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Maury Barbato
Posts:
792
From:
University Federico II of Naples
Registered:
3/15/05


Commutative Diagrams
Posted:
Jan 27, 2013 7:52 AM


Hello, 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_{n1} ... ° 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_{n1}°...f_1 = id).
The usual statement can be expressed in the following terms.
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



