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If $h_i:A_i\to A_{i+1}$ is a countable chain of morphisms in an abelian category $A$ that is AB3 then one can consider the (Bökstedt-Neeman) homotopy colimit of $A_i$ in $D^b(A)$. This is a two-term complex $\coprod A_i\stackrel{f}{\to} \coprod A_i$ with the corresponding "components" of $f$ being $id_{A_i}$, $-h_i$, and $0$, respectively. Clearly, the cokernel of $f$ is the colimit of $A_i$. My question is: which conditions can ensure that $f$ is a monomorphism?

$f$ is a colimit of monomorphisms; so it is a monomorphism if $A$ is AB5. Is $f$ monomorphic for $A$ being a general AB3 category (what about AB4?:))? Are there any alternative methods to study the kernel of $f$? Are there any references for this problem?

Upd. So, a nice example of prof. Rickard demonstrates that AB4 is not sufficient. However, I am (still) interested in "non-standard axioms" for $A$ that would ensure the injectivity of $f$.

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$\textrm{AB4}$ (and a fortiori $\textrm{AB3}$) is not enough, as it's not true for the opposite category of a module category, which is $\textrm{AB4}$.

Let $R$ be any ring, and consider the inverse system $$\dots\to R[x]\stackrel{\theta}{\to}R[x]\stackrel{\theta}{\to}R[x]\stackrel{\theta}{\to}R[x],$$ of $R$-modules, where the maps $\theta$ are multiplication by $x$.

The map $$\prod R[X]\stackrel{\text{id}-{\theta}}{\longrightarrow}\prod R[x]$$ is not surjective, as $(1,1,1,\dots)$ is not in the image.

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