Dan: great!

So, if \\(F : \mathbf{Set}^2 \to \mathbf{Set}\\) is a functor that discards the second set (or function), then a right adjoint \\(R : \mathbf{Set} \to \mathbf{Set}^2\\) attempts to restore this second set by choosing a _one-element set_. Category theorists like to use "1" to mean any one-element set, so I'd say

\[ R(S) = (S, 1) \]

for any set \\(S\\) and

\[ R(f) = (f, 1_1) \]

for any function \\(f: S \to T\\), where \\(1_1\\) means the identity function from our one-element set to itself.

(I'm just restating your answer in different language, to give people a second chance to read it and think about it.)

This should be compared to something I said in the lecture: if \\(F: \mathbf{Set} \to \mathbf{1}\\) is the functor that discards a set, a right adjoint of this functor attempts to restore this set by choosing a one-element set.

By the way, this isn't quite true:

> A functor \\(F\\) is left adjoint to a functor \\(R\\) if there is a one-to-one correspondence between the morphisms \\(F(A) \to B\\) and morphisms \\(A \to R(B)\\):

> \[ \mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) \]

What's true is this:

A functor \\(F\\) is left adjoint to a functor \\(R\\) if and only if there is a _natural_ one-to-one correspondence between the morphisms \\(F(A) \to B\\) and morphisms \\(A \to R(B)\\):

\[ \mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) .\]

This is where natural transformations come into the game. But I am deliberately urging people not to worry about this too much when they're just getting started with adjoints.

What you really used, I think, is this true fact, a consequence of the other one I just stated:

If a functor \\(F\\) is left adjoint to a functor \\(R\\), then there is a one-to-one correspondence between the morphisms \\(F(A) \to B\\) and morphisms \\(A \to R(B)\\):

\[ \mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) \]

Using this you can figure out some stuff about what a left adjoint must be like if it exists. In many cases, if you get a nice answer this way, the left adjoint will then exist, because your one-to-one correspondence will turn out to be natural.

So, if \\(F : \mathbf{Set}^2 \to \mathbf{Set}\\) is a functor that discards the second set (or function), then a right adjoint \\(R : \mathbf{Set} \to \mathbf{Set}^2\\) attempts to restore this second set by choosing a _one-element set_. Category theorists like to use "1" to mean any one-element set, so I'd say

\[ R(S) = (S, 1) \]

for any set \\(S\\) and

\[ R(f) = (f, 1_1) \]

for any function \\(f: S \to T\\), where \\(1_1\\) means the identity function from our one-element set to itself.

(I'm just restating your answer in different language, to give people a second chance to read it and think about it.)

This should be compared to something I said in the lecture: if \\(F: \mathbf{Set} \to \mathbf{1}\\) is the functor that discards a set, a right adjoint of this functor attempts to restore this set by choosing a one-element set.

By the way, this isn't quite true:

> A functor \\(F\\) is left adjoint to a functor \\(R\\) if there is a one-to-one correspondence between the morphisms \\(F(A) \to B\\) and morphisms \\(A \to R(B)\\):

> \[ \mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) \]

What's true is this:

A functor \\(F\\) is left adjoint to a functor \\(R\\) if and only if there is a _natural_ one-to-one correspondence between the morphisms \\(F(A) \to B\\) and morphisms \\(A \to R(B)\\):

\[ \mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) .\]

This is where natural transformations come into the game. But I am deliberately urging people not to worry about this too much when they're just getting started with adjoints.

What you really used, I think, is this true fact, a consequence of the other one I just stated:

If a functor \\(F\\) is left adjoint to a functor \\(R\\), then there is a one-to-one correspondence between the morphisms \\(F(A) \to B\\) and morphisms \\(A \to R(B)\\):

\[ \mathbf{Set}(F(A), B) \cong \mathbf{Set}^2(A, R(B)) \]

Using this you can figure out some stuff about what a left adjoint must be like if it exists. In many cases, if you get a nice answer this way, the left adjoint will then exist, because your one-to-one correspondence will turn out to be natural.