# Can natural section/retraction be checked pointwise?

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.

• This is a very nice counterexample! However I don't see how it implies that the obvious sufficient condition on $D$ is required (i.e. necessary). This would mean that every natural section/retraction is an isomorphism, which seems a bit strange to me. – Gnampfissimo Nov 22 '18 at 15:38
• ... and is wrong, since for any morphism $f: A \to B$ that is a proper section/retraction one gets a proper natural section/retraction by the morphism $f: \text{id}_A \to \text{id}_B$ in the arrow category. What did you mean with "required" then? – Gnampfissimo Nov 22 '18 at 15:50
• Sorry, that wasn't too clear--I meant that if you want to make the result true by only imposing conditions on $D$, then the obvious condition is necessary. Of course the result could also hold under other hypotheses (e.g., $C$ discrete). – Reid Barton Nov 22 '18 at 16:05

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.

• Even clearer than this, to me, is the similar example using $G$-sets (i.e. sets with $G$-action). This is mentioned in Reid’s answer, but spelling it out like in this answer: $G$-sets are the functor category $[BG,\mathrm{Set}]$. For any non-empty $G$-set $X$, the unique map $X \to 1$ certainly has a “pointwise” section — any element $x \in X$ gives a function $s_x : 1 \to X$. However, this section is natural just if $x$ is a fixpoint of the $G$-action. So for a $G$-set $X$ with no fixpoints, $X \to 1$ has a pointwise section but no (natural) section. – Peter LeFanu Lumsdaine Nov 23 '18 at 10:50

[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 $$f : \emptyset \to 1$$.

• Okay, thanks. This shows, that not every pointwise right inverse of a natural transformation is natural, but the natural transformation $\epsilon$ you gave actually does have a (natural) right inverse, so it's not a counter example to the main part of my question ("or even that such a choice always has to exist"), but only a confirmation of the first part ("I don't think that every choice of left inverses of the components yields a natural transformation"). – Gnampfissimo Nov 22 '18 at 14:47
• P.S.: I think in the end of the first paragraph you mean $\epsilon_X$, right? – Gnampfissimo Nov 22 '18 at 14:58
• I see. Perhaps you can make the question a bit more explicit then. – Andrej Bauer Nov 22 '18 at 15:11