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This question arose from the discussion in the comments here.

Let $F,G:\mathscr A\to\mathscr B$ be functors. If $A$ is an object of $\mathscr A$, then we say that the arrow $\alpha_A:F(A)\to G(A)$ is natural in $A$ if the family of arrows $(\alpha_A: F(A)\to G(A))_{A\in\mathscr A}$ forms a natural transformation (i.e., for all objects $B\in \mathscr B$ and all arrows $f:A\to B$, one has $\alpha_B\circ F(f)=G(f)\circ\alpha_B)$.

I'm trying to figure out whether the following claim is true.

Claim 1. Let $F,G:\mathscr A\to\mathscr B$ be functors and let $\alpha=(\alpha_A:F(A)\to G(A))_{A\in\mathscr A}$ be a famaily of arrows. Then $\alpha$ is a natural isomorphism iff (1) $\alpha_A$ is natural in $A$ for all $A\in\mathscr A$, and (2) $\alpha_A$ is an isomorphism for all $A\in \mathscr A$. Further, if $\mathscr A$ is nonempty, then condition (1) can be replaced with (1') $\alpha_A$ is natural in $A$ for some $A\in\mathscr A$.

The forward implication is clear: if $\alpha$ is a natural isomorphism, then it's a natural transformation, so $\alpha_A$ is natural in $A$ for all $A$ (hence for some $A$ if $\mathscr A$ is nonempty). The fact that $\alpha_A$ is an isomorphism follows from Lemma 1.3.11 (the proof is given here).

For the converse. If $\alpha_A$ is natural in $A$ for all $A$ (or even for some $A$), then $\alpha$ is a natural transformation by definition. Since $\alpha_A$ is an isomorphism for all $A$, each $\alpha_A$ has an inverse $\beta_A$. The conjecture is that $\beta$ is then the natural transformation that is an inverse of $\alpha$. It is clear that $\beta$ is an inverse of $\alpha$ because $(\beta\circ\alpha)_A=\beta_A\circ\alpha_A=1$ and similarly for the other composition. But is it true that $\beta$ is a natural transformation? I couldn't verify that $\beta_A$ is natural in $A$ for any (or even some) $A$. If $\beta$ is not a natural transformation, then would this modification of Claim 1 be true?

Claim 2. Let $F,G:\mathscr A\to\mathscr B$ be functors and let $\alpha=(\alpha_A:F(A)\to G(A))_{A\in\mathscr A}$ be a famaily of arrows. Then $\alpha$ is a natural isomorphism iff (1) $\alpha_A$ is natural in $A$ for all $A\in\mathscr A$, (2) $\alpha_A$ is an isomorphism for all $A\in \mathscr A$, and (3) the inverse of each $\alpha_A$ is natural in $A$ for each $A\in\mathscr A$. Further, if $\mathscr A$ is nonempty, then condition (1) can be replaced with (1') $\alpha_A$ is natural in $A$ for some $A\in\mathscr A$ and (3) can be replaced with (3') the inverse of each $\alpha_A$ is natural in $A$ for some $A\in\mathscr A$.


Just to spell it out, here is the right claim (thanks to @Max):

Claim 3. Let $F,G:\mathscr A\to\mathscr B$ be functors and let $\alpha=(\alpha_A:F(A)\to G(A))_{A\in\mathscr A}$ be a famaily of arrows. Then $\alpha$ is a natural isomorphism iff (1) $\alpha_A$ is natural in $A$, and (2) $\alpha_A$ is an isomorphism for all $A\in \mathscr A$.

Does this claim mean that the inverse of a natural transformation, it it exists, is a natural transformation?

Regarding the proof of Claim 3. Considering what I already wrote above, it remains to show that $\beta_A$ is natural in $A$. That is, if $f:A\to B$ is an arrow, then $$\beta_B\circ G(f)=F(f)\circ \beta_A.$$ The proof of this claim is actually contained here.

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    $\begingroup$ "is natural for some $A$" and "is natural for all $A$" don't mean anything, naturality is something global. Your claim 1, apart from the mention of condition (1') and without the first "for all $A$" is correct, you should try to prove that $\beta$ is natural $\endgroup$ – Maxime Ramzi Jun 30 '19 at 19:18
  • $\begingroup$ @Max Is the right claim you're referring to the same as saying that the inverse of a natural transformation is a natural transformation? I've also included the right claim in the question. $\endgroup$ – user634426 Jun 30 '19 at 19:22
  • $\begingroup$ Yes, claim 3 is correct. $\endgroup$ – Maxime Ramzi Jun 30 '19 at 19:23
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This is a semantic (but important) issue. Typically one says

"the arrow $f : Fc \to Gc$ is natural in $c$"

when it is clear from context that $f$ depends only of $c$, and so doing the same process for every object one could define a family $(f_x)_{x \in C}$ that assembles into a natural transformation. For example, the phrase

"Given a vector space $V$, the arrow $f^{**} : v \in V \mapsto ev_v \in V^{**}$ is natural in $V$."

means that for each vector space $V$, we can define

$$\alpha_V : v \in V \mapsto ev_v \in V^{**}$$

and this is a natural transformation between the identity functor and the double dual functor.

On the other hand, saying that $\alpha_V$ is natural in $V$ gives no additional information, as one is explicitly stating that this is the $V$-component of a natural transformation already.

As Max says in the comments, naturality is a global phenomenon. This abuse of notation is just to spare the trouble of defining some natural transformations one doesn't want to explicitly write, so we just assert that a certain assignment $x \rightsquigarrow f_x$ lets us produce a natural transformation $(f_x)_x$. As for claim $3$, I've included a proof in the linked post.

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  • $\begingroup$ So does this Claim 3 say, in particular, that the inverse of a natural transformation is always a natural transformation? (I.e., if $\alpha$ is a nat. transf. and $\beta_A$ is the inverse of $\alpha_A$, then $\beta_A$ is natural in $A$, so $(\beta_A)_{A\in\mathscr A}$ is itself a nat. transf.) $\endgroup$ – user634426 Jun 30 '19 at 20:00
  • $\begingroup$ By definition, yes: a natural transformation is an arrow $\alpha : F \Rightarrow G$ in the category of functors from $C$ to $D$, usually noted $D^C$. Thus, it is an isomorphism if and only if there exists another arrow (hence, natural transformation) $\beta : G \Rightarrow F$ such that $\beta \alpha = 1_F$ and $\alpha \beta = 1_G$. Moreover, as the proof I referenced shows, if every component of a natural transformation is invertible, the inverses assemble into a natural transformation and so the original n.t. is an iso. $\endgroup$ – guidoar Jun 30 '19 at 20:04

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