A graph (V, A, src, tgt) involves two sets and two functions. For two graphs to be comparable, their two sets and their two functions should be appropriately comparable. Let G = (V, A, src, tgt) and G′ = (V′, A′, src′, tgt′) be graphs. A graph homomorphism f from G to G′, denoted f : G → G′, consists of two functions f0: V → V′ and f1: A → A′ …the rest…
So this means that not all elements of A’ need to have a mapping from A and not all elements of V’ need to have a mapping from V. For example if G has m vertices and n edges and G’ has m'(>1) vertices and n'(>1) edges we could have all vertices of G go to one vertex of G'(say a) and all edges of G go to an edge from a to a(assuming that exists) in G’. That would be a valid homomorphism. Am I understanding that correctly ?
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Some will notice that we’ve slowed down a tad to accommodate the peleton who have been doing the diligence to keep up. Hopefully the last week “off” for the July 4th holiday in the states will have let everyone catch up a bit. The syllabus will be changing to meet our needs as we proceed.
For those with some general background in Category Theory, we’ll be getting into the serious material shortly.