# Definition of generator in an abelian category.

Let $$\mathcal A$$ be an abelian category. Let an object $$G$$ in $$\mathcal A$$ be such that $$Hom\left(G,\unicode{x2013} \right)$$ is a faithful functor from $$\mathcal A$$ to the category of sets.(I assume that I am working with a category where any family of objects has their coproduct existing inside the category)

Why is the above equivalent to the fact that every object $$X$$ in $$\mathcal A$$ admits an epimorphism $$G^I$$ to $$X$$? (where $$I$$ is the index category and is arbitrary)(where,$$G^I$$ is coproduct of copies of $$G$$ which exists in that category)

I do not see how injectivity of maps getting translated to epimorphism and vice versa? I guess I should start with appropriate short exact sequence and apply appropriate functor to make injectivity translated to epimorphism but I have no clue how to proceed practically, i.e how to make it precise.

Any help from anyone is welcome.

• I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects. – Lord Shark the Unknown Dec 3 '18 at 5:01
• @lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category. – HARRY Dec 3 '18 at 5:11
• $G^I$ is a weird notation for a possibly infinite coproduct – Max Dec 3 '18 at 8:11

## 1 Answer

Suppose there is an epimorphism $$p:G^I\to X$$ and suppose $$f:X\to Y$$ is a morphism which becomes $$0$$ after applying the functor $$\operatorname{Hom}(G,-)$$. This means that for every morphism $$g:G\to X$$, $$fg=0$$. In particular, by taking $$g$$ to be each of the inclusion maps $$G\to G^I$$ composed with $$p$$, this implies $$fp=0$$. Since $$p$$ is an epimorphism, this implies $$f=0$$. Thus if such an epimorphism $$p$$ exists for every $$X$$, $$\operatorname{Hom}(G,-)$$ is faithful.

Conversely, suppose $$\operatorname{Hom}(G,-)$$ is faithful and let $$X$$ be an object. Let $$I=\operatorname{Hom}(G,X)$$ and let $$p:G^I\to X$$ be the unique morphism such that for each $$i\in I$$, the composition of $$p$$ with the $$i$$th inclusion map $$G\to G^I$$ is $$i:G\to X$$. I claim $$p$$ is an epimorphism. To show this, it suffices to show that if $$f:X\to Y$$ is a morphism such that $$fp=0$$, then $$f=0$$. But given any such $$f$$, by composing with the inclusion maps $$G\to G^I$$ we see that $$fi=0$$ for all $$i:G\to X$$. This means that $$\operatorname{Hom}(G,-)$$ sends $$f$$ to $$0$$ and thus $$f=0$$ since $$\operatorname{Hom}(G,-)$$ is faithful.

• I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ? – Max Dec 3 '18 at 8:12
• @Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for. – HARRY Dec 3 '18 at 8:40
• @HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake) – Max Dec 3 '18 at 8:50