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Let $A$ be an AB3 abelian category. We will say that an object $M$ of $A$ is uniquely divisible if for any integer $n\neq 0$ the endomorphism $nid_M$ is invertible. We will say that $M'$ is ind-torsion if it belongs to the smallest Serre subcategory of $A$ that contains all torsion objects (that is, those $N\in A$ for which there exists $n\neq 0$ such that $nid_N=0$) and is closed with respect to $A$-coproducts.

My question is: which assumptions of $A$ are sufficient to ensure that there do not exist non-zero monomorphisms from a uniquely divisible $M$ into an ind-torsion $A'$? This statement is obvious for abelian groups, but I do not know how to generalize it further.

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If your category $A$ is AB5, the answer is positive, because the ind-torsion objects are only the objects $X$ of $A$ such that the canonical map from the colimit $_\infty X$ of $_n X$ to $X$ is an isomorphism, where $_n X$ is the kernel of multiplication by $n$ on $X$ and the (filtered) colimit is taken over positive integers $n$ ordered by divisibility. (The only thing to check is stability under extensions of this class of objects; thanks to AB5 the proof is the same as for abelian groups.) If $M$ is such that $n.Id_M$ is a mono for each integer $n>0$ (it is true for any subobject of a uniquely divisible object), then $M\to\, _n X$ is always zero, so $M\to\,_\infty X$ is always zero because $A$ is AB5.

I guess that AB4 is not enough.

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  • $\begingroup$ Sorry; after we dualize we will probably obtain an epimorphism from a uniquely divisible object and not a monomorphism; won't we? $\endgroup$
    – Mikhail Bondarko
    Commented May 18, 2020 at 7:13
  • $\begingroup$ Thanks for correcting me; I have corrected this. $\endgroup$
    – Aurélien Djament
    Commented May 18, 2020 at 8:01
  • $\begingroup$ Doesn't the map go in the wrong direction now? In the opposite category of abelian groups you have described a monomorphism from an ind-torsion object to a uniquely divisible one. However, $\mathbb{Q}$ is both ind-torsion in the opposite category of abelian groups and uniquely divisible, so $\text{id}_{\mathbb{Q}}$ works. $\endgroup$
    – Jeremy Rickard
    Commented May 18, 2020 at 9:36
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    $\begingroup$ The following should work for a counterexample in the opposite category of abelian groups: $\mathbb{Q}$ is a direct summand, so in particular a quotient, of a countable product of copies of $\mathbb{Q}/\mathbb{Z}$. I hope that now nothing is wrong with variances! $\endgroup$
    – Aurélien Djament
    Commented May 18, 2020 at 9:47
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    $\begingroup$ Yes, that works. In fact, in the opposite category of abelian groups, every object is ind-torsion. $\endgroup$
    – Jeremy Rickard
    Commented May 18, 2020 at 10:16

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