# Joyal's CatLab Functors of two variables

### Context

#### Basic category theory

Basic Category Theory

Category theory

# Contents

## Product of categories

The product of two categories $\mathbf{U}$ and $\mathbf{V}$ is the category $\mathbf{U}\,\times\, \mathbf{V}$ defined as follows:

• An object of the category $\mathbf{U}\,\times\, \mathbf{V}$ is a pair of objects $(A_1,A_2)\in Ob(\mathbf{U})\,\times\, Ob(\mathbf{V})$;

• A morphism $(A_1,A_2)\to (B_1,B_2)$ is a pair of morphisms $(u_1,u_2)\in \mathbf{U}(A_1,B_1)\,\times\, \mathbf{V}(A_2,B_2)$;

• If $(u_1,u_2):(A_1,A_2)\to (B_1,B_2)$ and $(v_1,v_2):(B_1,B_2)\to (C_1,C_2)$, then

$(v_1,v_2)(u_1,u_2)=(v_1u_1,v_2u_2):(A_1,A_2)\to (C_1,C_2).$

## Functors of two variables

• If $\mathbf{U},\mathbf{V}$ and $\mathbf{W}$ are categories, we shall often say that a functor $F:\mathbf{U}\,\times\, \mathbf{V} \to \mathbf{W}$ is a functor of two variables.

### Example

• The cartesian product $A\times B$ of two sets $A$ and $B$ is a functor of two variables $\mathbf{Set}\,\times \,\mathbf{Set}\to \mathbf{Set}$. The functor takes a pair of maps $f_1:A_1\to B_1$ and $f_2:A_2\to B_2$ to the map
$f_1\times f_2:A_1\times A_2\to B_1\times B_2$

defined by putting

$(f_1\times f_2)(x_1,x_2)=(f_1(x_1),f_2(x_2))$

for every $(x_1,x_2)\in A_1\times A_2$. If $g_1:B_1\to C_1$ and $g_2:B_2\to C_2$, then

$(g_1\times g_2)(f_1\times f_2)=(g_1f_1)\times (g_2f_2).$

## Notation

• From a functor of two variables $F:\mathbf{U}\,\times\, \mathbf{V} \to \mathbf{W}$ we obtain a functor of one variable
$F(A,-): \mathbf{V} \to \mathbf{W}$

for each object $A\in \mathbf{U}$ and a functor of one variable

$F(-,B): \mathbf{U} \to \mathbf{W}$

for each object $B\in \mathbf{V}$. By definition, the functor $F(A,-)$ takes the a morphism $g:B_1\to B_2$ in $\mathbf{V}$ to the morphism $F(1_A,g):F(A,B_1)\to F(A,B_2).$ Similarly, the functor $F(-,B)$ takes a morphism $f:A_1\to A_2$ in $\mathbf{U}$ to the morphism $F(f,1_B):F(A_1,B)\to F(A_2,B)$. If $f:A_1\to A_2$ is a morphism in $\mathbf{U}$ and $g:B_1\to B_2$ is a morphism in $\mathbf{V}$, then the following square commutes in the category $\mathbf{U}\times \mathbf{V}$ and its diagonal is the morphism $(f,g):(A_1,B_1)\to (A_2,B_2)$,

(1) 

Hence the following square commutes, square

(2) 

and we have

$F(f,g)=F(A_2,g)F(f,B_1)=F(f,B_2)F(A_1,g).$

## Exercise

Show that a functor of two variables $F:\mathbf{U}\,\times\, \mathbf{V} \to \mathbf{W}$ can be described as follows:

• for each pair of objects $A\in \mathbf{U}$ and $B\in \mathbf{V}$ we have an object $F(A,B)\in \mathbf{W}$;

• for each object $A\in \mathbf{U}$ we have a functor $F(A,-):\mathbf{V}\to \mathbf{W}$;

• for each object $B\in \mathbf{V}$ we have a functor $F(-,B):\mathbf{U}\to \mathbf{W}$;

such that the square (2) commutes for each pair of morphisms $(f,g):(A_1,B_1)\to (A_2,B_2)$ in $\mathbf{U}\,\times\, \mathbf{V}$.

Revised on November 20, 2020 at 16:38:24 by Dmitri Pavlov