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.


3 Answers 3


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.

  • 1
    $\begingroup$ This is the correct answer and should be accepted. $\endgroup$ Feb 26 at 13:54
  • $\begingroup$ @JannikPitt Edited! Thanks for the great answer - I never learned such aspects of abelian categories. I will think about the last paragraph. $\endgroup$ Feb 26 at 14:13
  • $\begingroup$ My answer was incorrect, so I have deleted it. This is a much better answer, thank you. $\endgroup$ Feb 26 at 17:48
  • $\begingroup$ One the paragraphs is slightly suspect, but this is a great answer otherwise. I think when you say: "being an Abelian category does not require ... you only need to check this is in fact a group law"; the addition is automatically a group law on any additive category! To promote your category from additive to Abelian, we need to check various equivalent conditions, e.g. all morphisms are "strict" (first isomorphism theorem holds, basically) or "all monomorphisms and epimorphisms are regular", but checking we have a group law is not one of those checks. $\endgroup$
    – FShrike
    Feb 27 at 10:59

For your first question about $\mathbf{Mod}_A$, it is a remarkable fact that being abelian is actually a property of a category and not some additional structure. That is, even though the straightforward definition of $\phi+\psi$ is via elements, this can be done without mentioning them, namely by declaring $\phi+\psi$ to be the morphism $$M\stackrel{(\phi,\psi)}{\to}N\oplus N \to N,$$ where the last morphism from $N\oplus N\to N$ is given by the identity map from both components. With quite some work you can then show that a category is additive if and only if it has all biproducts and this "addition" rule on parallel morphisms gives an abelian group structure on hom sets. The point being, that this can be done without ever looking inside the objects.

As for the second question, it may be also surprising that there are always several different valid "forgetful" functors $U:\mathbf{Mod}_A\to \mathbf{Set}$. In fact, if $A$ and $B$ are Morita equivalent, their module categories are equivalent and the standard forgetful functor from $\mathbf{Mod}_B$ will look rather different when precomposed with the equivalence to $\mathbf{Mod}_A$. As a consequence, the free modules from one perspective may not be free from the other. For a concrete instance, any $A$ is Morita equivalent to $M_n(A)$ with the equivalence being given by $M\mapsto M\otimes_A A^n$, so that the $A$-free module $A$ is sent to the $M_n(A)$-projective but not free $A^n$. Needless to say, the element structure is completely changed under such an equivalence.

As such, the labeling of any category as $\mathbf{Mod}_A$ is really the same thing as fixing one particular such forgetful functor $U$ from the outset.


It's very helpful to know in these kinds of circumstances whether you are viewing categories as a language/tool, or as an object of study in its own right. In the case of $A$ modules, I would say it's clear that we are really interested in properties of modules, so we know that there is a forgetful functor, we know what free modules are (useful!), we often have a tensor product, and this all helps us answer questions about modules (and rings). From this perspective, being an abelian category is something we could check for our category of interest $A-mod$, and then we would use general abelian category theory to assist us. Viewing $A-mod$ as a pure category, and forgetting all else we know could be an interesting idea, but for reaching qcoherent sheaves, cohomology, derived categories etc, it seems like an artificial restriction which doesn't help understanding.

However, the questions of how much is purely categorical can be very interesting! I'll give a few examples that I like, related to your questions, because they illustrate how this way of thinking leads to deep mathematics:

  1. Being an abelian category (with compatible preadditive structure and bi-products) is actually a property of a category, not an additional structure. Spelling this out, from the (non preadditive) axioms, the additive structure is determined by biproducts and zero objects.
  2. The forgetful functor to abelian groups is secretly a Hom functor, $Hom(A,\_)$, and this functor is conservative, and exact. If we had another object P with functor $Hom(P,\_)$ conservative and exact, one can show that $A-mod$ is isomorphic to $End(P)-Mod$, showing that by carefully considering the structure of "forgetful functors", one recovers morita theory, that different rings can have the same module categories!
  3. In the case of tannakian categories, one may show that we have a recognition principle, once we find a monoidal forgetful fuctor to Vect. It's an amazing (and not easy) theorem of Deligne that this forgetful functor is actually unique, and constructible intrinsically, giving an abstract tannakian reconstruction theorem.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .