## Elementary Category Theory

Recently I started learning CT, and I'm trying to express the property "set A is contained within set B" (in the Set Category) using CT language only, but I'm quite stuck.

This bothers me for some time now, so I was wondering if someone here has any leads?

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### CT Subset

Have you tried "there is a monomorphism from A to B" as a starting point?

If you are interested in a CT version of Set Theory, try "Conceptual Mathematics" by Lawvere and Schanuel or "Sets for Mathematics" by Lawvere and Rosebrugh.

### UTI

If I'm not wrong, this is the same as saying that there's a one-to-one function from A to B. What I can't get is how to ensure that f(A) = A.

Since I'm only a novice the choice of words may be lame, but it seems that I'm stuck at the "up-to-isomorphism" part.

Let A = {1, 2}, A' = {a, b}, B = {1, 2, 3}

Since we're talking in the function level, what's the difference between A and A'? We can always compose with the isomorphism {(1, a), (2, b)}

It seems to me that every property I try to use suffers from the same problem, how do I know that the monomorphism f keeps the elements within A and not in some isomorphic set?

### Iso Blues

how do I know that the monomorphism f keeps the elements within A and not in some isomorphic set?

Basically, the answer is that it doesn't matter, in CT or (with a little more subtlety) in Set Theory.

CT focuses on the mappings (morphisms) betweem objects rather than their contents. One of the insights one gets from the category Set is that the mapping behaviour of a set is determined up to iso by its cardinality.

In pure Set Theory (i.e. without distinguished ur-elements) this is equally true; whether the elements are labelled {a, b} or {1, 2} doesn't really change the structure of the set.

If you really needed to make a distinction for some reason, you could choose a distinguished set of labels and specify the mappings you want that preserve the labellings. Any set of labellings will still be iso any other with the same cardinality though.

### Moot

This is actually a subtle question and, as Marc says, it doesn't really matter in practice, so don't get hung up on it. If you describe a subset in pure CT', and then continue to use pure CT to manipulate your subsets (i.e., don't break the abstraction layer), then the question is moot anyway, since you can't distinguish isomorphic obejcts in pure CT (for a good definition of pure CT' :).

About the closest you can get in pure CT and using a generic notion of Set Theory (and, possibly intuitionism and Choice are also relevant), is that a subset is iso to a subobject', which is an equivalence class of monos. You can find this notion in Barr and Wells.

BTW, are your elements 1, 2, a and b ur-elements? Or are they encoded as sets? If the latter, you could force certain isos to be equality by representing 1 in the same way as a, for example. But you can't force every iso in Set to be an equality without breaking stuff like the Axiom of Foundation, I think.

### Look up "equalizers"

Everyone else has talked about the very important issue of categorical thinking being up to isomorphism, but no one has shown you the relevant construction after you internalize that. I'll try to remedy that gap.

So, let's suppose suppose that we're in the category Set. Your question is, how do we talk about subset relationships (E is a subset of A) categorically? In the categorical style, we very rarely do things directly with objects, preferring to work with arrows between them. So the obvious thing to look for is an arrow that contains the information we want, and functions in Set are defined on the elements of a set. So, how do you you say "E is a subset of A" in terms of the elements of E and A? Well, that's the same as saying "forall x. (x in E) => (x in A)". Reading that as a function, the existence of an inclusion i: E -> A which takes each element of E to the same element in A means that E is a subset of A.

This is actually not the direction category theorists normally work from, because it makes the construction of the relevant universal property seem obscure and mysterious. So I will advise you to look up equalizers, and try to read the definition thinking of an equalizer as an inclusion function.

### But what did equalizers add?

Since the equalizer is also unique up to isomorphism, I could have achieved the same thing with monomorphisms alone.

I understand that in CT differences up to isomorphism are neglectable, but the point of the excercise wasn't to use the construction but to excercise constructing. I'm still wondering if you can construct the argument "A is contained within B" in CT (even though it's pointless).

### Equalizers exactly express the property you want....

Instead, let's suppose that we are not allowed to talk about the elements of sets, but only about the functions and their properties. Now, with this restriction on our language, how do we express the idea that A is contained in B?

Well, suppose you've got a function h: A -> C with A as its domain. What happens if we try to extend it into a function B -> Cwith B as its domain? Obviously, this is an underconstrained choice -- since B is bigger you have many possible functions that agree on the subset A. Let's pick two of them and call them f and g. So, if you have an inclusion i: A -> B that projects values of A into B, then f and g have to be identical over that shared domain. IOW, f o i == g o i.

So, turning that around: the equalizer of two functions is the subset of their domain on which they agree, plus the inclusion function from the subset to the larger set. That's how you say "A is a subset of B" in category theory.

The idea of up-to-isomorphism shows up when you look at the universal property of this construction. Here's a picture:

  E --i--> A ===f,g===> C
|        ^
|        |
|        m
|        |
+---u--> O
`

The universal property says that, for any other object O, if you have an arrow m: O -> A, and f o m == g o m, then there is a unique arrow u: E -> O. Obviously, you can replace E with any isomorphic E', because you can compose u with the iso between E and E' to get a unique u': E' -> O. (In Set this is not very interesting, of course.)

### You're right

You're right. Strictly speaking, it's impossible in CT alone to express that A is a subset of B. CT doesn't give you any way to pick out the distinguished subobject that corresponds to subset.

### Category Theory for Dummies - slides available

For those interested in a brief introduction to category theory, James Cheney has recently posted some PDF slides titled Category Theory for Dummies on his home page.

In case anyone else goes looking, Cheney's list of presentations is now at this location; specifically, the "Category Theory for Dummies" slides are here.

### not even

Suppose we are talking about the category Set. Then the initial object {} of any set is unique in the sense of common language. There is only one initial object in Set.

In many categories this wouldn't hold up. Like Vect, where the 0 vector is in every subspace, so is initial, but the 0 vector is unique only up to isomorphism with other vector spaces. So then only knowing that there are vector spaces with linear transformations between them, it becomes impossible to establish which 0 vector is being referred to. On the other hand, the zero vector behaves the same on vector space 0 + a = a, there is an -a such that a + -a = 0,
a dot 0 = 0 (as in the real number this time), so the idea of the structure being impossible to distinguish is there.

Is my synopsis of this idea correct?

Then my next question is what would an equalizer look like on Vect?
I believe the diagram above may be incorrect as 'u' points the wrong way. Here

E--eq-->X-----f------>
^ ^-----g------> Y
| |
| |
| m
| |
+-u-----O

Then f and g are linear transformations that have some universal property being equalized by eq. If we push through this diagram starting with O, then on one hand we are have a linear transformation from O -> X, on the other hand some other linear transformation m takes us to E, which taken before eq commutes with m. Then at the same time the whole diagram commutes so taking the transfomation f of m is the same as taking the transformation g of eq of u. This on an intuitive level seems to be related to the null space of a transformation, so the kernel of a linear transformation. Saying that f and g are injective is too strong, but perhaps they are equalized by having the same kernel or the same nullity in their kernel. Am I way off here?

### so

my diagram looks nothing like what I typed. Wikipedia has better pictures anyway.