Localisation of categories, but instead of "isomorphisms" I want "morphisms with right inverse". Construction via calculus of fractions possible? Let $\mathcal{C}$ be a category and $S$ a collection of morphisms in $\mathcal{C}$. The localisation $\mathcal{C}[S^{-1}]$ is defined via a functor $F: \mathcal{C} \longrightarrow \mathcal{C}[S^{-1}]$ and the following universal property. For all $\varphi\in S$ the morphism $F(\varphi)$ is an isomorphism. Now let $G: \mathcal{C} \longrightarrow \mathcal{D}$ be another functor such that $G(\varphi)$ is an isomorphism for all $\varphi \in S$. Then $G$ factors via $HF$ for another functor $H: \mathcal{C}[S^{-1}] \longrightarrow \mathcal{D}$. One can explicitly construct $\mathcal{C}[S^{-1}]$ via the calculus of fractions under some conditions on $S$.
Now my question is the following: In the text above replace "isomorphism(s)" by "morphism(s) with right inverse". Are there similar conditions on $S$ such that we can do the same construction via something like calculus of fractions?
 A: I think there is a relatively good reason why such a thing shouldn't exists.
In general when you freely add right inverse or inverse, the general arrows of the resulting category will be zig-zag in the original category where the arrow in the wrong directions are all in S, and are there to represent composition with the freely added inverse.
The assumption for a calculus of fraction essentially provide a way to "move" these wrong direction arrow in S to one side of the zig-zag to make it just a span or cospan.
The problem in your situation is that you are not putting any relation between the various freely added right inverse that you add, contrary to the case of actual inverse where inverse being automatically unique any relation in the original category will give rise to relations between the inverse in the localized category.
Typically, you'll assume that starting from a cospan $A \to B \leftarrow C $ where the second arrow is in S, then you can complete it to a commutative square with one of the arrow also in S, which allow to replace the cospan by a span (or the other way around depending on if you do a  left or right calculus of fraction)
But when you'll try to do the same sort of rewrite with one sidded inverse to put all the arrow of S in a zig-zag on one side, you have no way to ensure that said rewrite is going to be equal to the zig-zag you started from (in the example above that going through the span or the cospan using the right inverse yields the same result).
Of course if you start adding a lot of relation between the inverses then some thing might become possible, but that will give you a very different universal property...
