11.9.7 Natural Isomorphisms

    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

    \begin{align*} \alpha ^{-1}\circ \alpha =\operatorname {\mathrm{id}}_{F},\\ \alpha \circ \alpha ^{-1}=\operatorname {\mathrm{id}}_{G}. \end{align*}

    Let $\alpha \colon F\Longrightarrow G$ be a natural transformation.

    1. 1.

      Characterisations. The following conditions are equivalent:

      1. (a)

        The natural transformation $\alpha $ is a natural isomorphism.

      2. (b)

        For each $A\in \operatorname {\mathrm{Obj}}\webleft (\mathcal{C}\webright )$, the morphism $\alpha _{A}\colon F_{A}\to G_{A}$ is an isomorphism.

  • 2.

    Componentwise Inverses of Natural Transformations Assemble Into Natural Transformations. Let $\alpha ^{-1}\colon G\Longrightarrow F$ be a transformation such that, for each $A\in \operatorname {\mathrm{Obj}}\webleft (\mathcal{C}\webright )$, we have

    \begin{align*} \alpha ^{-1}_{A}\circ \alpha _{A} & = \operatorname {\mathrm{id}}_{F\webleft (A\webright )},\\ \alpha _{A}\circ \alpha ^{-1}_{A} & = \operatorname {\mathrm{id}}_{G\webleft (A\webright )}. \end{align*}

    Then $\alpha ^{-1}$ is a natural transformation.

  • Item 1: Characterisations
    The implication Item 1a$\implies $Item 1b is clear, whereas the implication Item 1b$\implies $Item 1a follows from Item 2.

    Item 2: Componentwise Inverses of Natural Transformations Assemble Into Natural Transformations
    The naturality condition for $\alpha ^{-1}$ corresponds to the commutativity of the diagram
    for each $A,B\in \operatorname {\mathrm{Obj}}\webleft (\mathcal{C}\webright )$ and each $f\in \operatorname {\mathrm{Hom}}_{\mathcal{C}}\webleft (A,B\webright )$. Considering the diagram
    where the boundary diagram as well as Subdiagram (2) commute, we have

    \begin{align*} G\webleft (f\webright ) & = G\webleft (f\webright )\circ \operatorname {\mathrm{id}}_{G\webleft (A\webright )}\\ & = G\webleft (f\webright )\circ \alpha _{A}\circ \alpha ^{-1}_{A}\\ & = \alpha _{B}\circ F\webleft (f\webright )\circ \alpha ^{-1}_{A}. \end{align*}

    Postcomposing both sides with $\alpha ^{-1}_{B}$, we get

    \begin{align*} \alpha ^{-1}_{B}\circ G\webleft (f\webright ) & = \alpha ^{-1}_{B}\circ \alpha _{B}\circ F\webleft (f\webright )\circ \alpha ^{-1}_{A}\\ & = \operatorname {\mathrm{id}}_{F\webleft (B\webright )}\circ F\webleft (f\webright )\circ \alpha ^{-1}_{A}\\ & = F\webleft (f\webright )\circ \alpha ^{-1}_{A}, \end{align*}

    which is the naturality condition we wanted to show. Thus $\alpha ^{-1}$ is a natural transformation.


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