Question 1

How much of group/ring/lattice/... theory can be expressed in purely categorical terms (only using the notions object, morphism, identity morphism, and composition), that is, as properties of the category of groups/rings/lattices/...?

I know that the question is vague. Let me be a bit more precise.

For example, consider group theory. Whenever one proves a theorem $X$ about groups, one can ask: can $X$ be expressed purely in categorical terms? For instance, consider the following theorem $X$: "for each homomorphism $\varphi\colon G\to H$, there are homomorphisms

$$G\stackrel{\varphi_1}{\longrightarrow} G/\ker(\varphi)\stackrel{\varphi_2}{\longrightarrow}\text{im}(\varphi)\stackrel{\varphi_3}{\longrightarrow} H,$$

where $\varphi_1$ is surjective, $\varphi_2$ is an isomorphism, and $\varphi_3$ is injective." This theorem can be expressed purely in categorical terms, since each concept occurring in the theorem happens to have a categorical description: surjective --> epimorphism, injective --> monomorphism, quotient --> coequalizer, kernel, image. In contrast to, for example, Lagrange's theorem: it contains the notion of the "order" of a group, which I guess cannot be defined in terms of morphisms and composition.

In fact, one can (more) precisely define what I mean by "in purely categorical terms": consider the first-order language $L$ consisting of two sorts, objects and morphisms, two function symbols $\text{dom}$ and $\text{cod}$ (from morphisms to objects), and a ternary relation symbol between morphisms $f, g, h$ expressing that $h = g\circ f$. Then for each first-order (or higher-order) sentence $\varphi$ in the language $L$ and each category $\mathcal C$, one can ask whether $\varphi$ is satisfied in $\mathcal C$, $$\mathcal C\models \varphi.$$ In this sense, one has defined a notion of a "property" $\varphi$ of a category. For example, one can write down an $L$-sentence $\varphi_1$ expressing the associativity of composition. Then each category $\mathcal C$ satisfies $\varphi_1$. Also, one can write down an $L$-sentence $\varphi_2$ expressing that "for all objects $A, B$, there is an object $A\times B$ satisfying the universal property of the product". Then a category $\mathcal C$ satisfies $\varphi_2$ if and only if $\mathcal C$ has products.

Lawvere demonstrated that essentially all theorems of set theory can be expressed in this limited language (surprise: we don't need a binary relation $\in$) and derived from a small number of axioms (written down in the language), see ETCS.

Coming back to my question, I wonder how many theorems $X$ of group/ring/lattice/... theory can be expressed in this categorical language, in the sense that we can find an $L$-sentence $\varphi_X$ "expressing" $X$ (that is, $\mathcal C\models\varphi_X$, where $\mathcal C$ is the category of groups/rings/lattices/..., should be equivalent to $X$).

As another random example, can the spectral theorem of linear algebra, stating that an endomorphism $f$ is normal and triangularizable if and only if there is an orthonormal basis $B $consisting of eigenvectors of $f$, be expressed as a property of the category of finite-dimensional vector spaces?

Question 2

A related question that is coming to my mind is the following: sometimes in mathematics, it happens that two categories, say $\mathcal C$ and $\mathcal D$, studied in two different branches of mathematics, say $A$ and $B$, respectively, are shown to be equivalent. (For example, consider $A$ to be the study of commutative C*-algebras and $B$ the study of compact Hausdorff spaces.) How much does this say about the similarity of $A$ and $B$ as mathematical branches? If many questions and theorems can be expressed in the categorical language, this means that $A$ and $B$ are, as branches, essentially the same, since then each statement occurring in $A$ can be translated into the branch $B$ and vice versa. Otherwise (for example if some property of, say, Hausdorff spaces doesn't have a categorical analogue), this could mean that there can still be questions in the branch $B$ that cannot be translated into the branch $A$, therefore each of the two branches having its own right to exist.

  • $\begingroup$ Lagrange theorem follows from a statement saying that (left) cosets of $H<G$ partition $G$, and for each coset $gH$ there is a bijection between $gH$ and $H$. It does not look like this is impossible to do in a point-free way (the latter seems to be a common name for your "cathegorical langauge"). $\endgroup$ Oct 29, 2019 at 20:36
  • $\begingroup$ @Dima my understanding is that the Lagrange Theorem is true for finite dimensional Hopf Algebras (my understanding is that the dimension of a Hopf subalgebra divides the dimension of the algebra) and this implies that there is a "point-free" proof... whether that falls into Category Theory, I cannot say. $\endgroup$ Oct 29, 2019 at 21:26
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    $\begingroup$ Crossposted: math.stackexchange.com/q/3411075/660 $\endgroup$ Oct 29, 2019 at 21:29
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    $\begingroup$ It is a somewhat nontrivial result that every epimorphism of groups is surjective, and the same is not true for [associative unital commutative] rings. So you need to be very careful in the translation, and probably there are results which do not lend themselves to easy categorical reformulation. $\endgroup$ Oct 30, 2019 at 1:04
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    $\begingroup$ What is an answer supposed to look like? Do you expect someone to say "67.8% of group theory can be expressed this way?" Or a very long survey going through several books on undergraduate mathematics? I think the question is still too broad. $\endgroup$ Oct 30, 2019 at 7:07

3 Answers 3


I agree that the question is broad, but here's one sense in which the answer is "all of it". The functor $\rm Grp \to Set$ is represented by the object $\mathbb{Z}$ (the free group on one element); thus we can talk about "elements of a group $G$" in the language of the category $\rm Grp$ by talking about morphisms $\mathbb{Z}\to G$. If we have two such elements $g,h:\mathbb{Z}\to G$, they induce a map $\mathbb{Z}\ast\mathbb{Z}\to G$ from the free group on 2 elements, and there's a map $\mathbb{Z}\to \mathbb{Z}\ast\mathbb{Z}$ such that the composite $\mathbb{Z}\to \mathbb{Z}\ast\mathbb{Z}\to G$ corresponds to the product $g h$. In this way we can extract the entire group -- in the sense of a set with an operation on it -- from an object of $\rm Grp$. (Formally, $\mathbb{Z}$ is a cogroup object in $\rm Grp$, so homming out of it yields a group.) Since certainly all of group theory can be stated as theorems about sets-with-a-group-operation-on-them, we can therefore import it into the (possibly higher-order) language of $\rm Grp$ by rewriting it to refer to maps out of $\mathbb{Z}$ instead of elements. The same can be done for any other algebraic structure like rings, lattices, etc: there's always a co-structure object we can map out of to recover the underlying set with operations. Topology is trickier, but we can detect the points of a space by mapping out of a one-point space, and the opens of a space by mapping into the Sierpinski space.

So far we have to have parameters in our formulas, so our theorems are translated into statements not about $\rm Grp$ alone qua category but about it together with the cogroup object $\mathbb{Z}$. However, in many cases I expect that that object could be uniquely characterized, although I don't know offhand how generally that holds.

  • $\begingroup$ Let me add that there are two different cogroup structures on $\mathbb{Z}$, so your description alone doesn't fully recover $\mathrm{Grp} \to \mathrm{Set}$ in a categorical way. $\endgroup$ Jan 10, 2020 at 21:31
  • $\begingroup$ The answers here (mathoverflow.net/questions/352298/…) give a way to characterise the cogroup $\mathbb Z$. $\endgroup$ Jul 3, 2020 at 11:10

I think that your intended question is too vague and that your attempts to make it precise probably don't capture your intent (assuming I understand your intent correctly).

Suppose I do a gigantic (but finite) computer calculation to produce an approximate numerical solution to a PDE, or to solve the Boolean Pythagorean triples problem. That is certainly "concrete" mathematics in some sense of the word "concrete." Technically, the validity of the computation can be expressed in set theory or in any halfway-plausible candidate for a foundation of mathematics. I suspect, however, that even if it were possible to encode the gigantic finite computation as some incomprehensible formal string in the "language of category theory" in some technical sense, you wouldn't consider such a monstrosity to be a satisfactory category-theoretic "expression" of that piece of "concrete mathematics."

If I'm wrong, and you'd be satisfied with that, then it seems your question is whether it's possible to develop categorical foundations for mathematics without piggy-backing on standard foundations. That's a more standard question which has been much discussed.

On the other hand, if you're asking how much mathematics can be "naturally" expressed in category-theoretic language, then we're back to a vague question. There are certainly areas of mathematics where there's no obvious concept of a "morphism." Consider a proof that merge sort correctly sorts a list of items using $O(n\log n)$ comparisons. If you want to express this in category-theoretic terms, what's your category? What are the morphisms? Any attempt to shoehorn this basic result in the theory of algorithms into category-theoretic terms is going to seem artificial. This is perhaps an extreme example, but even for a less extreme example, I have to wonder why one would want to express something in category-theoretic terms if it doesn't seem natural to do so. And if you are not interested in any shoehorning, and just want to know how much stuff fits naturally into category theory, then we're back to Andrej Bauer's 67.8% comment.


In contrast to, for example, Lagrange's theorem: it contains the notion of the "order" of a group, which I guess cannot be defined in terms of morphisms and composition.

Perhaps the "order" cannot be defined, but "order divisible by a prime p" can be defined by: there is a non-trivial map from the cyclic group of order p. The Lagrange theorem then leads to a reformulation of the notion of a p-group in terms of weak factorisation systems/Quillen lifting property, see Thm.2.2(6) of (Formulating basic notions of finite group theory via the lifting property) for this and other reformulations of properties of finite groups in similar terms.

  • $\begingroup$ 'Order', and everything else, can be defined within the category of groups: mathoverflow.net/questions/352298/… $\endgroup$ Jul 3, 2020 at 11:08
  • $\begingroup$ I should have said 'defined by a relatively natural construction within category theory", not defined in the sense of say first order logic, which is presumably your meaning. $\endgroup$
    – user160570
    Jul 3, 2020 at 15:07
  • $\begingroup$ I think the construction in the accepted answer there also meets your criteria. The universal construction which maps each group to a cogroup object is a 'relatively natural construction within category theory', and two groups have the same order if and only if their corresponding cogroup objects are isomorphic. $\endgroup$ Jul 3, 2020 at 15:44

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