Lemma. Basic properties of incidence fields [fdip-00AC]
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  • November 3, 2026

Let \(R\) be a commutative unital ring, and let \(V,E,s,h\) and the field modules, field pullback and ring-valued incidence map be as in the two definitions above. The pointwise operations make both modules commutative unital \(R\)-algebras, the indicator fields are a basis of each, every field pullback preserves the operations, and \(d\) is \(R\)-linear. Call two vertices equivalent when they are equal or a finite sequence of edges has them as successive ends, ignoring orientation. The equivalence classes are the undirected components; a vertex that lies on no edge is a component of its own. Then \(df=0\) if and only if \(f\) is constant on each undirected component.

Proof. 

  • November 3, 2026

Every module and algebra identity follows by evaluation from the corresponding identity in \(R\). For each \(v\in V\), let \(\delta _v(u)\) be \(1\) if \(u=v\) and \(0\) otherwise. Evaluation at \(u\) proves \[f=\sum _{v\in V}f(v)\delta _v,\] and proves uniqueness of the coefficients. The same argument applies to edge fields. On an empty carrier the sum is empty and there is one field. Evaluation also proves every pullback law; for instance \(a^*(fg)(e)=f(a(e))g(a(e))\). Finally \[d(rf+g)(e)=r\bigl (f(h(e))-f(s(e))\bigr )+g(h(e))-g(s(e)),\] which proves linearity. These evaluations apply to the zero ring.

If \(df=0\), then \(f(h(e))=f(s(e))\) for every edge, so \(f\) takes the same value at the successive ends of any finite sequence of edges and is constant on each component. Conversely, the two ends of an edge lie in one component, and the two ends of a loop are the same vertex, so a field constant on each component has \(df=0\). The argument uses no division.