Session 26: Distributive categories and linear categories, Part 2

Section 5.31 Session 26: Distributive categories and linear categories, Part 2

Let’s pick up where I left off. My previous work on Exercise 1 is in dire need of revision.

Example 5.31.1. Exercise 1: (Part 2/2).

Continued from last week

Solution.

I’m starting to get the feeling that I’m overthinking this exercise and the real purpose was to connect the matrix product with composition. I’ve been basing my Python code around the properties of the adjacency matrix representation of a graph, but this idea really wasn’t explicitly made in the text anywhere.

If I’m just thinking of a traditional “matrix product”, I would evaluate

f \cdot g

as follows:

[\begin{matrix} f_{A X} & f_{A Y} \\ f_{B X} & f_{B Y} \end{matrix}] \cdot [\begin{matrix} g_{X U} & g_{X V} \\ g_{Y U} & g_{Y V} \end{matrix}] =

[\begin{matrix} f_{A X} \cdot g_{X U} + f_{A Y} \cdot g_{Y U} & f_{B X} \cdot g_{X U} + f_{B Y} \cdot g_{Y U} \\ f_{A X} \cdot g_{X V} + f_{A Y} \cdot g_{Y V} & f_{B X} \cdot g_{X V} + f_{B Y} \cdot g_{Y V} \end{matrix}]

My work from last week has a number of errors. Particularly, I mistakely had a duplicate

f_{X V}

where my

f_{Y V}

should have been, and the text defined

α

opposite to how I did. Specifically, they use

X \times Y \overset{α}{\to} X + Y

such that

α^{- 1} \circ α = 1_{X \times Y}

and

α \circ α^{- 1} = 1_{X + Y}

for

X + Y \overset{α^{- 1}}{\to} X \times Y .

Using that notation, can write our matrix product

f \cdot g

g \circ α \circ f .

This gives us the following:

[\begin{matrix} g_{X U} \circ α \circ f_{A X} + g_{Y U} \circ α \circ f_{A Y} & g_{X U} \circ α \circ f_{B X} + g_{Y U} \circ α \circ f_{B Y} \\ g_{X V} \circ α \circ f_{A X} + g_{Y V} \circ α \circ f_{A Y} & g_{X V} \circ α \circ f_{B X} + g_{Y V} \circ α \circ f_{B Y} \end{matrix}]

So I think the trink to this is that our category must have identity maps and an associative property. Given a product

f \cdot g = g \circ α \circ f,

we could theoretically substitute

f = α^{- 1}

g = α^{- 1} .

That gives us

α^{- 1} \cdot g = g \circ α \circ α^{1} = g

and

f \cdot α^{- 1} = α^{- 1} \circ α \circ f = f .

This would give us the following two equivalences:

[\begin{matrix} f_{A X} & f_{A Y} \\ f_{B X} & f_{B Y} \end{matrix}] = [\begin{matrix} f_{A X} & f_{A Y} \\ f_{B X} & f_{B Y} \end{matrix}] \cdot [\begin{matrix} 1_{X} & 0_{X Y} \\ 0_{Y X} & 1_{Y} \end{matrix}]

= [\begin{matrix} 1_{X} \circ α \circ f_{A X} + 0_{Y X} \circ α \circ f_{A Y} & 1_{X} \circ α \circ f_{B X} + 0_{Y X} \circ α \circ f_{B Y} \\ 0_{X Y} \circ α \circ f_{A X} + 1_{Y} \circ α \circ f_{A Y} & 0_{X Y} \circ α \circ f_{B X} + 1_{Y} \circ α \circ f_{B Y} \end{matrix}]

and

[\begin{matrix} g_{X U} & g_{X V} \\ g_{Y U} & g_{Y V} \end{matrix}] = [\begin{matrix} 1_{X} & 0_{X Y} \\ 0_{Y X} & 1_{Y} \end{matrix}] \cdot [\begin{matrix} g_{X U} & g_{X V} \\ g_{Y U} & g_{Y V} \end{matrix}]

= [\begin{matrix} g_{X U} \circ α \circ 1_{X} + g_{Y U} \circ α \circ 0_{X Y} & g_{X U} \circ α \circ 0_{Y X} + g_{Y U} \circ α \circ 1_{Y} \\ g_{X V} \circ α \circ 1_{X} + g_{Y V} \circ α \circ 0_{X Y} & g_{X V} \circ α \circ 0_{Y X} + g_{Y V} \circ α \circ 1_{Y} \end{matrix}]

Using the properties of our identity maps and zero maps, we can simplify those to the following:

[\begin{matrix} f_{A X} & f_{A Y} \\ f_{B X} & f_{B Y} \end{matrix}] = [\begin{matrix} α \circ f_{A X} + 0_{A X} & α \circ f_{B X} + 0_{B X} \\ 0_{A Y} + α \circ f_{A Y} & 0_{B Y} + α \circ f_{B Y} \end{matrix}]

[\begin{matrix} g_{X U} & g_{X V} \\ g_{Y U} & g_{Y V} \end{matrix}] = [\begin{matrix} g_{X U} \circ α + 0_{X U} & 0_{Y U} + g_{Y U} \circ α \\ g_{X V} \circ α + 0_{X V} & 0_{Y V} + g_{Y V} \circ α \end{matrix}]

Considering how we’ve defined a “sum” of maps, for any

A \overset{f}{\to} B

and

A \overset{g}{\to} B

there should be uniquely defined maps satisfying

[\begin{matrix} 1_{A A} & f \\ 0_{B A} & 1_{B B} \end{matrix}] \cdot [\begin{matrix} 1_{A A} & 0_{A B} \\ 0_{B A} & 1_{B B} \end{matrix}] = [\begin{matrix} 1_{A A} & f + 0_{A B} \\ 0_{B A} & 1_{B B} \end{matrix}] = f

and

[\begin{matrix} 1_{A A} & 0_{A B} \\ 0_{B A} & 1_{B B} \end{matrix}] \cdot [\begin{matrix} 1_{A A} & g \\ 0_{B A} & 1_{B B} \end{matrix}] = [\begin{matrix} 1_{A A} & 0_{A B} + g \\ 0_{B A} & 1_{B B} \end{matrix}] = g

In other words, we can add a zero map to any map and get the same map back. If follows that

[\begin{matrix} f_{A X} & f_{A Y} \\ f_{B X} & f_{B Y} \end{matrix}] = [\begin{matrix} α \circ f_{A X} & α \circ f_{B X} \\ α \circ f_{A Y} & α \circ f_{B Y} \end{matrix}]

and

[\begin{matrix} g_{X U} & g_{X V} \\ g_{Y U} & g_{Y V} \end{matrix}] = [\begin{matrix} g_{X U} \circ α & g_{Y U} \circ α \\ g_{X V} \circ α & g_{Y V} \circ α \end{matrix}]

are both “invariate” with respect to this map

α .

Our expression for the product above can then be simplified one step more:

[\begin{matrix} f_{A X} & f_{A Y} \\ f_{B X} & f_{B Y} \end{matrix}] \cdot [\begin{matrix} g_{X U} & g_{X V} \\ g_{Y U} & g_{Y V} \end{matrix}] =

[\begin{matrix} g_{X U} \circ α \circ f_{A X} + g_{Y U} \circ α \circ f_{A Y} & g_{X U} \circ α \circ f_{B X} + g_{Y U} \circ α \circ f_{B Y} \\ g_{X V} \circ α \circ f_{A X} + g_{Y V} \circ α \circ f_{A Y} & g_{X V} \circ α \circ f_{B X} + g_{Y V} \circ α \circ f_{B Y} \end{matrix}]

= [\begin{matrix} g_{X U} \circ f_{A X} + g_{Y U} \circ f_{A Y} & g_{X U} \circ f_{B X} + g_{Y U} \circ f_{B Y} \\ g_{X V} \circ f_{A X} + g_{Y V} \circ f_{A Y} & g_{X V} \circ f_{B X} + g_{Y V} \circ f_{B Y} \end{matrix}]

And that completes our proof.

Example 5.31.2. Exercise 2: (Part 1/?).

Prove that a category with initial and terminal objects...

Solution.

Assuming that our category has intial and terminal objects, any objects

S_{1}, S_{2}

that are initial must be isomorphic and any objects

T_{1}, T_{2}

that are terminal must be isomorphic. Without loss of generality, we can think of there being a unique objects

S, T

together with isomorphisms

S_{1} ⇆ S ⇆ S_{2}

and

T_{1} ⇆ T ⇆ T_{2} .

Since our category must also have sums and products, lets consider an arbitrary map

X \overset{f}{\to} Y .

By the definition of

S

as initial there’s precisely one map

S \to X

and by the defintion of

T

as terminal there’s precisely one map

Y \to T .

Since our category also has sums and products, consider the sum

S + X .

There should be set of 3 maps

j, j_{1}, j_{2}

satisfying the following commutative diagram:

$Commutative diagram of S+X$

Likewise, the definition of a product implies the existence of maps

p, p_{1}, p_{2}

sasifying the following:

$Commutative diagram of Y times T$

Thus, for any map

X \overset{f}{\to} Y

we have the following combined diagram:

$Commutative diagram of combined sum and product$

By the fact that there is precisely one map

S \to T,

the composition

p_{2} \circ p \circ f \circ j \circ j_{1}

must be that very same map. Since

f

can be any map, why not choose

X = Y

and

f = 1_{X} ?

That would give use the following diagram:

$Commutative diagram of combined sum and product for f = 1_X$

It follows from the uniqueness of the identity map that

1_{X} = p_{1} \circ p \circ j \circ j_{2} = 1_{X} \circ 1_{X} .

We’d also be able to define the unique map

S \to T

as the composition

p_{2} \circ p \circ j \circ j_{1} = S \to X \to T .

This all seems relevant, but considering that I haven’t invoked the definition of a “zero map” yet I must still be missing something.

I think I’m going to pause here and think about this some more.

🔗