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Functors of two variables

**Basic Category Theory** ## Contents * Categories * Functors * Natural transformations * Functors of two variables * Adjoint Functors and Monads

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Contents

Product of categories

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

(v 1,v 2)(u 1,u 2)=(v 1u 1,v 2u 2):(A 1,A 2)(C 1,C 2).(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

Example

(f 1×f 2)(x 1,x 2)=(f 1(x 1),f 2(x 2))(f_1\times f_2)(x_1,x_2)=(f_1(x_1),f_2(x_2))

for every (x 1,x 2)A 1×A 2(x_1,x_2)\in A_1\times A_2. If g 1:B 1C 1g_1:B_1\to C_1 and g 2:B 2C 2g_2:B_2\to C_2, then

(g 1×g 2)(f 1×f 2)=(g 1f 1)×(g 2f 2).(g_1\times g_2)(f_1\times f_2)=(g_1f_1)\times (g_2f_2).

Notation

(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).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:U×VWF:\mathbf{U}\,\times\, \mathbf{V} \to \mathbf{W} can be described as follows:

such that the square (2) commutes for each pair of morphisms (f,g):(A 1,B 1)(A 2,B 2)(f,g):(A_1,B_1)\to (A_2,B_2) in U×V\mathbf{U}\,\times\, \mathbf{V}.

Revised on July 28, 2012 at 14:43:36 by dsk