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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.

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I don't know...doesn't seem like there's even a map in general from $A\cup_X B \to A+B$, never mind one depending on $f$ and $g$. In topology the former is a quotient of the latter, unless I'm misunderstanding your notation. –  David White Oct 29 '13 at 22:00
    
It's true that such a morphism doesn't exist in general. It exists in $\mathbf{Set}$, and I was hoping for a broader class of categories (maybe something to do with presentability? well-pointedness?). There was a typo, I meant $a$ and $b$ monic, not $f$ and $g$ (already edited the question). I didn't intend to suggest that the desired morphism (a right inverse of $c$) $A \sqcup_X B \rightarrow A + B$ depends on $f$ and $g$ (though I have nothing against it). –  Maciej Piróg Oct 29 '13 at 22:28
    
Well, simplicial sets is a pretty nice category and I don't see how to do it even there. In particular, it's locally presentable –  David White Oct 29 '13 at 22:34
    
I have no idea whether this is relevant in your situation, but your pushout is a coproduct in $X/\mathbb C$. So if your $Y$ happens to be accompanied with an appropriate morphism $X\to Y$... –  მამუკა ჯიბლაძე Oct 31 '13 at 21:18
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2 Answers

up vote 2 down vote accepted

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.

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Quickly skimming through Diaconescu's proof: In a topos in which subobjects have complements (i.e., a boolean topos), a coequaliser of disjoint monos split. It is quite easy to see that if both $a$ and $b$ are monic, then $\mathsf{inl} \cdot a$ and $\mathsf{inr} \cdot b$ are disjoint, and the theorem gives me what I desire. Maybe it's not very surprising (the calculus of subobjects in boolean topoi sounds very $\mathbf{Set}$-like), but it has already allowed me to make my proof simpler and more elegant. Thanks. –  Maciej Piróg Oct 31 '13 at 21:36
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Claim: if $\mathbb{C}$ is a regular category with coproducts and such a factorization always exists, then regular epis split in $\mathbb{C}$.

Proof: In a regular category, any regular epi $e:A\to B$ is the quotient of its kernel pair $a,b:K\rightrightarrows A$, which is equivalently the pushout of $K \xleftarrow{\nabla} K+K \xrightarrow{(a,b)} A$. Thus, taking $C=B$ as well, the condition implies that the map $K+A \to B$ splits. Since this map factors through $e$, $e$ also splits.

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