DefEqualizer
In any category , given parallel morphisms , an equalizer of and consists of and , universal such that:That is, given with there is a unique with all as in the diagram:
e.g.
In , given functions , their equalizer is the injection into of the equationally defined subset . Since if for some , then for all . Then there is a unique with such that as is monic.
Prop In any category , if is an equalizer of some pair of morphisms, then is monic.
Prop
Prop Given a set , every subset is an equalizer for some pair of functions.
Proof Let set . Then consider the characteristic function :Then we have . So the following is an equalizer: $$$$Moreover, for every function we can form the equational subset: as an equalizer, in the same way.
Prop Write as power set of , we have in .
Proof Inherit the notation from the above proof, clearly we can derive that and are mutually inverse. Then indeed forms an isomorphism. Thus .
Prop In any category , if is an equalizer of some pair of morphisms, then is monic.
Proof Suppose is the equalizer of . Then for all object with morphisms and , we have Hence there exists unique with , that is .
Prop In the category of abelian groups, given , the arbitrary homomorphism forms an equalizer:
Coequalizers
DefCoequalizer
For any parallel morphisms in a category , a coequalizer consists of and , universal with the property , as in That is, given any and , if , then there exists a unique such that .
Prop If is a coequalizer of some pair of morphisms, then is epic.
Proof Observe by duality, we know that such a coequalizer in a category is an equalizer in , hence monic by the proposition, and so is epic in .
e.g.
Let be an equivalence relation on a set with are the two projections of the inclusion . The canonical projection defined by is then a coequalizer:
The coequalizer in of an arbitrary parallel pair of functions can be constructed by quotienting by the equivalence relation generated by the equations for all .
Prop For every monoid there are sets and and a coequalizer diagram, with and free, thus .