```Date: Jan 27, 2013 7:52 AM
Author: Maury Barbato
Subject: Commutative Diagrams

Hello,it is usually stated (see e.g. the introduction of Lang'salgebra) that a diagram commutes if every triangle orsquare of which it is made commutes. I looked for a general proof of this, but I didn't findit even in Bourbaki' works.The statement can be rigorously formulated in the followingterms. Let G be a directed planar graph made up of polygons.Let us suppose that each of such polygons has thefollowing property: (P) there exists two verteces A, B of the polygon suchthat the directed edges of the polygon define two oppositepaths, one from A to B, and the other from B to A (eventually A=B, and in this case all the edges forma clockwise or anticlockwise cycle).Now associate to every vertex a set, and to everydirected edge a map in the usual way. We define in such away a diagram D associated to G. As usual, wesay that D commutes if for every twopaths p_1,p_2,...,p_n and r_1,...,r_m beginning at the same vertex A and ending in the same vertex B,if f_i is the map associated to p_i, and g_i is themap associated to r_i, we havef_n ° f_{n-1} ... ° f_1 = g_m ° ... g_1 ,(if A=B, and p_1,..,p_n is a cycle starting andending in A, we require thatf_n°f_{n-1}°...f_1 = id).The usual statement can be expressed in the followingterms.The diagram D commutes if and only if every polygonin it commutes.Do you know a proof of this statement?Thank you very very much for your attention.My Best Regards,Maurizio Barbato
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