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.

    Definition 11.9.7.1.1 ▶ Natural Isomorphisms

    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*}

    Proposition 11.9.7.1.2 ▶ Properties of Natural Isomorphisms

    Let $\alpha \colon F\Longrightarrow G$ and $\beta \colon G\Rightarrow H$ be natural transformations.

    1. 1.

      Characterisations. The following conditions are equivalent:

      1. (a)

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

      2. (b)

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

    2. 2.

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

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

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

  • 3.

    Interaction With Vertical Composition. If $\alpha $ and $\beta $ are natural isomorphisms, then so is $\beta \circ \alpha $.

  • 4.

    Interaction With Horizontal Composition. Let $\alpha \colon F_{1}\Rightarrow G_{1}$ and $\beta \colon F_{2}\Rightarrow G_{2}$ be natural transformations.

    1. (a)

      If $\alpha $ and $\beta $ are natural isomorphisms, then so is $\beta \mathbin {\star }\alpha $.

    2. (b)

      If $\alpha $ is an isomorphism, then so is $\operatorname {\mathrm{id}}_{G_{1}}\mathbin {\star }\alpha \colon G_{1}\circ F_{1}\Rightarrow G_{1}\circ F_{2}$.

    3. (c)

      If $\beta $ is an isomorphism, then so is $\beta \mathbin {\star }\operatorname {\mathrm{id}}_{F_{1}}\colon G_{1}\circ F_{1}\Rightarrow G_{2}\circ F_{1}$.

    • Proof of Proposition 11.9.7.1.2.

    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 $A,B\in \operatorname {\mathrm{Obj}}(\mathcal{C})$ and each $f\in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)$. Considering the diagram
    where the boundary diagram as well as Subdiagram (2) commute, we have

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

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

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

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

    Item 3: Interaction With Vertical Composition
    This is a repetition of Item 5 of Proposition 11.9.4.1.2 and is proved there.

    Item 4: Interaction With Horizontal Composition
    This is a repetition of Item 5 of Proposition 11.9.5.1.3 and is proved there.

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