A natural transformation $\alpha \colon F\Longrightarrow G$ is a natural isomorphism if there exists a natural transformation $\alpha ^{-1}\colon G\Longrightarrow F$ such that
Let $\mathcal{C}$ and $\mathcal{D}$ be categories and let $F,G\colon \mathcal{C}\rightrightarrows \mathcal{D}$ be functors.
A natural transformation $\alpha \colon F\Longrightarrow G$ is a natural isomorphism if there exists a natural transformation $\alpha ^{-1}\colon G\Longrightarrow F$ such that
Let $\alpha \colon F\Longrightarrow G$ and $\beta \colon G\Rightarrow H$ be natural transformations.
Characterisations. The following conditions are equivalent:
Componentwise Inverses of Natural Transformations Assemble Into Natural Transformations. Let $\alpha ^{-1}\colon G\Longrightarrow F$ be a transformation such that, for each $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, we have
Then $\alpha ^{-1}$ is a natural transformation.
Interaction With Vertical Composition. If $\alpha $ and $\beta $ are natural isomorphisms, then so is $\beta \circ \alpha $.
Interaction With Horizontal Composition. Let $\alpha \colon F_{1}\Rightarrow G_{1}$ and $\beta \colon F_{2}\Rightarrow G_{2}$ be natural transformations.
If $\alpha $ and $\beta $ are natural isomorphisms, then so is $\beta \mathbin {\star }\alpha $.
If $\alpha $ is an isomorphism, then so is $\operatorname {\mathrm{id}}_{G_{1}}\mathbin {\star }\alpha \colon G_{1}\circ F_{1}\Rightarrow G_{1}\circ F_{2}$.
If $\beta $ is an isomorphism, then so is $\beta \mathbin {\star }\operatorname {\mathrm{id}}_{F_{1}}\colon G_{1}\circ F_{1}\Rightarrow G_{2}\circ F_{1}$.
Postcomposing both sides with $\alpha ^{-1}_{B}$, we get
which is the naturality condition we wanted to show. Thus $\alpha ^{-1}$ is a natural transformation.