# The Universe of Discourse

Sat, 08 Jun 2019

I have pondered category theory periodically for the past 35 years, but not often enough to really stay comfortable with it. Today I was pondering again. I wanted to prove that !!1×A \cong A!! and I was having trouble. I eventually realized my difficulty: my brain had slipped out of category theory mode so that the theorem I was trying to prove did not even make sense.

In most of mathematics, !!1\times A!! would denote some specific entity and we would then show that that entity had a certain property. For example, in set theory we might define !!1\times A!! to be some set, say the set of all Kuratowski pairs !!\langle \varnothing, a\rangle!! where !!a\in A!!:

$$1×A =_{\text{def}} \{ z : \exists a\in A : z = \{\{\varnothing\}, \{\varnothing, a\}\} \}$$

and then we would explicitly construct a bijection !!f:A\leftrightarrow 1×A!!:

$$f(a) = \{\{\varnothing\}, \{\varnothing, a\}\}$$

In category theory, this is not what we do. Everything is less concrete. !!\times!! looks like an operator, one that takes two objects and yields a third. It is not. !!1\times A!! does not denote any particular entity with any particular construction. (Nor does !!1!!, for that matter.) Instead, it denotes an unspecified entity, which happens to have a certain universal property, perhaps just one of many such entities with that property, and there is no canonical representative of the class. It's a mistake to think of it as a single thing, and it's also a mistake to ask the question the way I did ask it. You can't show that !!1×A!! has any specific property, because it's not a specific thing. All you can do is show that anything with the one required universal property must also have the other property. We should rephrase the question like this:

Let !!1×A!! be a product of !!1!! and !!A!!. Then !!1×A\cong A!!.

Maybe a better phrasing is:

Let !!1×A!! be some object that is a product of !!1!! and !!A!!. Then !!1×A\cong A!!.

The notation is still misleading, because it looks like !!1×A!! denotes the result of some operation, and it isn't. We can do a little better:

Let !!B!! be a product of !!1!! and !!A!!. Then !!B\cong A!!.

That it, that's the real theorem. It seems at first to be more difficult — where do we start? But it's actually easier! Because now it's enough to simply prove that !!A!! itself is a product of !!1!! and !!A!!, which is easily done: its projection morphisms are evidently !!! !! and !!{\text{id}}_A!!. And by a previous theorem that all products are isomorphic, any other product, such as !!B!!, must be isomorphic to this one, which is !!A!! itself.

(We can similarly imagine that any theorem that mentions !!1!! is preceded by the phrase “Let !!1!! be some terminal object.”)