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Question about graph homomorphism

According to the definition:

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 : GG′, consists of two functions f0: VV′ and f1: AA′ …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 ?

 

Exercise 4.1.1.7

Could someone explain Exercise 4.1.1.7?

Find an operation on the set M = \{1, 2, 3, 4\}, i.e., a legitimate function f : M \times M \rightarrow M, such that f cannot be the multiplication formula for a monoid on M. That is, either it is not associative or no element of M can serve as a unit.

Sincerely,
Max