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

it is usually stated (see e.g. the introduction of Lang's
algebra) that a diagram commutes if every triangle or
square of which it is made commutes.
I looked for a general proof of this, but I didn't find
it even in Bourbaki' works.
The statement can be rigorously formulated in the following
terms. Let G be a directed planar graph made up of polygons.
Let us suppose that each of such polygons has the
following property:

(P) there exists two verteces A, B of the polygon such
that the directed edges of the polygon define two opposite
paths, one from A to B, and the other from B to A
(eventually A=B, and in this case all the edges form
a clockwise or anticlockwise cycle).

Now associate to every vertex a set, and to every
directed edge a map in the usual way. We define in such a
way a diagram D associated to G. As usual, we
say that D commutes if for every two
paths 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 the
map associated to r_i, we have

f_n ° f_{n-1} ... ° f_1 = g_m ° ... g_1 ,

(if A=B, and p_1,..,p_n is a cycle starting and
ending in A, we require that
f_n°f_{n-1}°...f_1 = id).

The usual statement can be expressed in the following

The diagram D commutes if and only if every polygon
in it commutes.

Do you know a proof of this statement?

Thank you very very much for your attention.
My Best Regards,
Maurizio Barbato