Can the "structural" equivalence between partitions and equivalence relations be expressed as an isomorphism of categories?

Question

I have not taken a formal course in category theory, but as part of my abstract algebra class, I was given a brief introduction to subject of categories, so that I better understand and appreciate the recurring themes within abstract algebra (e.g. sub-objects, homomorphisms, isomorphisms, quotient-objects, etc.).

As part of my abstract algebra course, I learned about several instances of "isomorphism of categories", and came to appreciate them as essentially expressing the "structural" equivalence between two types of objects. For instance, I learned that the fact that abelian groups and $Z$-modules (where $Z$ is the ring of integers) have the same "structural" information can be expressed as an isomorphism of categories (i.e. between the category of abelian groups Ab and the category of $Z$-modules $Z$-Mod).

But I have also noticed similar types of "structural" equivalences occurring outside of abstract algebra, even in as simple of a subject as elementary set theory:

From elementary set theory, I know that equivalence relations and partitions are essentially the same, in that the information present within each equivalence relation is just enough to uniquely construct the corresponding partition, and vice versa. This seems very similar to the instances of "isomorphism of categories" I encountered within abstract algebra. Thus, I am wondering whether this is indeed another instance of an isomorphism of categories (between a "category" of equivalence relations and a "category" of partitions)?

Answer 1

Observe that every set is a (discrete) category (the objects are elements of the set and arrows are identities). Fix a set X. Consider the set of all equivalence relations on X as a discrete category E. Consider the set of all partitions of X as a discrete category P.

Then, there is an isomorphism of categories from E to P.

All set theoretic bijections can be thought of as isomorphisms of categories.

Answer 2

You can treat a set as a discrete category - the set's elements are the category's objects, and the only morphisms are the identity morphisms. You can then speak of when sets are isomorphic as categories, and the answer is when they are equinumerous as sets. But this view of partitions and relations is completely barren, naked, devoid of structure. It ignores where partitions and relations come from.

The better view is to treat the collection of partitions $P(X)$ and the collection of equivalence relations $E(X)$ on a set $X$ as a functor applied to a set $X$ (an endofunctor of $\sf Set$, to be precise). Your statement about the "information present" in a partition is enough to "uniquely construct" an equivalence relation and vice-versa translates to the fact that $P$ And $E$ are naturally isomorphic as functors.

The key word is "natural." It is possible for sets to be equinumerous, but not in a "natural" way, or more accurately, not in any way that is free from making arbitrary choices.

In the context of combinatorics, we speak of functors of finite sets as combinatorial species. Examples of species (set-valued functors applied to a finite set $X$) are: the power set $\mathcal{P}(X)$, the set of all $k$-subsets of $X$, the set of all graphs with vertex set $X$, the set of all group operations on $X$, etc. Basically, a species is a collection of all examples of some kind of "construction" which uses the elements of $X$. (More precisely, we might say a species is an endofunctor on the skeletal core of $\sf FinSet$.)

So, $P$ and $E$ are naturally isomorphic species. For contrast, consider the species $\mathrm{Lin}(X)$ of all linear orders on $X$ (ways of arranging its elements in some order) and the species of $\mathrm{Perm}(X)$ of all permutations of $X$ (self-bijections of $X$). If $|X|=n$, counting arguments show $|\mathrm{Lin}(X)|=|\mathrm{Perm}(X)|=n!$, so we know $\mathrm{Lin}(X)$ and $\mathrm{Perm}(X)$ are always equinumerous, but $\mathrm{Lin}$ and $\mathrm{Perm}$ are not isomorphic functors! (This is because $\mathrm{Lin}(X)$ and $\mathrm{Perm}(X)$ are not equivalent as $S_X$-sets; compare orbits.) There is no "natural" way to turn an linear ordering into a self-bijection or vice-versa, on an arbitrary set, without making some kind of arbitrary choice (like, picking a particular linear ordering and declaring all other orderings are permutations of it - similar to how a vector space has a distinguished origin but an abstract affine space does not).