11.5.4 The Natural Transformation Associated to a Functor

    Definition 11.5.4.1.1 ▶ The Natural Transformation Associated to a Functor

    Every functor $F\colon \mathcal{C}\to \mathcal{D}$ defines a natural transformation1

    called the natural transformation associated to $F$, consisting of the collection

    \[ \left\{ F^{\dagger }_{A,B} \colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B) \to \operatorname {\mathrm{Hom}}_{\mathcal{D}}(F_{A},F_{B}) \right\} _{(A,B)\in \operatorname {\mathrm{Obj}}(\mathcal{C}^{\mathsf{op}}\times \mathcal{C})} \]

    with

    \[ F^{\dagger }_{A,B} \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}F_{A,B}. \]

    1. 1This is the $1$-categorical version of Chapter 4: Constructions With Sets, Item 1 of Proposition 4.5.3.1.3.

    Proof of Definition 11.5.4.1.1.

    The naturality condition for $F^{\dagger }$ is the requirement that for each morphism

    \[ (\phi ,\psi ) \colon (X,Y) \to (A,B) \]

    of $\mathcal{C}^{\mathsf{op}}\times \mathcal{C}$, the diagram

    acting on elements as
    commutes, which follows from the functoriality of $F$.

    Proposition 11.5.4.1.2 ▶ Properties of Natural Transformations Associated to Functors

    Let $F\colon \mathcal{C}\to \mathcal{D}$ and $G\colon \mathcal{D}\to \mathcal{E}$ be functors.

    1. 1.

      Interaction With Natural Isomorphisms. The following conditions are equivalent:

      1. (a)

        The natural transformation $F^{\dagger }\colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}\Longrightarrow {\operatorname {\mathrm{Hom}}_{\mathcal{D}}}\circ {(F^{\mathsf{op}}\times F)}$ associated to $F$ is a natural isomorphism.

      2. (b)

        The functor $F$ is fully faithful.

    2. 2.

      Interaction With Composition. We have an equality of pasting diagrams

      in $\mathsf{Cats}_{\mathsf{2}}$, i.e. we have

      \[ (G\circ F)^{\dagger }=(G^{\dagger }\mathbin {\star }\operatorname {\mathrm{id}}_{F^{\mathsf{op}}\times F})\circ F^{\dagger }. \]
  • 3.

    Interaction With Identities. We have

    \[ \operatorname {\mathrm{id}}^{\dagger }_{\mathcal{C}}=\operatorname {\mathrm{id}}_{\operatorname {\mathrm{Hom}}_{\mathcal{C}}(-_{1},-_{2})}, \]

    i.e. the natural transformation associated to $\operatorname {\mathrm{id}}_{\mathcal{C}}$ is the identity natural transformation of the functor $\operatorname {\mathrm{Hom}}_{\mathcal{C}}(-_{1},-_{2})$.

    • Proof of Proposition 11.5.4.1.2.

    Item 1: Interaction With Natural Isomorphisms
    By Item 1, Proposition 11.9.7.1.2, $F^{\dagger }$ is a natural isomorphism if each $F^{\dagger }_{A,B}\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}F_{A,B}$ is an isomorphism in $\mathsf{Sets}$. By Chapter 4: Constructions With Sets, Item 1b of Item 1 of Proposition 4.7.3.1.2, this is the same as $F_{A,B}$ being bijective, i.e. injective and surjective. By Definition 11.6.1.1.1 and Definition 11.6.2.1.1, this is the case iff $F$ is fully faithful.

    Item 2: Interaction With Composition
    Given $f\in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)$, we have

    \begin{align*} (G\circ F)^{\dagger }_{A,B}(f) & = (G\circ F)_{A,B}(f)\\ & = G_{F(A),F(B)}(F_{A,B}(f))\\ & = G^{\dagger }_{F^{\mathsf{op}}(A),F(B)}(F_{A,B}(f))\\ & = (G^{\dagger }\mathbin {\star }\operatorname {\mathrm{id}}_{F^{\mathsf{op}}\times F})_{A,B}(F_{A,B}(f))\\ & = ((G^{\dagger }\mathbin {\star }\operatorname {\mathrm{id}}_{F^{\mathsf{op}}\times F})\circ F^{\dagger })_{A,B}(f). \end{align*}

    Hence the two natural transformations must be identical.

    Item 3: Interaction With Identities
    The component $\operatorname {\mathrm{id}}^{\dagger }_{\mathcal{C}|A,B}$ for $A,B\in \mathcal{C}$ is given by the morphism

    \[ \operatorname {\mathrm{id}}_{\mathcal{C}|A,B}\colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)\to \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B), \]

    which acts as the identity. Hence $\operatorname {\mathrm{id}}^{\dagger }$ is indeed the identity natural transformation.

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