Proof that grupoid G = G(*) has only trivial congruencies <=> For every grupoid G' and every homomorphism f: G -> G' applies that f is injective or |f(G)| = 1.
My solution is the following:
(=>) So we have only trivial congruencies on G and we want to proof that every homomorphism f: G -> G' is either injective or |f(G)| = 1
I am using a fact that (a CONGRUENT b) <=> (f(a) = f(b))
According to this theorem all those congruencies belong to ker(f). While having only trivial congurencies this means identity and GxG, so we proof this for those two separately.
id.: (a,a) belong to ker(f) (b,b) belong to ker(f) etc.
Every element from G is going to itself on G' (otherwise there would be a conflict with the definition of kernel. There would have to exist (a,b) belonging to ker(f), but that is not possible since we have an identity) So its injective. - OK
GxG: (a,a) belong to ker(f) (a,b) belong to ker(f) etc.
We use this fact about kernels ker(f) = {(a, b) belonging to G | f(a) = f(b)}
This means that every element is going to transform into one common element ( (a, a) : f(a) = f(a) = z (a, b) : f(a) = f(b) = z (a, c) : f(a) = f(c) = z (b, a) : f(b) = f(a) = z (b, b) : f(b) = f(b) = z ... )
Ergo: |f(G)| = 1 - OK
Now from the other side
(<=) So we say that for every grupoid G' and every homomorphism f: G -> G' for which applies that f is injective or |f(G)| = 1 applies that G must have only trivial congruencies.
For conflict we assume that G has a nontrivial congruency (such that its not either identity nor GxG). Then there must also exist (a,b) belonging to a wanted congruency and therefore: (a,b) belongs to ker(f) and f(a) = f(b). Which is in conflict with the fact that f should be injective.
For |f(G)| = 1, lets just take another pair (c,d) that may be in this nontrivial congruency. ( a != b != c != d, and there is no congruency relation betweet ab and cd). So now (c,d) also belongs to ker(f). But that would mean that |f(G)| != 1 which is also in conflict with presumption that |f(G)| = 1.
Therefore G has to contain trivial congruencies only.
Is it OK? Also I am using theorem which is defined for groups and not grupoids. (a CONGRUENT b) <=> (f(a) = f(b)) Can it be easily extended for this case or should a different aproach be taken? (if so, please show)