# Dual of axiom of choice

The (internal version of) axiom of choice, stated in the language of category theory says that every epimorphism splits - i.e., every epimorphism has a section.

Now, just like any other category theoretic statement, it has a dual statement which can be produced by inverting all of the arrows. Thus, dual of axiom of choice, we which we dub axiom of co-choice in this question, states that every monomorphism is split -- i.e. has a retraction.

Now, without any assumptions beyond ZF, this holds in the topos Set. However, like axiom of choice, it fails in Top. Off the top of my head, I don't know it failing in any other relevant category.

However, despite looking for some more information on it, there seems to be no literature studying axiom of co-choice. So, my question are these:

Is the axiom of co-choice not mathematically interesting? Is there any philosophical or mathematical justification for the dual of axiom of choice - a very interesting problem on its own - being boring? If it is truly a boring or trivial result, what classes of categories do satisfy it - i.e. what are some statements or requirements which are equivalent to it?

• It fails in Ab as well. The monomorphism $\Bbb Z\to\Bbb Z$ given by multiplication by $2$ doesn't split. – Arthur Jan 20 at 22:08
• It fails in Set: the monomorphism $\varnothing \to \{\star\}$ does not split. – Arturo Magidin Jan 20 at 22:09
• It doesn't apply to any topos, because the map $\emptyset \rightarrow *$ isn't split. I think I got a partial answer to my second question. – Ahraman Jan 20 at 22:27
• What I mean (and it might be a very stupid question) is that a monomorphism of sets $i :A \hookrightarrow B$ has a left inverse $g : B \to A$ defined sending an element $b$ in the image of $i$ to the unique $a$ such that $ia=b$, and an element on the complement of said image to some $a_0$. How do you choose such an $a_0$? – fosco Jan 21 at 13:07
• @Fosco Actually, that is an interesting point. It's a highlight of difference between constructive and classical mathematics. In classical mathematics, this is perfectly valid, because by law of excluded middle, we can rule out the complement of image of $i$ being empty, so by definition there is some $a_0$ in its complement. But in constructive mathematics, we cannot make this assertion, so injective functions don't necessarily have left inverses. – Ahraman Jan 21 at 13:38

"Every epimorphism splits" and "every monomorphism splits" turn out to be equivalent for abelian categories: both conditions are equivalent to the condition that every short exact sequence splits. You might call such an abelian category semisimple, although the discussion at this MO question suggests "split" instead. It is equivalent to the condition that $$\text{Ext}^n = 0$$ (as a functor) for $$n \ge 1$$. For example, the category $$\text{Mod}(R)$$ of $$R$$-modules satisfies this condition iff $$R$$ is semisimple.
It's also an interesting question in various settings to ask whether specific epimorphisms or monomorphisms split. In $$\text{Top}$$, for example, "does this epimorphism split?" specializes to asking whether a bundle has a section, and in $$\text{CRing}$$, "does this monomorphism split?" specializes to asking when a Diophantine equation has a solution. I wrote a blog post Topological Diophantine equations which discusses the analogy between these two cases in some more detail.
As another example, in abelian categories an object $$P$$ is projective iff every epimorphism with target $$P$$ splits, and dually an object $$I$$ is injective iff every monomorphism with source $$I$$ splits.