If $a$ is an object in a category $C$, then a subobject of $a$ is an isomorphism class of monomorphisms with codomain $a$. The subobjects of all the objects in $C$ partitions the class of monomorphisms in $C$ into a bunch of subclasses. My question is, is there a notion of a transversal, i.e. a function which picks out a representative from each of the subclasses?

In the case of a concrete category, where you have a forgetful functor $U:C\rightarrow Set$, there is an obvious choice of representative for a given subobject of an object $a$: the unique monomorphism $f$ in the subobject such that $U(f)$ is an inclusion map from some subset of $U(a)$ to $U(a)$. But is it possible to define a transversal even for non-concrete categories, in a such a way that the representatives are “compatible” with one another?

Is there already a name for this structure?

  • $\begingroup$ If your $U$ is not very nice, it may happen that no such $U(f)$ satisfies this. Isn't there a name for concrete categories where you can always lift structures along isomorphisms in the base category ? I seem to recall seeing that name in Joy of Cats. This isn't very important though $\endgroup$ Mar 11, 2019 at 13:45
  • $\begingroup$ @Max Sounds like isofibration to me, though Joy of Cats seems to use "transportable". $\endgroup$
    – Arnaud D.
    Mar 11, 2019 at 13:55
  • $\begingroup$ @ArnaudD. Thanks for the link ! $\endgroup$ Mar 11, 2019 at 14:45
  • $\begingroup$ @Max I posted a follow-up question based on your comment: math.stackexchange.com/q/3144177/71829 $\endgroup$ Mar 12, 2019 at 3:39

1 Answer 1


The answer, just as for so many questions which begin "in an arbitrary category, can I choose...", is that you need a very strong form of the axiom of choice to make things work.

First, note that not every category is well-powered. That is, there is no reason that we should be able to index the collection of subobjects of objects in a given category with any set, so if you're pedantic about wanting a function which does that indexing, you could have problems.

Even if you insist that your category be well-powered (or even small), it's easy to show that we need at least the axiom of choice. Given a cardinality $\kappa$, consider the category of sets of cardinality less than $\kappa$, with exactly one morphism between any pair of isomorphic sets, and morphisms from every set to $1$. Then all of the morphisms to the terminal object are monos since there are no non-trivial parallel pairs, and so an indexing of the subobjects is a choice of set of each cardinality less than $\kappa$.

Continuing in this vein, we can show that "every category has a transversal" is equivalent to the statement "every category has a skeleton". Given a category $C$, we can consider the category $D$ whose objects are objects of $C$, with exactly one morphism between any pair of isomorphic objects. Freely adjoin to this setup a terminal object. Then a transversal for $D$ provides a skeleton for $C$. Conversely, given $D$, we can consider the category $\mathrm{mono}(D)$ whose objects are monos in $D$ and whose morphisms are those completing triangles of monos with the same codomain; the isomorphism classes in this category are of course the subobjects of $D$, so a skeleton of $\mathrm{mono}(D)$ gives a transversal of $D$. You may be aware that "every category has a skeleton" is equivalent to the axiom of choice for proper classes; again, if we restrict ourselves to well-powered categories then the above construction doesn't hold water, but it should hopefully shed light on your question.


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