Understanding the definition of a natural isomorphism

Question

On Wikipedia I can read this:

If, for every object $X$ in $C$, the morphism $η_X$ is an isomorphism in $D$, then $η$ is said to be a natural isomorphism (or sometimes natural equivalence or isomorphism of functors).

But I don't understand this. I thought that, if $\eta: F \to G$ is a natural transformation; $F, G: C \to D$; and $X \in C$, this implies that $\eta_X \in C$. So I don't understand, how we can talk about $\eta_X$ is an isomorphism in $D$.

EDIT:

So does "$\eta$ is natural isomorphism" only imply that $\exists \varepsilon. \forall \eta_X. \exists \varepsilon_X. \eta_X \circ \varepsilon_X = 1_{G(X)} \wedge \varepsilon_X \circ \eta_X = 1_{F(X)}$, where $\circ$ is vertical composition of natural transformations? So is a natural isomorphism just a natural transformation, that happens to be an isomorphism (in the category of functors)?

Answer 1

A natural transformation $\eta$ maps objects of a category $\mathcal C$ to arrows of a category $\mathcal D$.
Specifically, if $F,G:\mathcal C\to\mathcal D,\ \ x\in Ob\mathcal C\ $ and $\eta:F\to G$, we have $\eta_x$ is an arrow $F(x)\to G(x)$ in $\mathcal D$.

A natural transformation $\eta$ is a natural isomorphism if

• each component $\eta_x$ is an isomorphism (invertible arrow in $\mathcal D$)
• equivalently, if $\eta$ itself is an isomorphism (invertible arrow) in the category $Fun(\mathcal C,\mathcal D)$ of functors $\mathcal C\to\mathcal D$ with natural transformations as arrows.

To see the equivalence, observe that composition of natural transformations in $Fun(\mathcal C,\mathcal D)$ is done 'componentwise', i.e. $(\eta\circ\psi)_x=\eta_x\circ\psi_x$, so that the inverse can also be taken componentwise.