Let $\mathrm{Set}$ be the category of all sets with morphisms being just functions between sets. For two sets $U,V$ there is the categorical notion of their coproduct which is just the disjoint union $U\sqcup V$ or alternatively it can be realised as a colimit.

Is there a categorical description for $U\cup V$, the usual set union? What I mean is a description which does not refer to the elements of $U$ and $V$.

  • $\begingroup$ In the category of sets, there is no such thing as the union of two objects: unions refer to elements in a way which makes them not invariant up to isomorphism. But there is such a thing as the union of two subobjects of an object. $\endgroup$ Oct 17, 2015 at 17:51

2 Answers 2


Yes. The union is the coproduct in the category of subsets of another set. You can construct it within the category of sets as a pushout, namely of the diagram $A\leftarrow A\cap B \to B$, which may or may not be satisfying since you can't define the intersection without elements any more easily than the union. But you shouldn't expect to be able to define the union without elements: categorically, there is no difference between sets of the same cardinality but with different elements, so you really have to specify some way in which you're going to identify elements. Writing both $A$ and $B$ as subsets of a fixed set, as a first mentioned, may be the best way to do that.


This is a tentative answer, I'm not sure:

Instead of looking for the union in $Set$, consider $Set_m$, here defined as the subcategory of $Set$ with all sets as elements and only monomorphisms as morphisms. (the singleton is not terminal in $Set_m$, as an unrelated sidenote)

Your $U$ and $V$ objects in $Set$ are also in $Set_m$ so consider them there and take their coproduct in $Set_m$. You get a set $U \cup V$ in $Set_m$ and two inclusion monomorphisms, and you can bring them back into $Set$ by the subcategory inclusion functor.


You must log in to answer this question.

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