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.

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.