0-chain Boundary

Question

Can anyone explain how adding two vertices in a connected graph to create a $0$-Chain is the boundary of some 1-dimensional chain? I know that the definition of boundary is the collection of $n+1$ faces, and each face is an $n-1$ simplice, but can someone explain the connection?

Answer 1

Yes. Suppose the two vertices are $u$ and $w$. Since the graph is connected, then there is a path $$u=v_0,v_1,v_2,\ldots,v_{k-2},v_{k-1},v_k=w$$ of vertices in the graph, connecting vertex $u$ to $w$, such that an edge exists in the graph between $v_i$ and $v_{i+1}$ for all $i$. Hence, $$[v_0,v_1]+[v_1,v_2]+\ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k]$$ is a 1-chain. Its boundary, with $\mathbb{Z}/2\mathbb{Z}$ coefficients, is \begin{align} &\partial([v_0,v_1]+[v_1,v_2]+\ldots+[v_{k-2},v_{k-1}]+[v_{k-1},v_k])\\ =&\partial[v_0,v_1]+\partial[v_1,v_2]+\ldots+\partial[v_{k-2},v_{k-1}]+\partial[v_{k-1},v_k]\\ =&(v_0+v_1)+(v_1+v_2)+\ldots+(v_{k-2}+v_{k-1})+(v_{k-1}+v_k)\\ =&v_0+2v_1+2v_1+\ldots+2v_{k-2}+2v_{k-1}+v_k\\ =&v_0+v_k \end{align} since we are using $\mathbb{Z}/2\mathbb{Z}$ coefficients. So we have found a 1-chain whose boundary is $v_0+v_k=u+w$.