What is a 1-chain with boundary zero in graph theory mean?

Question

I have attached an image from a page of Frank Harary's book. I don't understand what he meant by 1-chain with a boundary of zero. From what I understand, a 1-chain is a sum of edges, but the boundary operator δ was defined to be linear and if $x=uv$, then $\delta x=u+v$ where $u$ and $v$ are vertices so I can't connect the two concepts because how can two vertices add up to zero? Please define and recommend literature that dives deeper into those concepts. thank you]1

Answer 1

The boundary of an edge $x=uv$ is $u+v$, but this sum is evaluated in the vector space of formal $\mathbb{F}_2$-linear combinations of the vertices. In particular, for instance, if $x=uv$, $y=vw$, and $z=uw$, then the boundary of $x+y+z$ is $(u+v)+(v+w)+(u+w)=2u+2v+2w=0$ because $2=0$ in $\mathbb{F}_2$. More generally, whenever you have two copies of the same vertex, they cancel each other out. So if each vertex appears an even number of times in all the edges of your 1-chain, its boundary will be $0$.