Implication Details
Assumptions: additive, effective congruences
Conclusions: normal
Proof: Let be a monomorphism. Then we define a relation on via with maps defined by and . It is straightforward to check that and are jointly monomorphic. Now is a congruence because for generalized elements , factors through if and only if factors through . In other words, the relation on is exactly , which is an equivalence relation on (and in fact a congruence in ). Now by assumption, is the kernel pair of some morphism ; in other words, factors through if and only if . In particular, for , factors through if and only if factors through , which is equivalent to . We have thus shown that is the kernel of .