**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.