When a mathematician says that two categories are the same thing, they may mean there is an equivalence or an isomorphism between them. I am wondering if there is a precise way we can say that two categories that are isomorphic are different categories. I have no idea what I mean, but perhaps it has something to do with models of categories in Set. I just feel there may be justification sometimes when two mathematics feel that two isomorphic categories should be considered independently. Is this possible? Does it ever come up in discussion?

For example, consider this language on modules

"The concept of a Z-module agrees with the notion of an abelian groups."

This language "agrees" really means there is an isomorphism of categories. It seems like the theory of categories simply deletes any way of seeing them differently, but is there another way of seeing them where one might say"No, we consider them differently!"

I think this might have something to do with the way computer scientists talk about monads in a different way than mathematicians. If you've ever looked at these two literatures, they are nearly incompatible at times.

This may have something to do with the notion of "up to isomorphism" in category theory. When, in category theory, do we say, "no, I won't forget details that are deleted when we work 'up to isomorphism'"? Is there category theory that doesn't work "up to isomorphism"?

Which category theory concepts and constructions only work up to isomorphism and which, if any, go beyond this?

  • $\begingroup$ Hmm, I wonder... For example, even though you might be able to give a group isomorphism $S_3 \simeq GL_2(\mathbb{F}_2)$, you tend to use the two descriptions of the group differently: $S_3$ as acting canonically on $\{ 1, 2, 3 \}$ while $GL_2(\mathbb{F}_2)$ has a canonical two-dimensional representation over $\mathbb{F}_2$. And yes, there are relations between the two canonical representations, for example the representation of $GL_2(\mathbb{F}_2)$ permutes the three one-dimensional subspaces of $\mathbb{F}_2^2$. But still, we tend to keep the two distinct notations. $\endgroup$ Commented Sep 22, 2023 at 18:30
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    $\begingroup$ This is subtle because sometimes isomorphism is not good enough. We might want isomorphisms... as [insert object type here] and we might want those isomorphisms to be natural. Even that's not good enough sometimes - we might want the isomorphisms to be unique. Also it's common to have two distinct yet isomorphic groups/categories/spaces/... and maintain the two different perspectives on 'the same thing' that you get from the different presentations of that thing. Something something something, mathematics is about helping humans think about mathematics $\endgroup$
    – FShrike
    Commented Sep 22, 2023 at 18:58
  • $\begingroup$ For another example, but using equivalence instead of isomorphism, if $R$ is a commutative ring and $n \in \mathbb{N}^+$, consider the equivalence between left $M_n(R)$-modules and $R$-modules. It would be a bit weird to say that if $M$ is an $R$-module, then $M^n$ as an $M_n(R)$-module is exactly the same object as $M$. In practice, to use this equivalence, we'd most likely invoke the fact that the functor $M \mapsto M^n$ is essentially surjective, etc., rather than pushing the equivalence of categories into the realm of "we think of them as being different formulations of the same thing". $\endgroup$ Commented Sep 22, 2023 at 19:19
  • $\begingroup$ OK, maybe an easier example of what I was trying to get across in the last example: the duality equivalence $\mathbf{FinVect}_k \sim \mathbf{FinVect}_k^{op}$. $\endgroup$ Commented Sep 22, 2023 at 19:26
  • $\begingroup$ It is a theorem in some foundations that the categories $\textbf{AbGrp} = \mathbb Z\textbf{Mod}$, with an equality, not isomorphism or equivalence. $\endgroup$
    – Trebor
    Commented Sep 23, 2023 at 6:51

3 Answers 3


In an extremely literal Bourbakist sense, an abelian group is given by a pair $(A,\star)$ where $A$ is a set and $\star$ is a set of ordered pairs of the form $((a_1,a_2),a_3)$ such that all the $a_i$ are in $A$ and each element of $A\times A$ occurs as the first element of exactly one pair. Thus the category of abelian groups has as its class of objects a class all of whose elements are such pairs. (Of course not all such pairs form abelian groups, but given such a pair you can check whether it is an abelian group without providing any more structure.)

A $\mathbf Z$-module consists of a pair $((A,\star),\cdot)$ where $(A,\star)$ is as above and $\cdot$ is, to phrase it more colloquially than in the previous paragraph, a function $\mathbf Z\times A\to A$, again satisfying certain properties.

Thus the categories of abelian groups and of $\mathbf Z$-modules are certainly not equal: the class of pairs $(A,\star)$ and that of pairs $((A,\star),\cdot)$ are dramatically distinct. And we don't even need to go all the way to modules to get this kind of problem. For instance, why should it be a pair $(A,\star),$ rather than $(\star,A)$? Or perhaps I want a triple $(A,\star,e)$ rather than just an axiom saying that the identity $e$ exists? (Oh and by the way, have I defined ordered tuples so that $(A,\star,e)=((A,\star),e)$, or is it $(A,(\star,e)),$ or something else entirely?)

On the one hand, all of this is total nonsense and nobody cares; it is very hard to say that you're doing any kind of category theory if you're paying serious attention to the distinction between isomorphic but non-equal categories.

On the other hand, it does point to some non-nonsense (sense, I suppose?) as suggested in the other answer: rather than the rather old-fashioned and unmeaningful notion that we need to define an abelian group as some sort of ordered pair or triple, it is useful and important to consider that there are many different choices of syntax that you might use to compute in the theory of abelian groups. A very brief gloss of the idea of categorical semantics is that to give a syntax for abelian groups is to give a construction of the free category with finite products containing an abelian group object. All of the three-ish different syntaxes just mentioned give a distinct such construction of what is really the same category.


What you wrote down, the way you wrote it, goes right against the structuralist view of category theory.

What one can do is consider a syntax to talk about different ways to write stuff down. You can do this with a free monoid(oid). And with this you can consider a mapping (which will trivially be functorial) to the objects you say the syntax is talking about.

In your example you can consider a syntax for $\mathbb Z$-modules and another for abelian groups.

Through the interpretation functor from above, they should map to the same object (up to isomorphism), but at the syntax level they are clearly different by definition.

  • $\begingroup$ I must add that this is mostly a philosophical view on my part, and there might be others. I'm not sure there is a strict mathematical answer to this. Maybe a 'soft question' tag makes sense here. Cheers! $\endgroup$
    – Julián
    Commented Sep 22, 2023 at 18:46

I would say that the connotations may be different. For example, the additive groups of integers $\mathbb Z$ and even integers $2\mathbb Z$ are isomorphic, so there wouldn't be any group-theoretic difference at all between them. However it doesn't mean that you can replace any appearance of $2\mathbb Z$ with $\mathbb Z$. This is because most of the time $2\mathbb Z$ carries a connotation that it is to be considered as a subgroup $2\mathbb Z \hookrightarrow \mathbb Z$, which is different from the subgroup $\mathbb Z \hookrightarrow \mathbb Z$. The subgroup structures are different, but the group structures are the same. (This actually comes up in practice in category theory: here we have two equivalent categories, but they are intended to be interpreted as two non-equivalent subcategories of a common larger category.)

A more dramatic example of this is complete lattices and complete semi-lattices. The former is a lattice with arbitrary join and meet, and the latter with only arbitrary join. However, a complete semi-lattice already has arbitrary meet too! As mathematical objects, they are exactly the same: both are pairs $(L, \leq)$ satisfying exactly the same kinds of theorems. So why do we still use two different names? That is because the implied notion of homomorphisms between them are different. Homomorphisms between complete lattices are required to preserve all join and meet, which is not necessarily true for homomorphisms between complete semi-lattices.

So, a more straightforward answer to the question is: No, if you truly mean only those two isomorphic objects and nothing else, but that would be very far from actual mathematical practice.


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