Subsection 5.3.2 (01Q3): The Right Distributor

The right distributor of the product of sets over the coproduct of sets is the natural isomorphism

\[ \delta ^{\mathsf{Sets}}_{r} \colon \mathord {\times }\circ (\mathord {\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}}\times \operatorname {\mathrm{id}}_{\mathsf{Sets}}) \mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}\mathord {\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}}\circ (\mathord {(\times )}\times \mathord {(\times )})\circ \mathbf{\mu }^{\mathsf{Cats}_{\mathsf{2}}}_{4|\mathsf{Sets},\mathsf{Sets},\mathsf{Sets},\mathsf{Sets}}\circ ((\operatorname {\mathrm{id}}_{\mathsf{Sets}}\times \operatorname {\mathrm{id}}_{\mathsf{Sets}})\times \Delta _{\mathsf{Sets}}) \]

as in the diagram

whose component

\[ \delta ^{\mathsf{Sets}}_{r|X,Y,Z}\colon (X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y)\times Z\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }(X\times Z)\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}(Y\times Z) \]

at $(X,Y,Z)$ is defined by

\[ \delta ^{\mathsf{Sets}}_{r|X,Y,Z}(a,z)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} (0,(x,z)) & \text{if $a=(0,x)$,}\\ (1,(y,z)) & \text{if $a=(1,y)$} \end{cases} \]

for each $(a,z)\in (X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y)\times Z$.

Invertibility

The inverse of $\delta ^{\mathsf{Sets}}_{r|X,Y,Z}$ is the map

\[ \delta ^{\mathsf{Sets},-1}_{r|X,Y,Z}\colon (X\times Z)\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}(Y\times Z) \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }(X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y)\times Z \]

given by

\[ \delta ^{\mathsf{Sets},-1}_{r|X,Y,Z}(a)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} ((0,x),z) & \text{if $a=(0,(x,z))$,}\\ ((1,y),z) & \text{if $a=(1,(y,z))$} \end{cases} \]

for $a\in (X\times Z)\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}(Y\times Z)$. Indeed:

•
Invertibility I. The map $\delta ^{\mathsf{Sets},-1}_{r|X,Y,Z} \circ \delta ^{\mathsf{Sets}}_{r|X,Y,Z}$ acts on elements as

\begin{align*} & \mkern 40mu\mathclap {((0,x),z)}\mkern 40mu \mapsto \mkern 40mu\mathclap {(0,(x,z))}\mkern 40mu \mapsto \mkern 40mu\mathclap {(0,(x,z))}\mkern 35mu\mathrlap {,}\\ & \mkern 40mu\mathclap {((1,y),z)}\mkern 40mu \mapsto \mkern 40mu\mathclap {(1,(y,z))}\mkern 40mu \mapsto \mkern 40mu\mathclap {(1,(y,z))}\mkern 35mu\mathrlap {,} \end{align*}

but these are the two possible cases for elements of $(X \mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y) \times Z$. Hence the map is equal to the identity.
•
Invertibility II. The map $\delta ^{\mathsf{Sets}}_{r|X,Y,Z} \circ \delta ^{\mathsf{Sets},-1}_{r|X,Y,Z}$ acts on elements as

\begin{align*} & \mkern 40mu\mathclap {(0,(x,z))}\mkern 40mu \mapsto \mkern 40mu\mathclap {((0,x),z)}\mkern 40mu \mapsto \mkern 40mu\mathclap {(0,(x,z))}\mkern 35mu\mathrlap {,}\\ & \mkern 40mu\mathclap {(1,(y,z))}\mkern 40mu \mapsto \mkern 40mu\mathclap {((1,y),z)}\mkern 40mu \mapsto \mkern 40mu\mathclap {(1,(y,z))}\mkern 35mu\mathrlap {,} \end{align*}

but these are the two possible cases for elements of $(X \times Z) \mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}(Y \times Z)$. Hence the map is equal to the identity.

So $\delta ^{\mathsf{Sets}}_{r|X,Y,Z}$ is an isomorphism for all $X,Y,Z$.

Naturality

We need to show that, given functions

\begin{align*} f & \colon X \to X',\\ g & \colon Y \to Y',\\ h & \colon Z \to Z’ \end{align*}

the diagram

commutes. Indeed, this diagram acts on elements as

so it commutes and $\delta ^{\mathsf{Sets}}_{r}$ is a natural transformation.

Being a Natural Isomorphism

Since $\delta ^{\mathsf{Sets}}_{r}$ is natural and $\delta ^{\mathsf{Sets},-1}_{r}$ is a componentwise inverse to $\delta ^{\mathsf{Sets}}_{r}$, it follows from Chapter 11: Categories, Item 2 of Proposition 11.9.7.1.2 that $\delta ^{\mathsf{Sets},-1}_{r}$ is also natural. Thus $\delta ^{\mathsf{Sets}}_{r}$ is a natural isomorphism.

The Clowder Project

5.3.2 The Right Distributor