6
$\begingroup$

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).

$\endgroup$
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gnampfissimo Nov 22 '18 at 15:38
  • $\begingroup$ ... 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? $\endgroup$ – Gnampfissimo Nov 22 '18 at 15:50
  • 1
    $\begingroup$ 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). $\endgroup$ – Reid Barton Nov 22 '18 at 16:05
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter LeFanu Lumsdaine Nov 23 '18 at 10:50
2
$\begingroup$

[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 $f : \emptyset \to 1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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"). $\endgroup$ – Gnampfissimo Nov 22 '18 at 14:47
  • $\begingroup$ P.S.: I think in the end of the first paragraph you mean $\epsilon_X$, right? $\endgroup$ – Gnampfissimo Nov 22 '18 at 14:58
  • $\begingroup$ I see. Perhaps you can make the question a bit more explicit then. $\endgroup$ – Andrej Bauer Nov 22 '18 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.