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.
> 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.
Version 2.1.8p2
