> **Puzzle 150.** Figure out all the right adjoints of \$$F\$$, where \$$F : \mathbf{Set}^2 \to \mathbf{Set}\$$ is a functor that on objects it throws away the second set, \$$F(S,T) = S \$$, and on morphisms it throws away the second function \$$F(f,g) = f \$$.

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)\$$:

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

Since \$$A \in \mathbf{Set}^2\$$, then \$$A\$$ is a pair \$$(A_1, A_2)\$$:

$\mathbf{Set}(F(A_1, A_2), B) \cong \mathbf{Set}^2((A_1, A_2), R(B))$

We apply the defintion of the functor \$$F\$$ and write \$$R(B) \in \mathbf{Set}^2\$$ as a pair \$$(B_1, B_2)\$$:

$\mathbf{Set}(A_1, B) \cong \mathbf{Set}^2((A_1, A_2), (B_1, B_2))$

A function \$$(A_1, A_2) \to (B_1, B_2)\$$ is a pair of functions, one \$$A_1 \to B_1\$$, the other \$$A_2 \to B_2\$$:

$\mathbf{Set}(A_1, B) \cong \mathbf{Set}(A_1, B_1) \times \mathbf{Set}(A_2, B_2)$

If we pick \$$B_1 = B\$$ and \$$B_2 = \\{\bullet\\}\$$,
then we can map a function \$$f : A_1 \to B\$$ to the pair of functions \$$(f : A_1 \to B, ! : A_2 \to \\{\bullet\\})\$$, where \$$!\$$ denotes the unique function from any set to the singleton set \$$\\{\bullet\\}\$$.
Hence,
\begin{align} R(B) &= (B, \{\bullet\}) \\\\ R(f) &= (f, !). \end{align}