r]^{\phi_{n}} \ar[d]^{\partial_{n}}& B_{n}\ar[r]^{\psi_{n}}\ar[d]^{\partial_{n}'} & C_{n}\ar[r] \ar[d]^{\partial_{n}''} & 0\\ 0 \ar[r] & A_{n-1}\ar[r]^{\phi_{n-1}} \ar[d]^{\partial_{n-1}}& B_{n-1}\ar[r]^{\psi_{n-1}}\ar[d]^{\partial_{n-1}'} & C_{n-1}\ar[r] \ar[d]^{\partial_{n-1}''} & 0\\ & \vdots & \vdots & \vdots & }\]