Product V-categories

V-category product(1)

The \(\mathcal{V}\) product of two \(\mathcal{V}\) categories, \(\mathcal{X} \times \mathcal{Y}\)

This is also a \(\mathcal{V}\) category with:

  1. \(Ob(\mathcal{X}\times\mathcal{Y}) := Ob(\mathcal{X})\times Ob(\mathcal{Y})\)

  2. \((\mathcal{X} \times \mathcal{Y})((x,y),(x',y')) := \mathcal{X}(x,x') \otimes \mathcal{Y}(y,y')\)

Linked by

Cost-category product(1)
Exercise 2-75(2)

Let \(\mathcal{X} \times \mathcal{Y}\) be the \(\mathcal{V}\)-product of two \(\mathcal{V}\) categories.

  1. Check that for every object we have \(I \leq (\mathcal{X} \times \mathcal{Y})((x,y),(x,y))\)

  2. Check that for every three objects we have:

    • \((\mathcal{X} \times \mathcal{Y})((x_1,y_1),(x_2,y_2)) \otimes (\mathcal{X} \times \mathcal{Y})((x_2,y_2),(x_3,y_3)) \leq (\mathcal{X} \times \mathcal{Y})((x_1,y_1),(x_3,y_3))\)

Solution(1)
    • By axioms of \(\mathcal{V}\) categories: \(I \leq \mathcal{X}(x,x')=xx\) and \(I \leq \mathcal{Y}(y,y')=yy\)

    • By monotonicity: \(I \leq xx \land I \leq yy\) implies \(I = I \otimes I \leq xx \otimes yy\).

    • By the definition of a product category this last term can be written as \((\mathcal{X} \times \mathcal{Y})((x,y),(x,y))\)

    • By axioms of \(\mathcal{V}\) categories: \(\mathcal{X}(x_1,x_2) \otimes \mathcal{X}(x_2,x_3) \leq \mathcal{X}(x_1,x_3)\) and \(\mathcal{Y}(y_1,y_2) \otimes \mathcal{Y}(y_2,y_3) \leq \mathcal{Y}(y_1,y_3)\)

    • Therefore, by monotonicity, we have \((\mathcal{X}(x_1,x_2) \otimes \mathcal{X}(x_2,x_3)) \otimes (\mathcal{Y}(y_1,y_2) \otimes \mathcal{Y}(y_2,y_3)) \leq \mathcal{X}(x_1,x_3) \otimes \mathcal{Y}(y_1,y_3)\)

    • Desugaring the definiton of the hom-object in \(\mathcal{X}\times\mathcal{Y}\), the property we need to show is that \((\mathcal{X}(x_1,x_2) \otimes\mathcal{Y}(y_1,y_2)) \otimes (\mathcal{X}(x_2,x_3) \otimes\mathcal{Y}(y_2,y_3)) \leq (\mathcal{X}(x_1,x_3) \otimes\mathcal{Y}(y_1,y_3))\)

    • Given the associativity and commutitivity of the \(\otimes\) operator, we can rearange the order and ignore parentheses for the four terms on the LHS. Do this to yield the desired expression.

Exercise 2-78(2)
Solution(1)

The distance is the Manhattan/\(L_1\) distance: \(|5-(-1)| + |6-4| = 8\)