Reuben wrote:

> how do we make this definition of limit yield the notion of a product (which I understand is a kind of limit). People who have taken at least part of a category theory course may have encountered the definition of a product via a universal property ( how does this relate?

Anindya gave a very precise answer, so let me give a vague answer - sometimes it's nice to think a bit more vaguely.

The definition of product you're thinking about says roughly "a morphism from an object \\(A\\) to a product of objects \\(X \times Y\\) is the same as a morphism from \\(A\\) to \\(X\\) together with a morphism from \\(A\\) to \\(Y\\)".

(Experienced category theorists use "is the same as" to indicate a natural one-to-one correspondence obtained by a universal property. It needs to be fleshed out to be made precise, but it's nice when you're trying to talk quickly!)

In my lecture I only discussed limits in the category \\(\mathbf{Set}\\). This isn't strictly necessary, but let's stick with that case, and see how such limits can give products in the category \\(\mathbf{Set}\\).

For starters, let's say "a function from a set \\(A\\) to a product of sets \\(X \times Y\\) is the same as a function from \\(A\\) to \\(X\\) together with a function from \\(A\\) to \\(Y\\)".

Translating into a more mathematical style of talking, this says:

\[ \mathbf{Set}^2( (A, A) , (X,Y)) \cong \mathbf{Set}(A, X \times Y). \]

Get it?

If not:

* An element in the left-hand side is a morphism in \\(\mathbf{Set}^2\\) from the object \\( (A,A) \\) to the object \\( (X,Y) \\). Remember, an object in \\(\mathbf{Set}^2\\) is a pair of sets and a morphism is a pair of functions. So, an element in the left-hand side is a function from \\(A\\) to \\(X\\) together with a function from \\(A\\) to \\(Y\\).

* An element in the right-hand side is just a function from \\(A\\) to \\(X \times Y\\).

This more mathematical way of talking is nice because when you see this:

\\[ \mathbf{Set}^2( (A, A) , (X,Y)) \cong \mathbf{Set}(A, X \times Y). \\]

you should instantly guess that we're dealing with a left adjoint and a right adjoint. On the left we're taking \\(A\\) and "duplicating" it to get \\( (A,A)\\). On the right we're taking \\( (X,Y) \\) and "producting" it to get \\( X \times Y\\).

So, you should suspect that the left-hand side is talking about the "duplication" functor

\[ \Delta : \mathbf{Set} \to \mathbf{Set}^2 \]

which takes any set \\(A\\) and duplicates it to get \\( (A,A) \\),
while the right-hand side is talking about the "product" functor

\[ \times : \mathbf{Set}^2 \to \mathbf{Set} \]

which takes any pair of sets \\( (X,Y) \\) and forms their product \\( X \times Y\\).

In fact, you should suspect that

\[ \Delta : \mathbf{Set} \to \mathbf{Set}^2 \]

is the left adjoint of

\[ \times : \mathbf{Set}^2 \to \mathbf{Set} \]

And it's true!

In other words,

\[ \times : \mathbf{Set}^2 \to \mathbf{Set} \]

is the right adjoint of

\[ \Delta : \mathbf{Set} \to \mathbf{Set}^2 .\]

This is a cool, rather sophisticated way of defining the product. It was discovered by Lawvere.

But we can go a step further - perhaps too far - by thinking of this right adjoint as a right Kan extension. This is where Anindya starts his story. I'll tell it using slightly different notation. I'll let \\(\mathbf{1}\\) be the category with one object and one morphism, and \\(\mathbf{1} + \mathbf{1}\\) be the category with two objects and two morphisms. Note that

\[ \mathbf{Set} \cong \mathbf{Set}^{\mathbf{1}} \]


\[ \mathbf{Set}^2 \cong \mathbf{Set}^{\mathbf{1} + \mathbf{1}} .\]

Since \\(\mathbf{1}\\) is the terminal category there is a unique functor

\[ ! : \mathbf{1} + \mathbf{1} \to \mathbf{1} \]

so we get a functor

\[ \text{composition with } ! : \mathbf{Set}^{\mathbf{1}} \to
\mathbf{Set}^{\mathbf{1} + \mathbf{1}} \]

and if you take a look you'll see this just our old friend duplication

\[ \Delta : \mathbf{Set} \to \mathbf{Set}^2 \]

wearing fancy new clothes! The right adjoint of this functor is, by definition, right Kan extension along \\( !\\):

\[ \text{Ran}\_! : \mathbf{Set}^{\mathbf{1} + \mathbf{1}} \to \mathbf{Set}^{\mathbf{1}} \]

so this must be our old friend the right adjoint to duplication, namely

\[ \times : \mathbf{Set}^2 \to \mathbf{Set} .\]

I've been using \\(\mathbf{Set}\\) a lot here, but any category with binary products would do just as well. So, we can summarize and generalize as follows:

**Theorem.** If \\(\mathcal{C}\\) is a category with binary products, the functor \\( \times : \mathcal{C}^2 \to \mathcal{C} \\) is the right adjoint to the duplication functor \\(\Delta: \mathcal{C} \to \mathcal{C}^2\\).

We could also describe \\(\times\\) as a right Kan extension: it's just another way of saying the same thing.