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$\newcommand\Ab{\mathrm{Ab}}\newcommand\ab{\mathrm{ab}}\DeclareMathOperator\Ch{Ch}\DeclareMathOperator\Kom{Kom}\newcommand\ho{\mathrm{ho}}$Let $\Ab$ be the category of finitely generated abelian groups, $\ab$ be the subcategory with objects being the free ones, and $\Ch(\Ab)$ be the category of chain complexes over $\Ab$.

Finally, define $\Kom(\Ab) := \Ch(\Ab)/\ho$ to be the category whose objects are the same as those of $\Ch(\Ab)$, with morphisms space:

$$\Kom(\Ab)[X,Y] := \Ch(\Ab)[X,Y] \,/\, \{\text{chain-homotopy}\}.$$

I aim to tell if any two objects $X$ and $Y$ are isomorphic in $\Kom(\Ab)$.

One naïve invariant is the homology $H(-)$. Namely, if $H(X) \not \simeq H(Y)$ then $X \not\simeq Y$ in $\Kom(\Ab)$. However, this invariant is known to be "too weak" by algebraic topologists and algebraic geometers. In the context of algebraic topology, $X$ and $Y$ often are simplicial cochain complexes of some actual topological space $\tilde{X}$ and $\tilde{Y}$. There, one can study extra structures (e.g. cup products, Steenrod operations) on the homology $X$ and $Y$ to have a better chance to tell $X$ and $Y$ apart.

Question

Nevertheless, not all $X$ and $Y$ comes from an actual topological space. I wonder if there are invariants (other than homology) for objects in $\Kom(\Ab)$ (or $\Kom(\ab)$ if it is any easier).

Remarks

The same question for derived category is "trivial".

If classifying isomorphism classes in $\Kom$ turns out to be too hard, one may ask the same question for the derived categories $D(\Ab)$. Thanks to Fernando Muro's hint in the comment, I now know that taking homology is a complete invariant in our case. Essentially, this is because $\Ab$ is a hereditary category, i.e. a category $\mathcal{A}$ such that $\operatorname{Ext}_{\mathcal{A}}^2(-,-) = 0$ [3, section 1.3]. In abelian and hereditary categories $\mathcal{A}$, each bounded $X \in D(\mathcal{A})$ is isomorphic to $H(X)$ (see [3, Theorem 2.1] for a clear proof; a proof for unbounded $X$ is given in [4], but I found it confusing), and the converse is true [2].

For more general, non-hereditary categories $\mathcal{A}$, I wish to delegate the discussion to another thread "Invariants of objects in $D(\mathcal{A})$ for non-hereditary category $\mathcal{A}$" on MSE.

Related

Dror Bar-Natan's proof for that a certain chain complex (over a certain preadditive category) up to chain homotopy is an invariant of a tangle. [1]

Reference

[1] Khovanov's homology for tangles and cobordisms-[Dror Bar-Natan]

[2] Hanno's answer on "If the cohomology of two objects in the derived category are equal, are the objects isomorphic?"

[3] Hereditary Categories. Lectures 1 and 2 - Helmut Lenzing

[4] Henning Krause’s “Derived categories, resolutions, and Brown representability”

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    $\begingroup$ If you considered the derived category rather than the homotopy category, homology would suffice. The homotopy category is more complicated. $\endgroup$
    – Fernando Muro
    Commented Nov 27, 2022 at 20:18
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    $\begingroup$ It is not hard to see that every chain complex of free abelian groups splits as a direct sum of $2$-term chain complexes. From this one sees that in your $\operatorname{Kom}(ab)$ two objects are isomorphic iff their homologies are isomorphic. $\endgroup$
    – Achim Krause
    Commented Nov 27, 2022 at 21:33
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    $\begingroup$ Please, do not change standard notation, since this is a source of confusion. $\mathbf{Ab}$ (or $Ab$) is the category of all abelian groups. If you want to consider the subcategory of f.g. abelian groups, you can name it $\mathbf{Ab}_{\text{fg}}$ for example. Small letters such as $\mathbf{ab}$ actually often indicate finite versions, so again it's a bad idea for this to be the category of free abelian groups. Better write $\mathbf{Ab}_{\text{free}}$ (for example). Also, $K(\mathcal{A})$ is the established notation for the homotopy category of $\mathcal{A}$. $\endgroup$
    – Martin Brandenburg
    Commented Nov 27, 2022 at 22:25
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    $\begingroup$ @Student There's no hope to classify objects in the homotopy category of chain complexes of abelian groups, I believe. I think this would contain the classification of pairs $(A,B)$ where $A$ is an abelian group and $B\subset A$ is a subgroup. This is a wild problem of representation theory. $\endgroup$
    – Fernando Muro
    Commented Nov 28, 2022 at 0:08
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    $\begingroup$ @MartinBrandenburg Perhaps we should not be discouraging students from asking reasonable questions here by chastising them too much for not being familiar with our preferred notation, especially when they do such a clear job of establishing their notation for the questions. (And for what it's worth, I've certainly seen "Kom" before.) $\endgroup$
    – Greg Friedman
    Commented Nov 28, 2022 at 4:20

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