**Setup:** Let $\mathbb C$ be a category. Assume that the span $A \xleftarrow{a} X \xrightarrow{b} B$ has a pushout $A \xrightarrow{\mathsf{pinl}} A \sqcup\_X B \xleftarrow{\mathsf{pinr}} B$. Let $f : A \rightarrow C$ and $g : B \rightarrow C$ be such that $f \cdot a = g \cdot b$, which means that there exists a mediating arrow $[\![ f , g ]\!] : A \sqcup\_X B \rightarrow C$. Of course, the pushout can be defined as the codomain of the coequaliser $c = \mathsf{coeq}(\mathsf{inl} \cdot a, \mathsf{inr} \cdot b) : A+B \rightarrow A \sqcup\_X B$, and $[\![ f, g ]\!]$ appears as the coequaliser mediator for $[f,g] : A+B \rightarrow C$.


**Question:** Are there any natural general conditions under which $[\![ f,g ]\!]$ factors as $A \sqcup\_X B \xrightarrow{???} A+B \xrightarrow{[f,g]} C$? By "conditions" I mean, for example, $\mathbb C$ being a category of a particular kind.


It is the case when $\mathbb C = \mathbf{Set}$, one just needs to pick where to put the elements identified by the pushout. Can I hope for a condition more precise than puffed-up "all epis split"?


Would it help if I said that both $a$ and $b$ are monic? It helps in $\mathbf{Set}$, where for monic $a$ and $b$ I don't need the axiom of choice to construct such a factorisation.


**Motivation:** I'm working on Adamek *et al.*'s idealised completely iterative monads. In this context morphisms $Y \rightarrow A+B \xrightarrow{[f,g]} C$ (for specific $A,B,C,f,$ and $g$, of course) have some desired properties. However, I obtained some morphisms of the shape $Y \rightarrow A \sqcup\_X B \xrightarrow{[\![ f, g ]\!]} C$, and I'm trying to manipulate them to fit in the theory.



When does a pushout mediating arrow factor through the coproduct?
Taking $C$ to be the pushout, you're asking that the surjection
from the coproduct to the pushout be split.


There is no natural reason (in either the formal or informal sense)
why this should happen.


You don't say much about what sort of category $\mathbb C$ is supposed
to be, so we might as well switch the arrows and ask about pullbacks
being retracts of products instead.


The pullback of $p$ and $q$ is $\lbrace(x,y)\vert p(x)=q(y)\rbrace$,
so the splitting would take an arbitrary pair $(x,y)$ to one that 
satisfies the equation. Why should there be such a map?


Going back to the pushout, for simplicity in $\mathbf{Set}$ and in the
case where your maps $a$ and $b$ are mono, the pushout is the disjoint
union $A+B$ with some of the members of $A$ "identified with" those
of $B$. 


The corresponding members of the pushout therefore "don't know"
whether they came from $A$ or from $B$. However, your splitting
arbitrarily assigns each of them to one or the other. So the map
from $A+B$ to the pushout and back to $A+B$ leaves some points 
in place and swaps others between $A$ and $B$.


What you have here is therefore the idea behind Radu Diaconescu's
proof that Choice entails Excluded Middle in a topos.


An interesting exercise would be to translate this behaviour of
disjoint unions and pushouts in sets to "amalgamated products",
which I think are what coproducts and pushouts used to be called
in algebra.


