This might seem like a very vague question, but the details are really confusing me. So, for example, say we are studying the category of $A$-modules $\mathsf{Mod}_A$ where $A$ is a commutative unital ring.
At first, I was confused how $\mathsf{Mod}_A$ is an abelian category, as in the hom-sets having an abelian group structure. Sure, we know that for any $A$-module homomorphisms $\phi,\psi: M \to N$, we can add them to get another homomorphism $\phi + \psi$ in a natural way. But thinking categorically, doesn't $\mathsf{Mod}_A$ contain absolutely no information about the elements of modules? To define $\phi + \psi$ we must define in terms of elements, and I am not aware of how to construct some $\phi + \psi$ categorically. So, from what I have understood, the 'abelian category $\mathsf{Mod}_A$' contains not only information about $\mathsf{Mod}_A$, but additionally abelian group structures for the hom-sets $\operatorname{Hom}(M,N)$ for every object $M,N$. I guess this is related to something called an 'enriched category', but I have no knowledge on monoidal categories, so I would like an explanation in layman's terms.
Now, another fact about $\mathsf{Mod}_A$ is that we can equip it with a faithful functor $U : \mathsf{Mod}_A \to \mathsf{Set}$ and consider free objects, i.e. free modules. However I am still confused about the 'equip with a functor' part, since if we think in terms of elements, it seems quite obvious what $U$ should be (and I think everyone assumes $U$ to be contained in the data of $\mathsf{Mod}_A$).
So, to sum up, my question is about how much information a category has on itself, in this case $\mathsf{Mod}_A$. Do we have to add information whenever we are considering some specific construction on categories? If this is the case, then how can we distinguish the 'natural' $\mathsf{Mod}_A$ (where additional information is implicitly derived from our knowledge of modules) apart from an 'exotic' $\mathsf{Mod}_A$ equipped with some other, say, faithful functor to $\mathsf{Set}$? It seems like many authors assume categories like $\mathsf{Mod}_A$ to be equipped with every information derived from $A$-modules, but I don't think this is how categories work.
Best Answer
It indeed looks like you need extra information to answer the question whether a given category $\mathcal{C}$ is an abelian category, but this is actually not true. An abelian category is an additive category satisfying some extra properties, i.e. no extra structure is imposed once we have verified $\mathcal{C}$ is an additive category. A category $\mathcal{C}$ is additive if it is pre-additive and if finite coproducts and products coincide. The latter is again a property and not structure, because it asks for all objects $x,y\in\mathcal{C}$ for the canonical comparison map $x\sqcup y\to x\times y$ to be an isomorphism. However, in general, a category being a pre-additive category is not a property, but extra data: it is the data of an enrichment over abelian groups, informally a well-behaved choice of abelian group structure on each hom-set of your category. In this light, it looks like being an additive category is extra structure.
This is not true: the point is that, if your category $\mathcal{C}$ has biproducts (i.e. finite coproducts and products coincide), then any pre-additive category structure is unique, if $\mathcal{C}$ admits one in the first place. It is defined as follows: given $x,y\in\mathcal{C}$, the group law of the hom-set $\mathcal{C}(x,y)$ is given by $\mathcal{C}(x,y)\times\mathcal{C}(x,y)\cong\mathcal{C}(x,y\times y)\cong\mathcal{C}(x,y\sqcup y)\xrightarrow{\mathrm{fold}_*}\mathcal{C}(x,y)$, where $\mathrm{fold}\colon y\sqcup y\to y$ is the codiagonal map (given by the identity on $y$ on each component of the coproduct), and all the other isomorphisms are canonical maps. You can show that any other enrichment of $\mathcal{C}$ over abelian groups is necessarily equal to this one, via an Eckmann-Hilton argument if I am not mistaken.
In fact, an alternative definition of an additive category is that it is a semi-additive category in which the shear map $(\mathrm{pr}_1,\mathrm{fold})\colon x\oplus x\to x\oplus x$ is an isomorphism, where $x\oplus x$ denotes the biproduct (i.e. both product and coproduct) of $x$ with itself. Being a semi-additive category is again just a property, as you can see when you look up the definition. This definition as a semi-additive category with an extra property is equivalent to the one as a pre-additive category with an extra property.
All in all, being an abelian category does not require in advance the extra choice of group structure on all the hom sets, because if your category has biproducts (a property you need to check regardless), then there is only one possible multiplication map possible on the hom-sets, and then you only need to check that this is in fact a group law.
Edit: The following paragraph is not true in its entirity, but I am not quite sure where the mistake is. It's just that the conclusion is too strong to be completely true. I will keep it in here because it may contain interesting information, but be critical of it while reading.
There is a way to construct the forgetful functor $U\colon\mathsf{Mod}_A\to\mathsf{Set}$ without particular knowledge about modules themselves, in the sense that you do not need to know what modules are, and just need to know what the category of modules looks like. In general, you could make the argument that since $\mathsf{Mod}_A$ is equivalent to the category $\mathrm{Fun}^\times(\mathcal{T},\mathsf{Set})$ of finite-product preserving functors from the Lawvere theory $\mathcal{T}$ encoding $A$-modules into $\mathsf{Set}$, the forgetful functor just arises by restricting along the generating object $t\in\mathcal{T}$ that any Lawvere theory has. However, I am not sure if both $\mathcal{T}$ and the equivalence between $\mathsf{Mod}_A$ and $\mathrm{Fun}^\times(\mathcal{T},\mathsf{Set})$ are unique enough to then determine this forgetful functor $\mathsf{Mod}_A\to\mathsf{Set}$ up to equivalence. We can save the situation however: there is a canonical choice for $\mathcal{T}$, namely the finitely generated projective $A$-modules, with $t$ given by $A$ itself. So one way in which you can make the forgetful functor $\mathsf{Mod}_A\to\mathsf{Set}$ a consequence of categorical structure (and not of particular knowledge of modules) is as follows: given a general category $\mathcal{C}$ with finite products, we ask ourselves if there is an equivalence $\mathcal{C}\simeq\mathrm{Fun}^\times(\mathcal{T}^\mathrm{op}_\mathcal{C},\mathsf{Set})$, where $\mathcal{T}_\mathcal{C}$ is the full subcategory of $\mathcal{C}$ on compact projective objects, and where we require the equivalence to commute with the inclusion $\mathcal{T}_\mathcal{C}\to\mathcal{C}$ on the one side and the Yoneda embedding $\mathcal{T}_\mathcal{C}\to\mathrm{Fun}^\times(\mathcal{T}^\mathrm{op}_\mathcal{C},\mathsf{Set})$ on the other side. This equivalence, if it exists, is essentially unique. Now, $\mathcal{T}_\mathcal{C}$ can satisfy the property that its objets are generated under coproducts by a single object $t\in\mathcal{T}_\mathcal{C}$. This object $t$ is again essentially unique. If these two properties are satisfied, we can call a category $\mathcal{C}$ good. Now any good category $\mathcal{C}$ gives us a forgetful functor $\mathcal{C}\simeq\mathrm{Fun}^\times(\mathcal{T}^\mathrm{op}_\mathcal{C},\mathsf{Set})\xrightarrow{\mathrm{evaluation}\;\mathrm{at}\;t}\mathsf{Set}$, which, if $\mathcal{C}$ satisfies the property that it is good, is uniquely given by the pre-existing categorical structure of $\mathcal{C}$.
This is one way to construct the usual forgetful functor $\mathsf{Mod}_A\to\mathsf{Set}$, but in general there may be multiple forgetful functors into $\mathsf{Set}$, so all of this still requires the conscious decision that somehow the forgetful functor we constructed in the previous paragraph is the ''preferred'' one. This is however more of a meta-question, and in practice all forgetful functors $\mathcal{C}\to\mathbf{Set}$, where $\mathcal{C}$ is a category of some algebraic structure, are constructed in this way, so at least it it is very common to prefer this particular forgetful functor.