Can natural section/retraction be checked pointwise?

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

[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 $f : \emptyset \to 1$.