I've read somewhere that for a categorical diagram to commute, it is enough that all its polygonal subdiagrams commute.

I want a reference and a detailed proof of this.

Please also give a formal definition of polygonal subdiagrams.

  • $\begingroup$ Specifically pentagonal, or just polygonal? $\endgroup$ – Unit Jun 14 '15 at 14:39
  • $\begingroup$ @Unit "pentagonal" was a typo. Corrected $\endgroup$ – porton Jun 14 '15 at 14:43
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    $\begingroup$ Could you be more specific? It is true that in a monoidal category, if certain "basic" diagrams commute, then so do other more complicated diagrams constructed from them. For details see here $\endgroup$ – Matematleta Jun 14 '15 at 14:57
  • $\begingroup$ @Chilango I don't believe this is the point at all. It's more like this: in a diagram like this, it's enough that each square commute for the whole diagram to commute. And for a square to commute it's enough that both triangles commute. And so on. $\endgroup$ – Najib Idrissi Jun 15 '15 at 8:05
  • $\begingroup$ @porton You have read "somewhere"; where, exactly? And what is your precise definition, for a categorical diagram, of "to commute"? The answer will depend on the definition you have (because one could very well take "all polygonal subdiagrams commute" as a definition...). $\endgroup$ – Najib Idrissi Jun 15 '15 at 8:08

I myself have been recently looking for a citation and a proof, but here (I think) is the idea behind it.

First, to clear up the question, I will assume you are referencing the brief tidbit mentioned in the commutative diagram page on Wikipedia under "verifying commutativity" (https://en.m.wikipedia.org/wiki/Commutative_diagram). This means you are talking about a result concerning diagrams and commutativity generally, not just ones representing categories (at least conceptually). In a hand-wavey way, we assume each of the planar faces of subdiagrams commute and want to show the whole diagram commutes.

So assume all the polygonal subdiagrams in a diagram commute. Formally I think you should proceed by induction on the number of vertices. The base case is trivial. For the inductive step you pick any two vertices and show any two paths to those two vertices are the same. You can do this since any proper subpath is the same route by inductive hypothesis, extending a path and showing it's the same is fine, and paths which are equal to the same path are equal to one another.

Try working through a diagram with only four vertices and some triangles; the general procedure should become clear.


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