Do natural isomorphisms always respect composition of functors?

Question

Given functors $f,g \colon B \to C$, $h \colon A \to B$ and a natural isomorphism $\alpha \colon f \Rightarrow g$ when is $fh \simeq gh$ naturally? Sorry for not drawing a diagram but I don't know how to here.

Answer 1

Natural transformations can be horizontally composed (with other transformations, functors, objects, or even morphisms).

Any transformation $\alpha : f \Rightarrow g$ composed with a functor $h$ gives a natural transformation $\alpha h : fh \Rightarrow gh$. Its value at an object $x$ of $A$ can be computed by associativity:

$$ (\alpha h) x = \alpha (h x) $$

where I've used the notation of horizontal composition to refer to:

• The component of a natural transformation. $\alpha x$ means $\alpha_x$, the corresponding morphism $fx \to gx$.
• The value of a functor. $hx$ means $h(x)$, the corresponding object of $B$.

This identity is usually taken to be the definition of $\alpha h$.

If $\alpha$ is a natural isomorphism, then so is $\alpha h$. This could be verified on objects; $(\alpha h)x = \alpha (hx)$; since $\alpha$ is a natural isomorphism, $\alpha(hx)$ is an isomorphism.

Alternatively, the interchange law could be used:

$$ (\alpha h) \cdot (\alpha^{-1} h) = (\alpha \cdot \alpha^{-1}) (h \cdot h) = \mathrm{id}_gh $$

where I've used the convention that functors vertically compose as if they were the corresponding identity natural transformations. We can see that $\mathrm{id}_g h = \mathrm{id}_{gh}$. Similarly, $(\alpha^{-1} h) \cdot (\alpha h) = \mathrm{id}_fh$, so $\alpha^{-1} h$ is the inverse of $\alpha h$.

(with the above conventions in place, it is somewhat more convenient to write $\alpha \cdot \alpha^{-1} = g$ rather than $\alpha \cdot \alpha^{-1} = \mathrm{id}_g$)