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A split monomorphism is a morphism $m \colon A \to B$ for which there exists a morphism $e \colon B \to A$ such that $e \circ m = \mathrm{id}_A \colon A \to A$. Is there an elegant description of a universal construction that splits monomorphisms in a given category, i.e. that adds a new morphism $e$ for every monomorphism $m$?

I'm looking for adding one-sided inverses $e$, not two-sided inverses that make every monomorphism $m$ into an isomorphism. A variation on localization would work, but a concrete construction would be much more elegant. Is a variation on calculi of fractions known that might work?

By elegant construction, I'm thinking of something like the Karoubi envelope. An idempotent is a morphism $p \colon B \to B$ such that $p \circ p = p$. It splits when there exist $m$ and $e$ as above with $m \circ e = p$. The Karoubi envelope adds new morphisms $e$ and $m$, and a new object $A$, to every $p$, so in a sense is opposite to the question.

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    $\begingroup$ I assume that just freely adding retraction wouldn't qualify as elegant ? $\endgroup$ Commented Sep 20, 2017 at 15:31
  • $\begingroup$ If $\mathcal C$ is a category with coproducts, then there is a monad on the arrow category $\mathcal C^{\rightarrow}$ such that the category of algebras for that monad is exactly the category of split monomorphisms equipped with a chosen retraction. $\endgroup$ Commented Sep 20, 2017 at 17:30
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    $\begingroup$ An important difference between splitting monomorphisms and splitting idempotents is that splittings of idempotents, when they exist, have a universal property, and hence are unique up to unique isomorphism. Even more, their universal property uniquely determines both morphisms into and out of them. But splittings of monomorphisms are far from unique and there's no universal property in sight. $\endgroup$ Commented Sep 20, 2017 at 18:58

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Here are three possibilities:

First: you have the obvious 'universal solution':

You start from $C$ any category and $J$ a set of maps, you can consider the category $C'$ freely generated from $C$ by adding a retract $r_i$ to each arrow $i \in J$.

$C'$ has the same objects as $C$ and rather complicated arrows, they are described as formal sequence of arrow in $C$ and arrow of the form $r_i$, up to the obvious relation (composing the arrow in $C$ and $r_i \circ i = 1$).

this being said, I belive one can prove that the functor $C \rightarrow C'$ is faithfull if and only if all arrow in $J$ are monomorphisms. which might be interesting (I will give a proof below in the case where $C$ is small, but it should be true in general either, eventually using a large cardinal axiom, or by just refining the construction a little).

Second construction, less general, but closer to the 'category of fractions' you mentioned.

Let $C$ be a category with pullback (in fact, we only need pullback of subobjects). Then one can define a category $C'$ of ''partial map'' in $C$ :

An arrow $X \rightarrow Y$ in $C'$ is a sub-object $U \subset X$ together with a map $U \rightarrow Y$ (by subobject, I just mean a monomorphism $U \rightarrow X$, but two such maps are identified if there is an isomorphism $U \rightarrow U'$ wich makes everythings commute, such an isomorphism is unique if it exists).

Those maps are composed as follows: if $X \overset{(U,f)}{\rightarrow} Y \overset{(V,g)}{\rightarrow} Z$ then you pullback $V$ to $U$ along $f$ you get a $W \subset U \subset X$, the pullback of $f$ induce a map $W \rightarrow V$ which you can compose with $g$ to get your map $W \rightarrow Y$.

On easily check that:

1) this form a category.

2) the 'total arrow', i.e. those such that $U =X$ form a subcategory isomorphic to $C$.

3) for each monomorphism $i:X \rightarrow Y$, the partial arrow $r_i: Y \rightarrow X$ which is defined as the identity on $X \subset Y$ and is a retract of $i$.

4) every arrow in this category if of the form $f \circ r_i$ for $f \in C$ and $r_i$ one of the map above.

The thrid construction, only applies to a small category $C$ but with no other assumptions. You look at the category $\widehat{C}$ of presheaves on $C$. Monomorphism in $C$ stay monomorphisms in $\widehat{C}$, so one can look at the category whose objects are objects of $C$ and whose arrow are partial maps in $\widehat{C}$. As $\widehat{C}$ has pullbacks this also gives a solution to your problem.

This allows to proves the claim made in the first construction, at least for small category: for any small category $C$ and $J$ a set of monomorphisms in $C$, there exists a faithfull functor to a category $C'$ such that all arrow in $J$ have retraction in $C'$, hence the category obtained by universally adding the retraction is indeed a faithful extention of $C$.

Final remark:

None of the construction above, will produce a locally small category in full generality:

  • In the first if you start form a proper class of arrow $J$ you will almost certainly end up with a non locally small category.
  • In the second construction, the category of partial arrow will be locally small if and only if you started with a well-powered category.
  • For the third one, you need a small category, although it might works for certain non-small category. Also you can adapt the idea and use another an embeddings into a finitely complete category. For example any accessible category can be embedded in $prsh(C)$ of $C$ small.
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  • $\begingroup$ Just bumped into math.stackexchange.com/a/3605705/231353: the inclusion of the original category in the completion is not faithful in general. $\endgroup$ Commented Apr 8, 2021 at 15:13
  • $\begingroup$ I have to think about it because I wrote that four years ago, but the construction which I claimed is faithful only adds retract to the monomorphisms in $C$, so I don't see why the reslting category would be a group if one start from a cancelative monoid. I think, when applied to a cancelative monoid the construction described above corresponds to a known construction in the theory of inverse semi-group, I'm trying to find a reference... $\endgroup$ Commented Apr 8, 2021 at 15:37
  • $\begingroup$ (The post you link is trying to do this for both mono and epis at the same time, which I agree is impossible). $\endgroup$ Commented Apr 8, 2021 at 15:40

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