TITLE: Showing that a diagram commutes in the most economical way
QUESTION [1 upvotes]: Suppose that one had to consider (co)cones on a complicated diagram, with many arrows and objects and that one wished to prove that one of them is final/initial.
Given another (co)cone, one would construct a unique morphism such that the whole diagram commutes. Of course, one could always check commutativity for all "subdiagrams", but I assume this is far from necessary in most cases.
For example, for (co)cartesian squares, one only needs to show that two triangles commute.
In general, given a diagram, how would one determine the smallest number of equalities to check in order to show that it commutes?

REPLY [0 votes]: In general, without making further assumptions on the category in question, there are no shortcuts: take any 2 objects then any two connecting paths must be equal.
Nevertheless...you can somewhat optimize this procedure by considering first pairs of objects joint by multiple - non intersecting - paths. Non intersecting paths are paths which have only the starting and end objects in common, and nothing in between.
When you have k non intersecting paths joining 2 objects, then you need k-1 eqs. Once you have considered all such pairs of objects joint by multiple - non intersecting - paths, all other pairs are joint by intersecting paths - whose subpaths have been already considered and equated - plus maybe other non intersecting paths. You then simply add one more equation for each of these extra non intersecting path, and you are done.
In short: start with the shortest parallel paths first.
I hope I have been clear enough, otherwise I will draw a picture.
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