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 (https://en.wikipedia.org/wiki/Product_(category_theory)): 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

$\mathbf{Set} \cong \mathbf{Set}^{\mathbf{1}}$

and

$\mathbf{Set}^2 \cong \mathbf{Set}^{\mathbf{1} + \mathbf{1}} .$

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.