Every natural isomorphism decomposes through another

Question

I read a proof where the author implicitly used this:

Let $F,G$ be endofuctors of a category $\mathcal C$. If $\mu:F\Rightarrow G$ is a natural isomorphism, then the components of any other natural isomorphism $\psi:F\Rightarrow G$ can be written $\psi_X = \mu_X \theta _X$, with $\theta\in \text{Aut}(F)$.

This seems quite basic, but I do not see how this holds.

Answer 1

This has nothing to do with functors and natural transformations : let $C$ be a category, $x,y$ two objects and $f: x\to y$ an isomorphism. If $g$ is any other isomorphism $x\to y$, then it factors through $f$, i.e. for some $h\in Aut(x), g=fh$.

The proof is easy : if $g=fh$, then $f^{-1}g = h$. Thus if we start from $g,f$, we can put $h:= f^{-1}g$. As a composite of isomorphisms, it is an isomorphism, thus $h\in Aut(x)$. Now $fh = ff^{-1}g = id_y g = g$.

Answer 2

That is not true in general.

We have $\theta:\mathsf{id}_{\mathcal C}\to\mathsf{id}_{\mathcal C}$ and $\mu:F\to G$ so if $F\neq\mathsf{id}_{\mathcal C}$ then $\mu\circ\theta$ is not defined.

What we do have for every object $A$ is $$\psi_A=\mu_A\circ\mu_A^{-1}\circ\psi_A=\mu_A\circ\theta_A:F(A)\to G(A)$$where $\theta_A=\mu_A^{-1}\circ\psi_A:F(A)\to F(A)$.

Then $\theta:F\to F$ is invertible ($\theta_A^{-1}=\psi_A^{-1}\circ\mu_A$) and can be recognized as a natural isomorphism $F\to F$.