Peter wrote:

> Nobody with a solution to Puzzle 150/151? I was trying to solve it but wasn't yet successful so far. We need bijections \\( [S \times T,X] \rightarrow [(S,T),G(X)] \\) and \\( [G(X),(S,T)] \rightarrow [X,S \times T] \\) for a certain functor \\( G: \textbf{Set} \rightarrow \textbf{Set}^2 \\).

I'm glad you're trying these. I'm a little worried: it looks here like you're trying to make the same functor \\(G\\) into both the left and right adjoint. That won't work. Maybe you're not really wanting this, but I think other readers will be less confused if you said:

> For \\( G: \textbf{Set} \rightarrow \textbf{Set}^2\\) to be a right adjoint of \\(\times: \textbf{Set}^2 \to \textbf{Set}\\), we need a bijection \\( [S \times T,X] \rightarrow [(S,T),G(X)] \\). For \\(H\\) to be a left adjoint of \\(\times: \textbf{Set}^2 \to \textbf{Set}\\), we need a bijection \\( [H(X),(S,T)] \rightarrow [X,S \times T] \\)...

... where, clearly, you are using square brackets \\([A,B]\\) to denote the set of morphisms from an object \\(A\\) to an object \\(B\\).

I will now read further into your comment: I just had to get this off my chest.

But I have a hint that may be helpful. In my formulation of the problem, only _one_ of \\(G\\) and \\(H\\) will actually _exist!_ The functor \\(\times: \textbf{Set}^2 \to \textbf{Set}\\) does not have both a left and right adjoint. It has just one of these.