let $K= \{\sigma_1...\sigma_n\}$ be a simplicial complex with n simplices with faces of the ith simplex having lower index than i. Let the field of coefficients be $\mathbb{Z}_2$
recall the boundary matrix is defined as:
$\partial[i,j]$= (1 if $\sigma_i$ is in boundary of $\sigma_j$ for $i<j$; 0 otherwise) ;
can some one give a counter example or show that:
the square of the boundary matrix of a simplicial complex K is zero.
certainly if we restrict to a fixed dimension (say 2), then the composition of sub matrices is zero: $d_1 \circ d_2 =0$ where $d_k: K^k \rightarrow K^{k-1}$
To check that $\partial^2=0$, you just have to check that $\partial^2(\sigma_i)=0$ for each $i$. But this is trivial: if $\sigma_i$ has dimension $m$, then $\partial(\sigma_i)$ is a sum of simplices of dimension $m-1$ and so $\partial(\partial(\sigma_i))=0$ since we are in the case of a fixed dimension which you say you understand.