Item 4 (018B) — The Clowder Project

11.9.5 Horizontal Composition of Natural Transformations

Definition 11.9.5.1.1 ▶ Horizontal Composition of Natural Transformations

The horizontal composition^1,2 of two natural transformations $\alpha \colon F\Longrightarrow G$ and $\beta \colon H\Longrightarrow K$ as in the diagram

of $\alpha $ and $\beta $ is the natural transformation

\[ \beta \mathbin {\star }\alpha \colon (H\circ F)\Longrightarrow (K\circ G), \]

as in the diagram

consisting of the collection

\[ \left\{ (\beta \mathbin {\star }\alpha )_{A} \colon H(F(A)) \to K(G(A)) \right\} _{A\in \operatorname {\mathrm{Obj}}(\mathcal{C})}, \]

of morphisms of $\mathcal{E}$ with

¹Further Terminology: Also called the Godement product of $\alpha $ and $\beta $.
²Horizontal composition forms a map
\[ \mathbin {\star }_{(F,H),(G,K)}\colon \operatorname {\mathrm{Nat}}(H,K)\times \operatorname {\mathrm{Nat}}(F,G)\to \operatorname {\mathrm{Nat}}(H\circ F,K\circ G). \]

Proof of Definition 11.9.5.1.1.

First, we claim that we indeed have

This is, however, simply the naturality square for $\beta $ applied to the morphism $\alpha _{A}\colon F(A)\to G(A)$. Next, we check the naturality condition for $\beta \mathbin {\star }\alpha $, which is the requirement that the boundary of the diagram

commutes. Since

•
Subdiagram (1) commutes by the naturality of $\alpha $.
•
Subdiagram (2) commutes by the naturality of $\beta $.

so does the boundary diagram. Hence $\beta \circ \alpha $ is a natural transformation.¹

¹Reference: Proposition 1.3.4 of [Borceux, Handbook of Categorical Algebra I].

Definition 11.9.5.1.2 ▶ Whiskering of Functors With Natural Transformations

Let

be a diagram in $\mathsf{Cats}_{\mathsf{2}}$.

1.
The left whiskering of $\alpha $ with $G$ is the natural transformation¹

\[ \operatorname {\mathrm{id}}_{G}\star \alpha \colon G\circ \phi \Longrightarrow G\circ \psi . \]
2.
The right whiskering of $\alpha $ with $F$ is the natural transformation²

\[ \alpha \star \operatorname {\mathrm{id}}_{F}\colon \phi \circ F\Longrightarrow \psi \circ F. \]

¹Further Notation: Also written $G\alpha $ or $G\mathbin {\star }\alpha $, although we won’t use either of these notations in this work.
²Further Notation: Also written $\alpha F$ or $\alpha \mathbin {\star }F$, although we won’t use either of these notations in this work.

Proposition 11.9.5.1.3 ▶ Properties of Horizontal Composition of Natural Transformations

Let $\mathcal{C}$, $\mathcal{D}$, and $\mathcal{E}$ be categories.

1.
Functionality. The assignment $(\beta ,\alpha )\mapsto \beta \mathbin {\star }\alpha $ defines a function

\[ \mathbin {\star }_{(F,G),(H,K)}\colon \operatorname {\mathrm{Nat}}(H,K)\times \operatorname {\mathrm{Nat}}(F,G)\to \operatorname {\mathrm{Nat}}(H\circ F,K\circ G). \]
2.
Associativity. Let
be a diagram in $\mathsf{Cats}_{\mathsf{2}}$. The diagram
commutes, i.e. given natural transformations
we have

\[ (\gamma \mathbin {\star }\beta )\mathbin {\star }\alpha =\gamma \mathbin {\star }(\beta \mathbin {\star }\alpha ). \]
3.
Interaction With Identities. Let $F\colon \mathcal{C}\to \mathcal{D}$ and $G\colon \mathcal{D}\to \mathcal{E}$ be functors. The diagram
commutes, i.e. we have

\[ \operatorname {\mathrm{id}}_{G}\mathbin {\star }\operatorname {\mathrm{id}}_{F}=\operatorname {\mathrm{id}}_{G\circ F}. \]

4.
Middle Four Exchange. Let $F_{1},F_{2},F_{3}\colon \mathcal{C}\to \mathcal{D}$ and $G_{1},G_{2},G_{3}\colon \mathcal{D}\to \mathcal{E}$ be functors. The diagram
commutes, i.e. given a diagram
in $\mathsf{Cats}_{\mathsf{2}}$, we have

\[ (\beta '\mathbin {\star }\alpha ')\circ (\beta \mathbin {\star }\alpha )=(\beta '\circ \beta )\mathbin {\star }(\alpha '\circ \alpha ). \]

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

(a)
If $\alpha $ and $\beta $ are natural isomorphisms, then so is $\beta \mathbin {\star }\alpha $.
(b)
If $\alpha $ is an isomorphism, then so is $\operatorname {\mathrm{id}}_{G_{1}}\mathbin {\star }\alpha \colon G_{1}\circ F_{1}\Rightarrow G_{1}\circ F_{2}$.
(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 11.9.5.1.3.

Item 1: Functionality

There is nothing to prove.

Item 2: Associativity

Write $G_{2}G_{1}\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}G_{2}\circ G_{1}$, and similarly for other compositions. For each $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, we have

\begin{align*} ((\gamma \mathbin {\star }\beta )\mathbin {\star }\alpha )_{A} & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\gamma \mathbin {\star }\beta )_{G_{1}(A)}\circ F_{3}F_{2}(\alpha _{A})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\gamma _{G_{2}G_{1}(A)}\circ F_{3}(\beta _{G_{1}(A)}))\circ F_{3}F_{2}(\alpha _{A})\\ & = \gamma _{G_{2}G_{1}(A)}\circ F_{3}(\beta _{G_{1}(A)})\circ F_{3}F_{2}(\alpha _{A})\\ & = \gamma _{G_{2}G_{1}(A)}\circ F_{3}(\beta _{G_{1}(A)}\circ F_{2}(\alpha _{A}))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\gamma _{G_{2}G_{1}(A)}\circ F_{3}((\beta \mathbin {\star }\alpha )_{A})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\gamma \mathbin {\star }(\beta \mathbin {\star }\alpha ))_{A}. \end{align*}

Thus we have $(\gamma \mathbin {\star }\beta )\mathbin {\star }\alpha =\gamma \mathbin {\star }(\beta \mathbin {\star }\alpha )$, showing horizontal composition to be associative.

Item 3: Interaction With Identities

We have

\begin{align*} (\operatorname {\mathrm{id}}_{G}\mathbin {\star }\operatorname {\mathrm{id}}_{F})_{A} & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\operatorname {\mathrm{id}}_{G})_{F_{A}}\circ G_{(\operatorname {\mathrm{id}}_{F})_{A}}\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{id}}_{G_{F_{A}}}\circ G_{\operatorname {\mathrm{id}}_{F_{A}}}\\ & = \operatorname {\mathrm{id}}_{G_{F_{A}}}\circ \operatorname {\mathrm{id}}_{G_{F_{A}}}\\ & = \operatorname {\mathrm{id}}_{G_{F_{A}}}\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\operatorname {\mathrm{id}}_{G\circ F})_{A} \end{align*}

for each $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, showing the desired equality.

Item 4: Middle Four Exchange

Let $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$ and consider the diagram

The top composition

is given by $((\beta '\circ \beta )\mathbin {\star }(\alpha '\circ \alpha ))_{A}$, while the bottom composition

is given by $((\beta '\mathbin {\star }\alpha ')\circ (\beta \mathbin {\star }\alpha ))_{A}$. Now, Subdiagram (1) corresponds to the naturality condition

for $\beta \colon G_{1}\Longrightarrow G_{2}$ at $\alpha '_{A}\colon F_{2}(A)\to F_{3}(A)$, and thus commutes. Thus we have

\[ ((\beta '\circ \beta )\mathbin {\star }(\alpha '\circ \alpha ))_{A} = ((\beta '\mathbin {\star }\alpha ')\circ (\beta \mathbin {\star }\alpha ))_{A} \]

for each $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$ and therefore

\[ (\beta '\mathbin {\star }\alpha ')\circ (\beta \mathbin {\star }\alpha )=(\beta '\circ \beta )\mathbin {\star }(\alpha '\circ \alpha ). \]

This finishes the proof.

Item 5: Interaction With Natural Isomorphisms

Omitted.

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