Is there a procedure to find the coefficients of the boundary operator I'm reading the second volume of A course in mathematics for students of physics, by Sternberg and Bamberg. The first section is an exposition of methods from algebraic topology to solve electrical systems. We consider sets of points (nodes of the ensemble) and lines connecting those points (wires) with some orientation, and the vector spaces that are generated from associating vectors to those wires and nodes (for example if there are two points A and B connected by a path $\alpha$ then we associate two different vector spaces, the two dimensional one generated by $(1,0)$ and $(0,1)$ (that is, A and B) and the one dimensional one which is just $(1)$ because there is only one path. We are then introduced to a linear operator $\partial$ which acts on the space of paths to produce a set of two points, that is the boundary of said path. When computing the coefficients of the operator in some basis, the authors usually just look at the image, as far a I can tell. My question is if there isn't some procedure or algorithm that I can use to figure it out between two random vector spaces.  In the example given above, the boundary would obviously be $(-1,1)=B-A$. Basically what I want to know is how do I formulate this operator between two general vector spaces ( I figure they have some extra structure that the drawings fill out, but I'm not sure). Any help would be appreciated, even if you could just reroute me to some material I would be thankful.
 A: If the lines between points are oriented, as you indicate, then the standard convention is that the boundary of a line going from $v$ to $w$ is $w-v$. So you choose a basis for one vector space to correspond with the vertices, a basis for the other to correspond to the oriented paths, and then this formula gives you the boundary.
In more detail: if you have $n$ vertices, then you form a vector space of dimension $n$. If you're comfortable with abstract vector spaces, then use the vertices $v_1, v_2, \dots, v_n$ as the basis. If you prefer to work with $\mathbb{R}^n$, then associate the $i$th standard basis vector $e_i$ with $i$th vertex $v_i$. Do the same with the edges $p_1, p_2, \dots, p_m$. The boundary map is determined by what it does on a basis for this vector space, and if the $j$th basis element corresponds to $p_j$ and if $p_j$ goes from $v_i$ to $v_k$ for some $i$ and $k$, then the boundary map on the $j$th basis element is $e_k - e_i$. If you represent the boundary map as a matrix with one row for each vertex, one column for each path, then every column will have one $1$, one $-1$, and the rest $0$s.
