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, $c\in\ker\partial_r=Z_r(K)$.