Qfyilyi

Analogously to
this old question,
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).

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.

[Note: this post does not answer the question, which was whether it is possible to give $epsilon$ which has a right inverse, but every right inverse is non-natural.]

To provide a counter-example, let us consider the category of sets and the constant functors $F(X) = 2 = {0,1}$, $G(X) = 1 = {0}$. There is precisely one transformation $epsilon : F Rightarrow G$, namely $epsilon_X(x) = 0$, and it is natural.

Every transformation $eta : G Rightarrow F$ is a right inverse of $epsilon$, but not every such $eta$ is natural. For instance, take
$$eta_X(0) =
begin{cases}
0 & text{if $X = emptyset$} \
1 & text{otherwise}.
end{cases}
$$
Then naturality of $eta$ fails for the map $f : emptyset to 1$.

Here's an example I find easier to think about than the examples given so far. Let $G$ be a group and $k$ a field, let $C = BG$ be the category with one object with automorphisms $G$, and let $D = text{Vect}(k)$ be the category of $k$-vector spaces. Then the functor category $[C, D]$ is the category of linear representations of $G$ over $k$.

Your question in this special case, for retracts, is equivalent to asking whether every subrepresentation $V subseteq W$ is a direct summand as a representation (it is always a direct summand as a vector space). This is true iff the group algebra $k[G]$ is semisimple, which is true iff $G$ is finite and the characteristic of $k$ does not divide $|G|$. So for an explicit counterexample we can either take $G = mathbb{Z}$ and consider a nontrivial Jordan block, or take $G = C_p, k = mathbb{F}_p$ and, well, again consider a nontrivial Jordan block.