\classheader{2012-10-08} Homotopy was invented to study the topological invariance of homology groups. \begin{proposition} Let $X$ be a $\Delta$-complex with path components $X_1,\ldots,X_r$. Then $H_0(X)\cong\Z^r$ and furthermore, for all $i\geq 0$, \[H_i(X)=\bigoplus_{j=1}^d H_i(X_j).\] \end{proposition} \begin{proof} We'll show that $X$ connected implies that $H_0(X)\cong\Z$. Then we just have to prove the general part. Note that \[H_0(X)=Z_0(X)/B_0(X)=C_0(X)/\im(\partial_1).\] Pick a vertex $v_0$, and let $u$ be any other vertex. \begin{center} \begin{tikzpicture}[scale=1.5] \coordinate[label=left:$v_0$] (a) at (0,0); \coordinate (b) at (1,0); \coordinate (c) at (.5,.86); \coordinate (d) at (1.5,.86); \coordinate (e) at (2,0); \coordinate (f) at (2.5,.86); \coordinate[label=right:$u$] (g) at (3,0); \coordinate (h) at (2.5,-.86); \begin{scope}[thick] \draw[postaction={decorate}] (a) -- (b); \draw[postaction={decorate}] (b) -- (c); \draw[postaction={decorate}] (c) -- (a); \draw[postaction={decorate}] (d) -- (c); \draw[postaction={decorate}] (b) -- (d); \draw[postaction={decorate}] (d) -- (e); \draw[postaction={decorate}] (d) -- (f); \draw[postaction={decorate}] (e) -- (f); \draw[postaction={decorate}] (b) -- (e); \draw[postaction={decorate}] (f) -- (g); \draw[postaction={decorate}] (e) -- (g); \draw[postaction={decorate}] (g) -- (h); \draw[postaction={decorate}] (h) -- (e); \end{scope} \end{tikzpicture} \end{center} We can see that $[u]=[v_0]\in H_0(X)$ because $X$ is path-connected, so there exists a path from $v_0$ to $u$, which we can homotope to a path of edges $[v_0v_1]\cup [v_1v_2]\cup\cdots[v_nu]$. \begin{center} \hfill \begin{tikzpicture}[scale=1.5] \coordinate[label=left:$v_0$] (a) at (0,0); \coordinate (b) at (1,0); \coordinate (c) at (.5,.86); \coordinate (d) at (1.5,.86); \coordinate (e) at (2,0); \coordinate (f) at (2.5,.86); \coordinate[label=right:$u$] (g) at (3,0); \coordinate (h) at (2.5,-.86); \begin{scope}[thick] \draw[postaction={decorate}] (a) -- (b); \draw[postaction={decorate}] (b) -- (c); \draw[postaction={decorate}] (c) -- (a); \draw[postaction={decorate}] (d) -- (c); \draw[postaction={decorate}] (b) -- (d); \draw[postaction={decorate}] (d) -- (e); \draw[postaction={decorate}] (d) -- (f); \draw[postaction={decorate}] (e) -- (f); \draw[postaction={decorate}] (b) -- (e); \draw[postaction={decorate}] (f) -- (g); \draw[postaction={decorate}] (e) -- (g); \draw[postaction={decorate}] (g) -- (h); \draw[postaction={decorate}] (h) -- (e); \draw [style={shorten >=3pt,shorten <=3pt},red,line width=2pt,line cap=round] plot [smooth,samples=20] coordinates {(0,-0.05) (1,0.5) (2,0.1) (2.5,0.3) (3,-0.05)}; \end{scope} \end{tikzpicture}\hfill \begin{tikzpicture}[scale=1.5] \coordinate[label=left:$v_0$] (a) at (0,0); \coordinate (b) at (1,0); \coordinate[label=above:$v_1$] (c) at (.5,.86); \coordinate[label=above:$\cdots$] (d) at (1.5,.86); \coordinate (e) at (2,0); \coordinate[label=above:$v_n$] (f) at (2.5,.86); \coordinate[label=right:$u$] (g) at (3,0); \coordinate (h) at (2.5,-.86); \begin{scope}[thick] \draw[postaction={decorate}] (a) -- (b); \draw[postaction={decorate}] (b) -- (c); \draw[postaction={decorate}] (c) -- (a); \draw[postaction={decorate}] (d) -- (c); \draw[postaction={decorate}] (b) -- (d); \draw[postaction={decorate}] (d) -- (e); \draw[postaction={decorate}] (d) -- (f); \draw[postaction={decorate}] (e) -- (f); \draw[postaction={decorate}] (b) -- (e); \draw[postaction={decorate}] (f) -- (g); \draw[postaction={decorate}] (e) -- (g); \draw[postaction={decorate}] (g) -- (h); \draw[postaction={decorate}] (h) -- (e); \draw [style={shorten >=1pt,shorten <=1pt},red,line width=2pt,line cap=round] (a) -- (c) -- (f) -- (g); \end{scope} \end{tikzpicture}\hfill$\text{}$ \end{center} Let \[\sigma=[v_0v_1]+[v_1v_2]+\cdots+[v_nu],\] so that \[\partial\sigma=(v_1-v_0)+(v_2-v_1)+\cdots+(u-v_n).\] Thus $u-v_0=\partial \sigma\in \im(\partial_1)$, and thus $[u]=[v_0]\in H_0(X)$. For any $w=\sum a_iv_i$, we therefore have that $[w]=\sum a_i[v_i]=\left(\sum a_i\right)[v_0]$. Thus, $[v_0]$ generates $H_0(X)$. To prove that $[v_0]$ has infinite order in $H_0(X)$, we'll use contradiction. Suppose it had order $n$; then $n[v_0]=0\in H_0(X)$, so that there is some $\sigma\in C_1(X)$ such that $\partial \sigma=n[v_0]$. Now we define a homomorphism $\psi:C_0(X)\to\Z$ by sending every generator of $C_0(X)$ to 1. Thus, for any 1-simplex $[uv]$, \[\psi(\partial[uv])=\psi(v-u)=\psi(v)-\psi(u)=0\] but $n=n\psi(v_0)=\psi(nv_0)=\psi(\partial\sigma)=0$, which is a contradiction. Thus, we've shown that $H_0(X)\cong\Z$ for $X$ connected. Now we want to prove that \[H_i(X)=\bigoplus_{j=1}^d H_i(X_j).\] Let \[\phi_n:\bigoplus_{i=1}^d C_n(X_i)\to C_n(X)\] be defined by sending $\sigma$ to $\sigma$ (since $X_i\subseteq X$, a chain on $X_i$ is literally a chain on $X$) and we can get an inverse map by noting that, for any simplex $\sigma$, its image $\sigma(\Delta^n)$ is connected, and therefore can only lie in one connected component. Thus $\phi_n$ is an isomorphism. Note that ``$\phi$ commutes with $\partial$'', i.e. \[\phi\circ\partial_n^+=\partial_n\circ \phi\] where $\partial_n^+=\oplus \partial_n^{X_i}$ (this fact is clear). Now we will need a homological algebra fact. For any chain complexes $\cal{C}=\{C_n,\partial_n\}$ and $\cal{C}'=\{C_n',\partial_n'\}$, a chain map $f:\cal{C}\to\cal{C}'$ is a collection of maps $f_n:C_n\to C_n'$ such that $f_n\circ\partial_n=\partial_n'\circ f_n$, or as a diagram, \[\xymatrix@=0.5in{C_n \ar[r]^{\partial_n} \ar[d]_{f_n} & C_{n-1} \ar[d]^{f_{n-1}}\\ C_n'\ar[r]_{\partial_n'} & C_{n-1}'}\] The homological algebra fact is that any chain map $f:\cal{C}\to\cal{C}'$ \textbf{functorially} induces a homomorphism $f_*:H_n(\cal{C})\to H_n(\cal{C}')$. Here is a proof. Given a $\sigma\in Z_n(\cal{C})$, or equivalently $\partial_n\sigma=0$, we get that \[\partial_n(f_n\circ \sigma)=f_n(\partial_n\sigma)=f_n(0)=0\] so $f_n(\sigma)\in Z_n(\cal{C}')$. We map $Z_n(\cal{C})$ to $H_n(\cal{C})$ by sending $\sigma$ to $[\sigma]$, and likewise for $\cal{C}'$. To show $f$ induces a compatible map on homology (the dashed arrow) \[\xymatrix@=0.5in{Z_n(\cal{C}) \ar[r] \ar[d] & Z_n(\cal{C}') \ar[d]\\ H_n(\cal{C})=Z_n(\cal{C})/B_n(\cal{C}) \ar@{-->}[r] & H_n(\cal{C}')=Z_n(\cal{C}')/B_n(\cal{C}')}\] note that for any $\tau\in B_n(\cal{C})$, we have $\partial'(f(\tau))=f(\partial\tau)=f(0)=0$, so that $f(\tau)\in B_n(\cal{C}')$. Thus $f_*$ is well-defined. Now, to finish our proof, use the functoriality of the above map to see that, because $\phi$ is an isomorphism, it induces an isomorphism on homology. \end{proof} Our next topic is the topological invariance of $H_n(X)$. Our goal is to show that if $X$ and $Y$ are $\Delta$-complexes and $f:X\to Y$ is a homeomorphism, then $H_i(X)\cong H_i(Y)$ for all $i$. The main problem is that if $\Delta_n\xrightarrow{\sigma} X$ is a simplex of $X$, then the composition $\Delta_n\xrightarrow{\sigma}X\xrightarrow{f}Y$ may not be a simplex of $Y$. \begin{center} \begin{tikzpicture}[scale=1.5] \coordinate (a) at (0,0); \coordinate (b) at (1,0); \coordinate (c) at (.5,.86); \coordinate (xa) at (2,0); \coordinate (xb) at (3,0); \coordinate (xc) at (2.5,.86); \coordinate (xd) at (3.5,.86); \coordinate (xe) at (4,0); \coordinate (xf) at (4.5,.86); \coordinate (xg) at (5,0); \coordinate (xh) at (4.5,-.86); \begin{scope}[thick] \draw [thick, draw=black, fill=gray, opacity=1] (7,0.15) ellipse (0.5cm and 0.3cm); \draw [dashed] plot [smooth] coordinates {(3.5,0.29) (5,0.5) (7,0)}; \draw [thick, draw=black, fill=gray, opacity=1] (3,0) -- (4,0) -- (3.5,0.86) -- cycle; \draw[postaction={decorate}] (a) -- (b); \draw[postaction={decorate}] (b) -- (c); \draw[postaction={decorate}] (c) -- (a); \draw[postaction={decorate}] (xa) -- (xb); \draw[postaction={decorate}] (xb) -- (xc); \draw[postaction={decorate}] (xc) -- (xa); \draw[postaction={decorate}] (xd) -- (xc); \draw[postaction={decorate}] (xb) -- (xd); \draw[postaction={decorate}] (xd) -- (xe); \draw[postaction={decorate}] (xd) -- (xf); \draw[postaction={decorate}] (xe) -- (xf); \draw[postaction={decorate}] (xb) -- (xe); \draw[postaction={decorate}] (xf) -- (xg); \draw[postaction={decorate}] (xe) -- (xg); \draw[postaction={decorate}] (xg) -- (xh); \draw[postaction={decorate}] (xh) -- (xe); \draw [dashed] plot [smooth] coordinates {(0.5,0.29) (2,0.5) (3.5,0.29)}; \draw [thick, draw=black, fill=gray, opacity=1] (0,0) -- (1,0) -- (0.5,0.86) -- cycle; \path (7,0) coordinate (origin); \path (0:1cm) ++(7,0) coordinate (P0); \path (1*72:1cm) ++(7,0) coordinate (P1); \path (2*72:1cm) ++(7,0) coordinate (P2); \path (3*72:1cm) ++(7,0) coordinate (P3); \path (4*72:1cm) ++(7,0) coordinate (P4); % Draw the edges of the pentagon \draw (P0) -- (P1) -- (P2) -- (P3) -- (P4) -- cycle; % Add "spokes" \draw (origin) -- (P0) (origin) -- (P1) (origin) -- (P2) (origin) -- (P3) (origin) -- (P4); \end{scope} \end{tikzpicture} \end{center} There are two main strategies for overcoming this. We can either work out the simplicial approximation theorem, in which we show that we can homotope our map sufficiently and subdivide our simplices sufficiently so that any map can be realized as a simplical map, or we can come up with a homology theory that we can't compute but know is isomorphic to what we want.