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.