Analogously to
[this old question](https://mathoverflow.net/questions/17953/can-epi-mono-for-natural-transformations-be-checked-pointwise),
I was asking myself if it is possible to describe left/right invertible natural transformations by their components. Obviously this property is inherited by the components of a transformation.


For the converse, I don't think that every choice of left inverses of the components yields a natural transformation, or even that such a choice always has to exist. My thoughts until now have been:


Given a natural transformation $\varepsilon: F \Rightarrow G$ between functors $F, G: C \to D$, if its components are right invertible with sections $\eta\_C: GC \to FC$, naturality of $\eta$ comes down to $Gc \circ \eta\_C = p\_{C'} \circ Gc \circ \eta\_C$ for any morphism $c: C \to C'$, where I defined the idempotents $p\_C := \eta\_C \circ \varepsilon\_C$. At this point I don't see under which conditions there is a choice of $\eta\_C$ and $\eta\_{C'}$ such that the above equation is fulfilled (apart from $p\_C = \text{id}\_{FC}$, i.e. natural isomorphisms, of course).


I also don't see a way to apply the solution of the question mentioned above, since being a section/retraction is not a property that can be detected using (co)limits as far as I know (apart from "boring" categories, where the sections are exactly the effective monomorphisms, or even all monomorphisms).


Another approach might be that the split monics/epics are exactly the absolute monics/epics, but I didn't get very far this way...


If it turns out that not every natural transformation with left/right invertible components is left/right invertible, I'd be interested if there is an additional criterion on the components that guarantees the existence of a left/right inverse of the transformation (and is ideally equivalent to it).



Can natural section/retraction be checked pointwise?
No, you cannot check the property of being a section/retraction pointwise. Take $C = {\cdot \to \cdot}$ so that the category of functors from $C$ to $D$ is the arrow category of $D$. In the arrow category, the property of an object being an isomorphism (as a morphism of $D$) is closed under retracts (exercise). Now let $f : A \to B$ be a morphism of $D$ which admits a retraction but is not an isomorphism. Then there is an obvious morphism in the arrow category of $D$ from the object $(f : A \to B)$ to the object $(1 : B \to B)$ and it is pointwise the inclusion of a retract. But it cannot be the inclusion of a retract in the arrow category, as then the original map $f$ would be an isomorphism.


This example shows that the obvious sufficient condition on $D$ (namely, every retraction is already an isomorphism) is required. But perhaps you prefer some concrete examples for intuition. Then consider, in the category of $G$-sets, the map $G \to \*$, which has no section; or in the category of simplicial sets, the map $\partial \Delta^1 \to \Delta^1$, which is not the inclusion of a retraction.


