What conditions on an Abelian category allow members of a direct sum to be determined entirely by their components?

Question

EDIT: In comments, with thanks to Maxime Ramzi, this question has a good answer in that what I want to be true is true when $\mathscr{A}$ satisfies axiom $\mathsf{AB}5$, that $\mathscr{A}$ is cocomplete and filtered colimits are exact. I suppose a new question is: is the truth of my question for $\mathscr{A}$ equivalent to exactness of filtered colimits in $\mathscr{A}$?

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$\newcommand{\A}{\mathscr{A}}\newcommand{\tot}{\operatorname{Tot}^{\oplus}}$Weibel claims that if $\A$ is an Abelian category then, for a bounded below complex (of homological type) $C$ and any Cartan-Eilenberg resolution $Q$ of $C$, the map $\tot Q\to C$ is a quasiisomorphism. This is not too hard to show by a chase. However, as a side remark he mentions that we need $\A$ to be cocomplete to form $\tot Q$ in general - which I of course agree with - but the exact phrasing suggested he also believes $\tot Q\to C$ is a quasiisomorphism in that case. I don't believe this is true, and it was a bit ambiguous for me in the text.

Specifically, the obstruction is this: (letting $C$ be an unbounded complex) diagram chasing needs to be handled with care in Abelian categories. There is the formalism of what Mac Lane calls generalised members and we can use these to do a lot but it is important to remember these are not really elements of sets. The difficulty in the unbounded case is, in showing homological injectivity and demonstrating if a cycle of $\tot Q$ is a boundary in $C$ it must be a boundary in $\tot Q$ (forcing us to actually construct the boundary member):

Suppose two members $x,y\in_m\bigoplus_{i\in J}X_i$ have the "same" components, meaning $\pi_jx\equiv\pi_j y$ for all $j\in J$. Must it follow that $x\equiv y$?

I believe this boils down to the following subquestion:

Under what conditions on $\A$ is it true that if $a,b:Y\to\bigoplus_{i\in J}X_i$ are two arrows with $\pi_ja=\pi_jb$ for all $j$ we have $a=b$?

More technically it would suffice to show this is true for a certain class of $Y$, so long as this class represents all members up to equivalence and is closed under direct sum or something like that. Essentially, I'm asking:

For which Abelian categories is $\bigoplus$ formed by some kind of "finite sum" construction?

I say that because the finite sum construction makes my question hold for categories of modules.

Answer 1

In B. Mitchell. Theory of categories (1965) three properties of abelian categories were introduced.

An abelian category is termed $C_1$ if it is cocomplete with exact coproducts, $C_2$ if it is bicomplete and natural transformation $\eta_{-, X}: \coprod_X (-_x) \to \prod_X (-_x)$ is mono for every set $X$, and $C_3$ if it is cocomplete with exact filtered colimits.

In bicomplete abelian categories the property in the OP is equivalent to $C_2$.

It's obvious that complete $C_3$ category is $C_2$, and $C_2$ always implies $C_1$ (it's in the 3rd chapter of Mitchell's book).

In the paper Exactness of direct limits for abelian categories with an injective cogenerator by L. Positselski and J. Stovicek arXiv one of the results is the following.

Theorem. Let $\mathrm A$ be a complete abelian category with an injective cogenerator $W$. (It is automatically cocomplete by special adjoint functor theorem.) Following conditions are equivalent:

• filtered colimits in $\mathrm A$ are exact;
• $\eta_{W, X}: \coprod_X W \to \prod_X W$ is mono;
• every morphism $f: W^{(X)} \to W$ factors through $\eta_{W, X}$;
• summation morphism $\sigma_{W, X}: W^{(X)} \to W$ factors through $\eta_{W, X}$.

As far as I know, there's no known example without injective cogenerator, $\eta$ being mono and non-exact filtered colimits. If there are no products (and, necessarily, no injective cogenerator), I suspect that this "local finiteness" condition on coproducts behaves quite patologically; it seems very unlikely to imply exactness of filtered colimits. Construction of a cocomplete but not complete abelian category is already quite involved; I haven't checked (yet) whether this construction by J. Rickard serves as a counterexample to the updated question.