# A claim about a family of arrows forming a natural isomorphism

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.

• "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 – Maxime Ramzi Jun 30 '19 at 19:18
• @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. – user634426 Jun 30 '19 at 19:22
• Yes, claim 3 is correct. – Maxime Ramzi Jun 30 '19 at 19:23

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.

• 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.) – user634426 Jun 30 '19 at 20:00
• 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. – guidoar Jun 30 '19 at 20:04