# Homology 3: Cycle Groups and Boundary Groups

Let us use $\langle\cdots\rangle$ for an unoriented simplex and $(\cdots)$ for an oriented simplex.

Examples. 1. $(p_0p_1)=-(p_1p_0)$.

2. \begin{eqnarray*}\sigma_2&=&(p_0p_1p_2)=(p_2p_0p_1)=(p_1p_2p_0)\\-(p_0p_2p_1)&=&-(p_2p_1p_0)=-(p_1p_0p_2).\end{eqnarray*}

Let $K=\{\sigma_\alpha\}$ be an $n$-dimensional simplicial complex of oriented simplexes.

Definition. The $r$-chain group $C_r(K)$ of a simplicial complex $K$ is a free abelian group generated by $r$-simplexes of $K$. If $r>\dim K$, $C_r(K):=0$. An element of $C_r(K)$ is called an $r$-chain.

Let there be $N_r$ $r$-simplexes in $K$. Denote them by $\sigma_{r,i}$ ($1\leq i\leq N$). Then $c\in C_r(K)$ is expressed as $$c=\sum_{i=1}^{N_r}c_i\sigma_{r,i},\ c_i\in\mathbb Z.$$ The integers $c_i$ are called the coefficients of $c$. The addition of two $r$-chains $\sum_ic_i\sigma_{r,i}$ and $c’=\sum_ic_i’\sigma_{r,i}$ is $$c+c’=\sum_i(c_i+c_i’)\sigma_{r,i}.$$ The unit element is $0=\sum_i0\cdot\sigma_{r,i}$. The inverse element of $c$ is $-c=\sum_i(-c_i)\sigma_{r,i}$. Hence we see that $C_r(K)$ is a free abelian group of rank $N_r$ $$C_r(K)\cong\stackrel{N_r}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}.$$

Denote the boundary of an $r$-simplex $\sigma_r$ by $\partial_r\sigma_r$. Since a 0-simplex has no boundary, $$\partial_0p_0=0.$$ For a 1-simplex $(p_0p_1)$, $$\partial_1(p_0p_1):=p_1-p_0.$$ Let $\sigma_r=(p_0\cdots p_r)$ ($r>0$) be an oriented $r$-simplex. The boundary $\partial_r\sigma_r$ of $\sigma_r$ is an $(r-1)$-chain defined by $$\partial_r\sigma_r:=\sum_{i=0}^r(-1)^i(p_0p_1\cdots\hat{p}_i\cdots p_r)$$ where the point $p_i$ under $\hat{}$ is omitted. For example, \begin{eqnarray*}\partial_2(p_0p_1p_2)&=&(p_1p_2)-(p_0p_2)+(p_0p_1),\\\partial_3(p_0p_1p_2p_3)&=&(p_1p_2p_3)-(p_0p_2p_3)+(p_0p_1p_3)-(p_0p_1p_2).\end{eqnarray*} The boundary $\sigma_r$ defines a homomorphism called the boundary operator $$\partial_r: C_r(K)\longrightarrow C_{r-1}(K);\ c=\sum_i c_i\sigma_{r,i}\longmapsto\partial_rc=\sum_ic_i\partial_r\sigma_{r,i}.$$

Let $K$ be an $n$-dimensional simplicial complex. Then there exists a sequence of free abelian groups and homomorphisms $$0\stackrel{i}{\hookrightarrow}C_n(K)\stackrel{\partial_n}{\longrightarrow}C_{n-1}(K)\stackrel{\partial_{n-1}}{\longrightarrow}\cdots\stackrel{\partial_2}{\longrightarrow}C_1(K)\stackrel{\partial_1}{\longrightarrow}C_0(K)\stackrel{\partial_0}{\longrightarrow}0.$$ This sequence is called the chain complex associated with $K$ and is denoted by $C(K)$.

Definition. $Z_r(K):=\ker\partial_r\subset C_r(K)$ is called the $r$-cycle group. The elements of $Z_r(K)$ are called $r$-cycles. If $c\in Z_r(K)$, i.e. if $c$ is an $r$-cycle, $\partial_rc=0$. If $r=0$, $\partial_rc=0$ for all $c\in C_0(K)$, so $C_0(K)=Z_0(K)$.

Definition. Let us consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)$ and let $c\in C_r(K)$. If there exists $d\in C_{r+1}(K)$ such that $c=\partial_{r+1}d$, then $c$ is called an $r$-boundary. The set of $r$-boundaries $B_r(K)$ ($=\partial_{r+1}C_{r+1}(K)={\rm Im}\partial_{r+1}$) is a subgroup of $C_r(K)$ called the $r$-boundary group. If $K$ is an $n$-dimensional simplicial complex, $B_n(K)=0$.

Consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)\stackrel{\partial_r}{\longrightarrow}C_{r-1}(K)$. Then the following lemma holds.

Lemma. The composite map $\partial_r\partial_{r+1}:C_{r+1}(K)\longrightarrow C_{r-1}(K)$ is a zero map.

Proof. Since $\partial_r$ is a linear operator on $C_r(K)$, it suffices to prove the identity $\partial_r\partial_{r+1}=0$ for the generators of $C_{r+1}(K)$. If $r=0$, $\partial_0\partial_1=0$ since $\partial_0$ is a zero operator. Let us assume that $r>0$. Take $\sigma=(p_0\cdots p_rp_{r+1})\in C_{r+1}(K)$. \begin{eqnarray*}\partial_r\partial_{r+1}\sigma&=&\partial_r\sum_{i=0}^{r+1}(-1)^i(p_0\cdots \hat{p}_i\cdots p_{r+1})\\&=&\sum_{i=0}^{r+1}(-1)^i\partial_r(p_0\cdots \hat{p}_i\cdots p_{r+1})\\&=&\sum_{i=0}^{r+1}(-1)^i\{\sum_{j=0}^{i-1}(-1)^j(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\&&\sum_{j=i+1}^{r+1}(-1)^{j-1}(p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\}\\&=&\sum_{i=0}^{r+1}\sum_{j=0}^{i-1}(-1)^{i+j}(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\&&\sum_{i=0}^{r+1}\sum_{j=i+1}^{r+1}(-1)^{i+j-1}p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\\&=&0.
\end{eqnarray*}

Theorem. $B_r(K)\subset Z_r(K)$ or equivalently $\mathrm{Im}\partial_{r+1}\subset\ker\partial_r$.

Proof. Let $c\in B_r(K)$ Then $c=\partial_{r+1}d$ for some $d\in C_{r+1}(K)$. By Lemma, $\partial_rc=\partial_r\partial_{r+1}d=0$. Hence, $c\in\ker\partial_r=Z_r(K)$.

# Homology 2: Simplexes and Simplicial Complexes

Definition. A 0-simplex $\langle p_0\rangle$ is a point or a vertex. A 1-simplex $\langle p_0p_1\rangle$ is a line or an edge. A 2-simplex $\langle p_0p_1p_2\rangle$ is a triangle with its interior included. A 3-simplex $\langle p_0p_1p_2p_3\rangle$ is a solid tetrahedron.

A 0-simplex $\langle p_0\rangle$ may be simply written as $p_0$.

Note that in order for an $r$-simplex to represent an $r$-dimensional object, the vertices $p_i$ must be geometrically independent, i.e. no $(r-1)$-dimensional hyperplane contains all the $r+1$ points. Let $p_0,\cdots,p_r$ be points geometrically independent in $\mathbb R^m$ ($m\geq r$). The $r$-simplex $$\sigma_r=\{x\in\mathbb R^m: x=\sum_{i=0}^r c_ip_i,\ c_i\geq 0,\ \sum_{i=0}^r c_i=1\}$$ has the points $p_0,\cdots,p_r$ as its vertices. The ordered $r+1$-tuple $(c_0,c_1,\cdots,c_r)$ is called the barycentric coordinate of $x$. The 3-simplex $\langle p_0p_1p_2p_3\rangle$ four 0-faces (vertices) $p_0,p_1,p_2,p_3$; six 1-faces (edges) $\langle p_0p_1\rangle$, $\langle p_0p_2\rangle$, $\langle p_0p_3\rangle$, $\langle p_1p_2\rangle$, $\langle p_1p_3\rangle$, $\langle p_2p_3\rangle$; four 2-faces (faces) $\langle p_0p_1p_2\rangle$, $\langle p_0p_2p_3\rangle$, $\langle p_0p_1p_3\rangle$, $\langle p_1p_2p_3\rangle$.

Let $K$ be a set of finite number of simplexes in $\mathbb R^m$. If these simplexes are nicely fitted together, $K$ is called a simplicial complex. By nicely fitted together we mean that:

1. An arbitrary face of a simplex of $K$ belongs to $K$.
2. If $\sigma$ and $\sigma’$ are two simplexes of $K$, $\sigma\cap\sigma’$ is either empty or a face of $\sigma$ and $\sigma’$.

The dimension of a simplicial complex is defined to be the maximum dimension of simplexes in $K$.

Let $\sigma_r$ be an $r$-simplex and $K$ be the set of faces of $\sigma_r$. Then $K$ is an $r$-dimensional simplicial complex.For example, take $\sigma_3=\langle p_0p_1p_2,p_3\rangle$. Then $$\begin{array}{c}K=\{p_0,p_1,p_2,p_3,\langle p_0p_1\rangle,\langle p_0p_2\rangle,\langle p_0p_3\rangle,\langle p_1p_2\rangle,\langle p_1p_3\rangle,\langle p_2p_3\rangle,\\\langle p_0p_1p_2\rangle,\langle p_0p_1p_3\rangle,\langle p_0p_2p_3\rangle,\langle p_1p_2p_3\rangle,\langle p_0p_1p_2p_3\rangle\}.\end{array}$$

Definition. Let $K$ be a simplicial complex of simplexes in $\mathbb R^m$. The union of all the simplexes of $K$ is a subset of $\mathbb R^m$ called the polyhedron $|K|$ of a simplicial complex $K$. Note that $\dim |K|=\dim K$.

Let $X$ be a topological space. If there is a simplicial complex $K$ and a homeomorphism $f:|K|\longrightarrow X$, $X$ is said to be triangulable and the pair $(K,f)$ is called a triangulation of $X$.

Example. The following picture shows a triangulation of $S^1\times [0,1]$.

Example. The following example is not a triangulation of $S^1\times [0,1]$.

Let $\sigma_2=\langle p_0p_1p_2\rangle$ and $\sigma_2’=\langle p_2p_3p_0\rangle$. Then $\sigma_2\cap\sigma_2’=\langle p_0\rangle\cup\langle p_2\rangle$. This is neither $\emptyset$ nor a simplex.

# Homology 1: Free Abelian Groups

Before we discuss homology groups, we review some basics of abelian group theory.

The group operation for an abelian group is denoted by $+$. The unit element is denoted by $0$.

Let $G_1$ and $G_2$ be abalian groups. A map $f: G_1\longrightarrow G_2$ is said to be a homomorphism if $$f(x+y)=f(x)+f(y),\ x,y\in G_1.$$ If $f$ is also a bijection (i.e one-to-one and onto), $f$ is called an isomorphism. If there is an isomorphism $f: G_1\longrightarrow G_2$, $G_1$ is said to be isomorphic to $G_2$ and we write $G_1\stackrel{f}{\cong} G_2$ or simply $G_1\cong G_2$.

Example. Define a map $f: \mathbb Z\longrightarrow\mathbb Z_2=\{0,1\}$ by $$f(2n)=0\ \mbox{and}\ f(2n+1)=1.$$ Then $f$ is a homomorphism.

A subset $H\subset G$ is a subgroup if it is a group with respect to the group operation of $G$.

Example. For any $k\in\mathbb N$, $k\mathbb Z=\{kn: n\in\mathbb Z\}$ is a subgroup of $\mathbb Z$.

Example. $\mathbb Z_2=\{0,1\}$ is not a subgroup of $\mathbb Z$.

Let $H$ be a subgroup of $G$. Define a relation on $G$ by $$\forall x,y\in G,\ x\sim y\ \mbox{if}\ x-y\in H.$$ Then $\sim$ is an equivalence relation on $G$. The equivalence class of $x\in G$ is denoted by $[x]$, i.e. \begin{eqnarray*}[x]&=&\{y\in G: y\sim x\}\\&=&\{y\in G: y-x\in H\}.\end{eqnarray*} Let $G/H$ be the quotient set $$G/H=\{[x]: x\in G\}.$$ Define an operation $+$on $G/H$ by $$[x]+[y]=[x+y],\ \forall [x],[y]\in G/H.$$ Then $G/H$ becomes an abelian group with this operation.

Example. $\mathbb Z/2\mathbb Z=\{[0],[1]\}$. Define $\varphi: \mathbb Z/2\mathbb Z\longrightarrow\mathbb Z_2$ by $$\varphi([0])=0\ \mbox{and}\ \varphi([1])=1.$$ Then $\mathbb Z/2\mathbb Z\cong\mathbb Z_2$. In general, for every $k\in\mathbb N$, $\mathbb Z/k\mathbb Z\cong\mathbb Z_k$.

Lemma 1. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then

(a) $\ker f=\{x\in G_1: f(x)=0\}=f^{-1}(0)$ is a subgroup of $G_1$.

(b) ${\mathrm im}f=\{f(x): x\in G_1\}$ is a subgroup of $G_2$.

Theorem 2 [Fundamental Theorem of Homomorphism]. Let $f: G_1\longrightarrow G_2$ be a homomorphism. Then $$G_1/\ker f\cong{\mathrm im}f.$$

Example. Let $f: \mathbb Z\longrightarrow\mathbb Z_2$ be defined by $$f(2n)=0,\ f(2n+1)=1.$$ Then $\ker f=2\mathbb Z$ and ${\mathrm im}f=\mathbb Z_2$. By Fundamental Theorem of Homomorphism, $$\mathbb Z/2\mathbb Z\cong\mathbb Z_2.$$

Take $r$ elements $x_1,x_2,\cdots,x_r$ of $G$. The elements of $G$ of the form $$n_1x_1+n_2x_2+\cdots+n_rx_r\ (n_i\in\mathbb Z,\ 1\leq i\leq r)$$ form a subgroup of $G$, which we denote $\langle x_1,\cdots,x_r\rangle$. $\langle x_1,\cdots,x_r\rangle$ is called a subgroup of $G$ generated by the generators $x_1,\cdots,x_r$. If $G$ itself is generated by finite lelements, $G$ is said to be finitely generated. If $n_1x_1+\cdots+n_rx_r=0$ is satisfied only when $n_1=\cdots=n_r=0$, $x_1,\cdots,x_r$ are said to be linearly independent.

Definition. If $G$ is fintely generated by $r$ linearly independent elements, $G$ is called a free abelian group of rank $r$.

Example. $\mathbb Z$ is a free abelian group of rank 1 generated by 1 (or $-1$).

Example. Let $\mathbb Z\oplus\mathbb Z=\{(m,n):m,n\in\mathbb Z\}$. The $\mathbb Z\oplus\mathbb Z$ is a free abelian group of rank 2 generated by $(1,0)$ and $0,1)$. More generally, $$\stackrel{r\ \mbox{copies}}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}$$ is a free abelian group of rank $r$.

Example. $\mathbb Z_2=\{0,1\}$ is fintely generated by 1 but is not free. $1+1=0$ so 1 is not linearly independent.

If $G=\langle x\rangle=\{0,\pm x,\pm 2x,\cdots\}$, $G$ is called a cyclic group. If $nx\ne 0$ $\forall n\in\mathbb Z\setminus\{0\}$, it is an infinite cyclic group. If $nx=0$ for some $n\in\mathbb Z\setminus\{0\}$, it is a finite cyclic group. Let $G=\langle x\rangle$ and let $f:\mathbb Z\longrightarrow G$ be a homomorphism defined by $f(k)=kx$, $k\in\mathbb Z$. $f$ is an epimorphism (i.e. onto homomorphism), so by Fundamental Theorem of Homomorphism, $$G\cong\mathbb Z/\ker f.$$ If $G$ is a finite group, then there exists the smallest positive integer $N$ such that $Nx=0$. Thus $$\ker f=\{0,\pm N,\pm 2N,\cdots\}=N\mathbb Z.$$ Hence $$G\cong\mathbb Z/N\mathbb Z\cong\mathbb Z_N.$$ If $G$ is an infinite cyclic group, $\ker f=\{0\}$. Hence, $$G\cong\mathbb Z/\{0\}\cong\mathbb Z.$$

Lemma 3. Let $G$ be a free abelian group of rank $r$, and let $H$ be a subgroup of $G$. Then one may always choose $p$ generators $x_1,\cdots,x_p$ out of $r$ generators of $G$ so that $k_1x_1,\cdots,k_px_p$ generate $H$. Hence, $$H\cong k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z$$ and $H$ is of rank $p$.

Theorem 4 [Fundamental Theorem of Finitely Generated Abelian Groups] Let $G$ be a finitely generated abelian group with $m$ generators. Then $$G\cong\stackrel{r}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus \mathbb Z_{k_1}\oplus\cdots\oplus\mathbb Z_{k_p}$$ where $m=r+p$. The number $r$ is called the rank of $G$.

Proof. Let $G=\langle x_1, \cdots,x_m\rangle$ and let $f: \mathbb Z\oplus\cdots\oplus\mathbb Z\longrightarrow G$ be the surjective homomorphism $$f(n_1,\cdots,n_m)=n_1x_1+\cdots +n_mx_m.$$ Then by Fundamental Theorem of Homomorphism $$\mathbb Z\oplus\cdots\oplus\mathbb Z/\ker f\cong G.$$ $\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}$ is a free abelian group of rank $m$ and $\ker f$ is a subgroup of $\mathbb Z\oplus\cdots\oplus\mathbb Z$, so by Lemma 3 $$\ker f\cong k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z.$$ Define $\varphi:\stackrel{p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus \cdots\oplus k_p\mathbb Z\longrightarrow\mathbbÂ Z/k_1\mathbb Z\oplus\cdots\oplus\mathbb Z/k_p\mathbb Z$ by $$\varphi((n_1,\cdots,n_p)+k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z)=(n_1+k_1\mathbb Z,\cdots,n_p+k_p\mathbb Z).$$ Then $$\stackrel{p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z\stackrel{\varphi}{\cong}\mathbb Z/k_1\mathbb Z\oplus\cdots\oplus\mathbb Z/k_p\mathbb Z.$$ Hence, \begin{eqnarray*}G&\cong&\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/\ker f\\&\cong&\stackrel{m}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}/k_1\mathbb Z\oplus\cdots\oplus k_p\mathbb Z\\&\cong&\stackrel{m-p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus\mathbb Z/k_1\mathbb Z\oplus\cdots\oplus Z/k_p\mathbb Z\\&\cong&\stackrel{m-p}{\overbrace{\mathbb Z\oplus\cdots\oplus\mathbb Z}}\oplus\mathbb Z_{k_1}\oplus\cdots\oplus\mathbb Z_{k_p}.\end{eqnarray*}