> **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,
\[
\begin{align}
R(B) &= (B, \{\bullet\}) \\\\
R(f) &= (f, !).
\end{align}
\]