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Let $\mathcal A$ be an additive category. Then there is a free abelian category $F(\mathcal A)$ on $\mathcal A$. I'm aware of two constructions in the literature, and I'd like to relate them.

The second construction is Adelman's. This describes $F(\mathcal A)$ as $\mathcal A^{[2]} / \sim$. So the objects are pairs $A \to B \to C$ of composable morphsims in $\mathcal A$, morphisms $(\alpha, \beta, \gamma) : (A \xrightarrow f B \xrightarrow g C) \to (A' \xrightarrow{f'} B' \xrightarrow{g'} C')$ are represented by triples of maps forming two commutative squares, and the relevant congruence is obtained when we mod out by morphisms of the form $(\delta f, f'\delta, \gamma)$ and morphisms of the form $(\alpha, \epsilon g, g' \epsilon)$ for $\delta : B \to A'$, $\epsilon : C \to B'$.

The first construction is Freyd's. This says that

$$F(\mathcal A) = Fun^\oplus_\omega(Fun^\oplus_\omega(\mathcal A, Ab),Ab).$$

Here $Fun^\oplus_\omega(\mathcal B, \mathcal C)$ is the category of finitely-presentable additive functors $\mathcal B \to \mathcal C$. Without too much fuss, one works out that $Fun^\oplus_\omega(\mathcal B, Ab) = (\mathcal B^{op})^{[1]} / \sim$, where the relevant congruence mods out by those $(\psi, \phi) : (Y' \xleftarrow{f'} X') \leftarrow (Y \xleftarrow{f} X)$ such that $\phi = f'\chi$ for some $\chi : Y' \leftarrow X$. Equivalently, one mods out by pairs of the form $(\chi f, f'\chi)$ as well as pairs of the form $(\psi, 0)$.

Thus, Freyd ends up describing $F(\mathcal A)$ as the quotient by a congruence on $(((\mathcal A^{op})^{[1]})^{op})^{[1]} = \mathcal A^{[1]\times[1]}$, the category of commutative squares in $\mathcal A$.

So that's the funny thing -- Adelman and Freyd are describing $F(\mathcal A)$ as quotients of different categories -- pairs of composable morphisms versus commutative squares. I'd like to resolve the combinatorial discrepancy.

Question: What does the canonical equivalence between Adelman's model for $F(\mathcal A)$ and Freyd's model for $F(\mathcal A)$ do?

For instance, does it lift to some kind of functor between $\mathcal A^{[2]}$ and $\mathcal A^{[1] \times [1]}$?

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  • $\begingroup$ I hope that Freyd and Adelman provide proofs of the universal property? Applied to this case, it will give the equivalence. If their proofs are constructive, this should help. Sorry for the trivial comment, but I am curious if you tried that already. $\endgroup$ Commented Sep 25, 2023 at 16:34

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