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Let $\mathcal{C}$ be a category, $F:\mathcal{C}\to \mathcal{C}$ be a covariant functor, and $G:\mathcal{C} \to \mathcal{C}$ be a contravariant functor. Let $X \in \mathcal{C}$ and consider the objects $Y = F(X)$ and $Z = G(X)$. Is it true that every isomorphism between $Y$ and $Z$ is unnatural? If so, how do we prove this?

Wikipedia gives the definition of an unnatural isomorphism as one which can't be extended to a natural transformation on the entire category.

The motivation for the problem comes from the idea that a finite-dimensional vector space isn't naturally isomorphic to it's dual, but I'd like to treat it abstracted away from this example if possible.

Brendan Langfield
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    Surely your generality is too great. For example, let $\mathcal C$ be the category with one object and one arrow ... – GEdgar Mar 07 '21 at 17:26
  • What is an "unnatural isomorphism" between two single objects? – Thorgott Mar 07 '21 at 17:44
  • @GEdgar It very well may be too general. Can you suggest further constraints to add so that the question makes sense? – Brendan Langfield Mar 07 '21 at 18:55
  • @Thorgott The nlab page for unnatural isomorphism (https://ncatlab.org/nlab/show/unnatural+isomorphism) has a subsection titled "between objects", but I was using the wikipedia definition: "Conversely, a particular map between particular objects may be called an unnatural isomorphism (or "this isomorphism is not natural") if the map cannot be extended to a natural transformation on the entire category." This is from here: https://en.wikipedia.org/wiki/Natural_transformation#Unnatural_isomorphism – Brendan Langfield Mar 07 '21 at 18:56
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    Maybe I'm misunderstanding, but it seems to me that the answer is rather trivial then: no isomorphism between $Y$ and $Z$ can be extended to a natural transformation between $F$ and $G$, since there are no natural transformations between a covariant and a contravariant functor. If you relax natural transformation to a more appropriate notion such as dinatural transformation, the claim is of course false. – Thorgott Mar 07 '21 at 19:09
  • @Thorgott See here: https://math.stackexchange.com/questions/622589/in-categorical-terms-why-is-there-no-canonical-isomorphism-from-a-finite-dimens I agree with the OP of this question in hoping there is a better explanation than "the definition of this construct doesn't apply". It seems dinatural transformations allow for such an explanation. – Brendan Langfield Mar 07 '21 at 21:58
  • So a follow up question would be how do we prove the following: "If the only dinatural transformation between a covariant functor $F:\mathcal{C} \to \mathcal{C}$ and a contravariant functor $G:\mathcal{C} \to \mathcal{C}$ is $0$, then for every $X \in \mathcal{C}$, all isomorphisms from $F(X)$ to $G(X)$ are unnatural." Or something like that. It seems like we need an equivalent definition of "unnatural isomorphism" involving dinatural transformations. As @GEdgar pointed out, this might be too general. – Brendan Langfield Mar 07 '21 at 22:02
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    Well, if the only dinatural transformation between these functors is the zero transformation, then, by hypothesis, you cannot extend an isomorphism between two such objects to a natural transformation between the functors, because it wouldn't be the zero transformation by virtue of being an isomorphism on at least one object. I'm not actively trying to be unproductive, for the record, but I cannot think of a meaningful way to recast your question in a way that doesn't just become tautology. Perhaps someone else can. – Thorgott Mar 07 '21 at 22:05
  • Maybe what I asked is trivial. In that case, is the motivating question trivial? How do we prove that there's no natural isomorphism between $V$ and its dual? It seems to me that an answer of "because the dual functor is contravariant and natural transformations don't apply to contravariant functors" is indicative of the need for something which does apply to contravariant functors. But as you pointed out above, working with dinatural transformations doesn't seem to get us to a meaningful statement, unless I'm misunderstanding something. – Brendan Langfield Mar 07 '21 at 22:24

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