Lecture 11: Adding and squishing

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In this lecture, we’re going to talk about how to compose objects in a category. A priori, this is weird; aren’t we supposed to be composing morphisms in a category, not objects? But it turns out that there is a pretty sensible way of doing this.

The idea is the following. In order to compose objects in \mathsf{C}, we have to first give a “specification” for what the properties of the composed object should be. This “specification” comes in the form of a \mathsf{C}-set (a functor from \mathsf{C} to \mathsf{Set}); recall that a \mathsf{C}-set abstractly describes the “morphisms out” of a yet-unspecified object in \mathsf{C}. We then pick an object in \mathsf{C} and show that it has the right properties, by showing that it is a representative of the \mathsf{C}-set we constructed. It turns out that it doesn’t matter the specific object we choose, because any two objects that have the same “properties” are isomorphic.

One important thing to emphasize is that given an object x and a functor F, there could be several different isomorphisms y_\mathsf{C}(x) \cong F. When we say “x is a representative for F”, we really mean that we have picked a specific isomorphism.

Coproducts of types in Julia

Let’s start with an example.

Coproducts in “a category of types” correspond to sum types, or tagged unions. In Julia:

struct Left{T}
  val::T
end

struct Right{T}
  val::T
end

const Coproduct{S,T} = Union{Left{S}, Right{T}}

Now, suppose we want to write a function foo out of Coproduct{Int,String} into Int that takes Left(x) to x^2 and Right(x) to length(x). We can use Julia’s dispatch system to help us here:

foo(l::Left{Int}) = l.val^2
foo(r::Right{String}) = length(r.val)

(foo(Left(4)), foo(Right("hello")))

(16, 5)

How did we write this function? Well, we wrote two functions, one out of Int and one out of String.

In general, a function from Coproduct{S,T} to A is given by a function from S to A and a function from T to A.

The description of coproducts as the representatives of a certain \mathsf{C}-set expresses this property exactly.

Namely, if we work in \mathsf{Jul}, the category of Julia types and functions between them, then the functor F \colon \mathsf{Jul}\to \mathsf{Set} which sends A to the set of pairs of functions S -> A and T -> A is represented by the type Coproduct{S,T}.

Note that we could have use types other than Coproduct{S,T} and gotten the same property; for instance we could have used

struct TaggedUnion{S,T}
  tag::Bool
  val::Union{S,T}
  function TaggedUnion{S,T}(tag::Bool, x::S) where {S,T}
    @assert tag == true
    new(tag, x)
  end
  function TaggedUnion{S,T}(tag::Bool, y::T) where {S,T}
    @assert tag == false
    new(tag, y)
  end
end

Different choices of representing object can have different performance characteristics, but their “external interface” is precisely the same: a map out of the coproduct is equivalent to a map out of each of the two summands.

Angles and squishing

Very often in programming, we talk about angles. We represent an angle by a real number, measured in radians or degrees. But there’s a problem: \theta and \theta + 2\pi represent the same angle! That is, if we rotate something in real life by \theta or if we rotate it by \theta + 2\pi, we get the same result at the end (assuming that it can rotate freely, and isn’t winding up a spring or something).

One solution for this is to only work with angles in the range [0,2\pi). But this is not ideal for some scenarios; for instance if we are working with a pendulum where the resting angle is 0, then it is more convenient to have angles in the range [-\pi,\pi), because the pendulum might swing forwards or backwards.

Why can we do this; why are [0,2\pi) and [-\pi,\pi) equally good representations for angles? Can we find a characterization of the properties that [0,2\pi) and [-\pi,\pi) have?

Well, let’s go back to thinking about angles as real numbers. A “well-behaved” program which takes an angle \theta as an input should return the same value on \theta and \theta + 2\pi. So now we can characterize the functions out of \mathbb{R} that are well behaved as a functor F from \mathsf{Set} to \mathsf{Set}. This functor sends a set X to the set of functions f \colon \mathbb{R}\to X such that f(\theta) = f(\theta + 2\pi) for all \theta \in \mathbb{R}, which we call the set of functions with period 2\pi.

F(X) = \{ f \colon \mathbb{R}\to X \mid \text{$f$ has period $2\pi$} \}

The action of F on morphisms g \colon X \to Y is given by

F(g)(f \colon \mathbb{R}\to X) = g \circ f \colon \mathbb{R}\to Y

One can check that if f has period 2\pi then g \circ f also does.

Then [0,2\pi) and [-\pi,\pi) are both representatives for F. More specifically, we can construct isomorphisms between y_{\mathsf{Set}}([0,2\pi)) and F, and also between y_{\mathsf{Set}}([-\pi,\pi)) and F. This is because a function out of [0,2\pi) can be extended to a function out of \mathbb{R} with period 2\pi, a function out of \mathbb{R} with period 2\pi can be restricted to [0,2\pi), and going back and forth gets to to the same place. The same holds for functions out of [-\pi,\pi).

We can now do this more abstractly.

Consider two functions 1_\mathbb{R}, p \colon \mathbb{R}\to \mathbb{R}. 1_\mathbb{R} is the identity, so 1_\mathbb{R}(\theta) = \theta, and then p(\theta) = \theta + 2\pi. A function f \colon \mathbb{R}\to X has period 2\pi if and only if f \circ 1_{\mathbb{R}} = f \circ p. Thus, we could equivalently characterize F by

F(X) = \{ f \colon \mathbb{R}\to X \mid f \circ 1_{\mathbb{R}} = f \circ p \}

The advantage of this description is that it relies purely on categorical structure, i.e. composition of morphisms. This leads to the following definition.

Let \mathsf{C} be a category, and A,B be objects in \mathsf{C}. Then the coequalizer of a pair of morphisms p,q \colon A \to B is a representing object for

F(X) = \{ f \colon B \to X \mid f \circ p = f \circ q \}

If coproducts allow you to “add” together objects in a category, coequalizers allow you to “squish down” objects in a category, like how we squished \{\ldots, \theta - 2\pi, \theta, \theta + 2\pi, \ldots\} to a single point because they all represent the same angle.

It’s kind of inconvenient that whenever we have a representative of a functor, we have to lug around a whole natural isomorphism. It turns out that for coequalizers, we can get by with much less data.

If C is the coequalizer of p,q \colon A \to B, then there is a map e \colon B \to C such that for any map f \colon B \to X with f \circ p = f \circ q, there is a unique map \tilde{f} \colon C \to X such that

commutes.

Moreover, any such map e gives rise to an isomorphism between the functor

F(X) = \{ f \colon B \to X \mid f \circ p = f \circ q \}

and y_{\mathsf{C}}(C).

We call e the coequalizing morphism.

Proof. The basic idea is that the map e \colon B \to C corresponds to the identity 1_C \in y_\mathsf{C}(C)(C) when we take the isomorphism F(C) \cong y_{\mathsf{C}}(C)(C).

We will return to this in more detail in a future lecture.

So when we give an instance of a coequalizer, we just have to give an object and coequalizing morphism, which is often more convenient than writing out the whole natural isomorphism.

Returning back to angles, there’s a third representative for the coequalizer of 1_\mathbb{R} and p: the circle S^1 defined by

S^{1} = \{z \in \mathbb{C} \mid |z| = 1 \}

This is “nicer” because there’s no “jump”, i.e. when we move off the right side of the interval [0,2\pi), we jump back to the left side, but in the circle we just smoothly move around. The coequalizing morphism is the map e \colon \mathbb{R}\to S^{1}, given by \theta \mapsto e^{i\theta}.

As sets, there’s no difference between [0,2\pi) and S^{1}. However, if we instead consider the category of \mathsf{Top} of topological spaces and continuous maps between them (continuous roughly means “no jumps”), then S^{1} and [0,2\pi) are not the same!

Specifically, the map \mathbb{R}\to [0,2\pi) is not continuous, because we have the “jump” from 2\pi-\epsilon to 0. However, the map \mathbb{R}\to S^{1} is. So the coequalizer of 1_\mathbb{R} and p in \mathsf{Top} is S^{1}, not [0,2\pi).

Examples to think about if we have time

In \mathsf{Mat}, the category where the objects are natural numbers and the morphisms are matrices, consider the following setup, where M is an arbitrary matrix and 0 is the zero matrix.

Recall that the rank of M is the dimension of the range of M, and let k be the rank. Then the coequalizer of this diagram is m - k, with the following coequalizing matrix m \to (m - k).

Let v_{1},\ldots,v_{k} be an orthonormal basis for the range of M. Then extend this to an orthonormal basis v_{1},\ldots,v_{m} for all of \mathbb{R}^{m}. Then define a matrix E \colon m \to (m-k) by using the row vectors v_{k+1},\ldots,v_{m}.

By orthogonality, E(\alpha_{1} v_{1} + \cdots + \alpha_{k} v_{k}) = 0. Thus, EM = 0 = E0. Moreover, any other matrix E' \colon m \to l such that E'M = 0 factors through E.

If we have time, we will prove this and also livecode a Julia implementation using the LinearAlgebra library to find orthonormal bases.