# understanding natural transformations that are not natural isomorphisms

What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?

A natural equivalence or natural isomorphism is a natural transformation $$\eta : F \stackrel{nt}{\leftrightarrow} G$$ where all the arrows in the "image" (what's the actual term?) of $$\eta$$ are isomorphisms. So, for every $$X$$ in the object of the domain of our functors $$F : C \stackrel{ftr}{\to} D$$ and $$G : C \stackrel{ftr}{\to} D$$, $$\eta_X : D_{\text{arr}}$$ and $$\eta_X^{-1}$$ both exist.

This means that the ordinary natural transformation law

$$\eta_Y \circ F(f) = G(f) \circ \eta_x$$

can be written as

$$F(f) = \eta_Y^{-1} \circ G(f) \circ \eta_X$$

or, using $$\triangleright$$ as reverse composition ...

$$F(f) = \eta_X \triangleright G(f) \triangleright \eta_Y^{-1}$$

In my opinion, writing it this way makes it clear just how strong a claim the existence of $$\eta$$ is making. We can apply the functor $$F$$ to an arrow by applying $$G$$ instead and then "correcting it" by adding $$\eta_X$$ in front and $$\eta_Y^{-1}$$ behind it.

However, $$\eta_X$$, crucially, does not depend on $$f$$ at all. It isn't enough that an arrow exists from the source of $$F(f)$$ to the source of $$G(f)$$ , it has to be possible to pick that arrow "uniformly" and make the same choice for every arrow in $$f$$'s homset.

If $$\eta$$ were just a natural transformation, what's the right way of thinking about what that means for the relationship between $$F$$ and $$G$$ ?

## 1 Answer

What's the right way to think about what a natural transformation that is not a natural isomorphism is? How strong of a claim is it making about the relationship between the two functors it's related to?

No stronger a claim than a morphism between two objects makes about the two objects (in fact natural transformations are a special case of this, in the functor category). For example, if $$A$$ is an abelian category and $$F, G : C \to A$$ are two functors, then there is always a natural transformation $$\eta : F \to G$$, namely the zero natural transformation, which tells you nothing.

A nice concrete example here is if $$A = \text{Vect}$$ and $$C = BG$$ is the one-object category with automorphisms a group $$G$$. Then the functor category $$[BG, \text{Vect}]$$ is the category of linear representations of $$G$$, with natural transformations given by $$G$$-equivariant maps.