Simplicial chain complex ($C_{\bullet }(K)$)

The simplicial chain group $C_n(K)$ is freely generated by oriented $n$-simplices, subject to reversal of orientation changing sign. Its boundary is

For an abstract simplicial complex $K$, the integer simplicial chain group $C_n(K;\mathbb{Z})$ is the free abelian group on its oriented $n$-simplexes, with reversal of orientation changing sign. Thus a chain is a finite formal integer sum of oriented simplices. After choosing one orientation for each simplex, those chosen simplices form a basis of this abelian group.

For a filled triangle with vertices $0,1,2$, $C_0\cong {\mathbb{Z}}^3$, $C_1\cong {\mathbb{Z}}^3$ and $C_2\cong \mathbb{Z}$. Its oriented two-simplex has boundaryTaking another boundary gives $([2]-[1])-([2]-[0])+([1]-[0])=0$. This cancellation makes the simplicial chain complex possible: boundaries are cycles, and simplicial homology measures cycles modulo boundaries. With coefficients in an abelian group $A$, use $C_n(K;\mathbb{Z}){\otimes }_{\mathbb{Z}}A$.^[ref]

A simplicial $q$-chain is a finite formal linear combination of oriented $q$-simplices of a simplicial complex, with coefficients in a specified abelian group or ring. With integer coefficients,where reversing a simplex's orientation changes its sign. The chains form the simplicial chain group. In characteristic two the sign distinction disappears.

The boundary operator is extended linearly fromFor a filled triangle, $\partial [0,1,2]=[1,2]-[0,2]+[0,1]$. Applying the boundary again gives $(2-1)-(2-0)+(1-0)=0$, where each vertex label denotes its basis zero-chain. The cancellation exemplifies ${\partial }^2=0$.

A chain with zero boundary is a simplicial cycle; a chain that is the boundary of a higher chain is a chain boundary. Their quotient is the relevant simplicial homology group.

In an augmented chain complex of a simplex with vertices $v_0,\dots ,v_n$, put $s(1)=[v_0]$ and $s[v_{i_0},\dots ,v_{i_q}]=[v_0,v_{i_0},\dots ,v_{i_q}]$ when $v_0$ is absent, and zero when it is present. Direct expansion of the boundary operator gives $ds+sd=1$. This proves that every positive-degree chain cycle is a chain boundary, without invoking cellular homology.