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.

  • $\begingroup$ I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects. $\endgroup$ – Angina Seng Dec 3 '18 at 5:01
  • $\begingroup$ @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. $\endgroup$ – HARRY Dec 3 '18 at 5:11
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    $\begingroup$ $G^I$ is a weird notation for a possibly infinite coproduct $\endgroup$ – Maxime Ramzi Dec 3 '18 at 8:11

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.

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  • $\begingroup$ 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 ? $\endgroup$ – Maxime Ramzi Dec 3 '18 at 8:12
  • $\begingroup$ @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. $\endgroup$ – HARRY Dec 3 '18 at 8:40
  • $\begingroup$ @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) $\endgroup$ – Maxime Ramzi Dec 3 '18 at 8:50

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