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In a category $\mathcal C$, let $f: X \rightarrow Y$ and $g: Y \rightarrow X$ such that $g \circ f = id_X$. Is there a general criterion on $\mathcal C$ such that the following holds: there is an object $Z$ such that $Y = X \sqcup Z$ (the coproduct of $X$ and $Z$)?

Thanks for any answer.

(retitled the question as per the comment below)

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  • $\begingroup$ It's certainly true if C is abelian, but I guess you're looking for something a bit more interesting. $\endgroup$
    – Hugh Thomas
    Commented May 21, 2010 at 3:07
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    $\begingroup$ As a terminological point, it is usual in category theory to say that a "split monomorphism" is one with a retract, whereas one which satisfies the conclusion of your question is "complemented." So your question would be less confusing if it were titled "when are all split monomorphisms complemented?" $\endgroup$
    – Mike Shulman
    Commented May 21, 2010 at 18:30
  • $\begingroup$ I retitled it accordingly. $\endgroup$
    – Jakob
    Commented May 22, 2010 at 7:42

2 Answers 2

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In the setting of the question, the composite $f\circ g$ is idempotent. Thus one hypothesis on $\mathcal C$ that will guarantee the existence of the desired splittings is that $\mathcal C$ be Karoubian, or (same concept, alternative name) pseudo-abelian, which is to say: $\mathcal C$ is pre-additive (the Hom-sets are abelian groups) and all idempotents have kernels (and hence cokernels).

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This is true in boolean categories (extensive + terminal object + (T : 1 → 1 + 1) is subobject classifier), but for any monic, not just split monics. There's quite a nice and easy read on extensive categories from 93:

Carboni et al. Introduction to extensive and distributive categories. Journal of Pure and Applied Algebra (1993) vol. 84 pp. 145-158

However, modulo axiom of choice, all monics in Set split, and I suspect this could be true for any boolean category.

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  • $\begingroup$ Even in Set, not all monics split: consider any function from an empty set to a nonempty one. (-: $\endgroup$
    – Mike Shulman
    Commented May 21, 2010 at 15:01
  • $\begingroup$ Also, I think the more usual definition of "Boolean category" includes regularity but not extensivity: ncatlab.org/nlab/show/Boolean%20category $\endgroup$
    – Mike Shulman
    Commented May 21, 2010 at 18:48
  • $\begingroup$ That sneaky empty set. :) $\endgroup$
    – Aleks Kissinger
    Commented Jul 16, 2010 at 15:38

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