**Puzzle 149.** Again suppose \\(F: \mathbf{Set} \to \mathbf{1}\\) is the functor that sends every set to \\(\star\\) and every function to \\(1_\star\\). A _left_ adjoint \\(L : \mathbf{1} \to \mathbf{Set} \\) is a functor for which there's a natural one-to-one correspondence between functions

\[ m: L(\star) \to S \]

and morphisms

\[ n: \star \to F(S) \]

for every set \\(S\\). On the basis of this, try to figure out all the left adjoints of \\(F\\).

Keith wrote:

> Demanding that \\(L(\star)\to S\\), for all \\(S\\) is the same thing as asking for a set \\(T\\) such that, for all \\(S\\), \\(T \to S\\) holds.

> The only such set is the empty set, so the left adjoint to \\(F: \mathbf{Set} \to \mathbf{1}\\) is a functor taking \\(\mathbf{1}\\) to the empty set.

Right!

So we see the beginnings of a nice pattern:

* a right adjoint of the unique functor \\(F: \textbf{Set} \to \textbf{1} \\) sends the one object of \\(\textbf{1}\\) to a set with one element, sometimes called '1' Such a set is a **terminal object** in \\(\mathbf{Set}\\), meaning that there's exactly one function _from_ any set _to_ this set.

* a left adjoint of the unique functor \\(F: \textbf{Set} \to \textbf{1} \\) sends the one object of \\(\textbf{1}\\) to the set with no elements, sometimes called '0'. Such a set is an **initial object** in \\(\mathbf{Set}\\), meaning that there's exactly one function _to_ any set _from_ this set.

This makes some sense because right adjoints are defined using morphisms _to_ them, as are terminal objects. Left adjoints are defined using morphisms _from_ them, as are initial objects.

Finally, a discussion of your argument showing that \\(L(\star)\\) is empty.

Mathematicians don't say "\\(T \to S\\) holds". We say "X holds" when X is a proposition that can be true or false, and X happens to be true. There is no proposition "\\(T \to S\\) ". In fact "\\(T \to S\\)" doesn't mean anything by itself. What makes sense are functions \\(f: T \to S\\). Given two sets \\(T\\) and \\(S\\), we can have one, more than one, or no functions \\(f: T \to S\\).

Here's how we prove \\(L(\star)\\) must be empty:

If \\(L\\) is the left adjoint of \\(F\\), there must be a one-to-one correspondence between morphisms \\(n: \star \to F(S)\\) and functions \\(m: L(\star)\to S\\), for all \\(S\\). Since \\(F(S) = \star\\), there is exactly one morphism \\(n: \star \to F(S)\\) . Thus, there must be _exactly one_ function \\(m: L(\star) \to S\\) for any set \\(S\\).

This forces \\(L(\star)\\) to be the empty set: there's exactly one function from the empty set to any set... and the empty set is the _only_ set with this property.

\[ m: L(\star) \to S \]

and morphisms

\[ n: \star \to F(S) \]

for every set \\(S\\). On the basis of this, try to figure out all the left adjoints of \\(F\\).

Keith wrote:

> Demanding that \\(L(\star)\to S\\), for all \\(S\\) is the same thing as asking for a set \\(T\\) such that, for all \\(S\\), \\(T \to S\\) holds.

> The only such set is the empty set, so the left adjoint to \\(F: \mathbf{Set} \to \mathbf{1}\\) is a functor taking \\(\mathbf{1}\\) to the empty set.

Right!

So we see the beginnings of a nice pattern:

* a right adjoint of the unique functor \\(F: \textbf{Set} \to \textbf{1} \\) sends the one object of \\(\textbf{1}\\) to a set with one element, sometimes called '1' Such a set is a **terminal object** in \\(\mathbf{Set}\\), meaning that there's exactly one function _from_ any set _to_ this set.

* a left adjoint of the unique functor \\(F: \textbf{Set} \to \textbf{1} \\) sends the one object of \\(\textbf{1}\\) to the set with no elements, sometimes called '0'. Such a set is an **initial object** in \\(\mathbf{Set}\\), meaning that there's exactly one function _to_ any set _from_ this set.

This makes some sense because right adjoints are defined using morphisms _to_ them, as are terminal objects. Left adjoints are defined using morphisms _from_ them, as are initial objects.

Finally, a discussion of your argument showing that \\(L(\star)\\) is empty.

Mathematicians don't say "\\(T \to S\\) holds". We say "X holds" when X is a proposition that can be true or false, and X happens to be true. There is no proposition "\\(T \to S\\) ". In fact "\\(T \to S\\)" doesn't mean anything by itself. What makes sense are functions \\(f: T \to S\\). Given two sets \\(T\\) and \\(S\\), we can have one, more than one, or no functions \\(f: T \to S\\).

Here's how we prove \\(L(\star)\\) must be empty:

If \\(L\\) is the left adjoint of \\(F\\), there must be a one-to-one correspondence between morphisms \\(n: \star \to F(S)\\) and functions \\(m: L(\star)\to S\\), for all \\(S\\). Since \\(F(S) = \star\\), there is exactly one morphism \\(n: \star \to F(S)\\) . Thus, there must be _exactly one_ function \\(m: L(\star) \to S\\) for any set \\(S\\).

This forces \\(L(\star)\\) to be the empty set: there's exactly one function from the empty set to any set... and the empty set is the _only_ set with this property.