Nobody with a solution to puzzle 150/151:? I was trying to solve it but wasn't yet successfull 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 think we can use the universal property of the product: If I choose a set X together with maps \$$h_S: X \rightarrow S \$$ and \$$h_T: X \rightarrow T \$$ then there is exactly one \$$h: X \rightarrow S \times T \$$ such that \$$h_S = pr_S \circ h \$$ and \$$h_T = pr_T \circ h \$$ whereafter the \$$pr_i \$$ are the respective projection from the product to its components.

If I choose any X and maps to S and T then \$$G(X):=(X_1,X_2) \$$ with \$$X_1 := h_S^{-1}(S) \$$, \$$X_2 := h_T^{-1}(T) \$$. There is exactly one \$$h: X \rightarrow S \times T \$$ that maps under G to \$$h'=(h'_1,h'_2): G(X) \rightarrow (S,T) \$$. So there are so much left adjoints of SxT as there are pairs of maps from X to S and T, for every X. ??? A lot of adjoints....

In the other direction, if we start with a pair \$$(X_1,X_2) \$$ in \$$\textbf{Set}^2 \$$ and a map \$$(h'_1,h'_2) \$$, we get a map \$$F((h'_1,h'_2)): X_1 \times X_2 \rightarrow S \times T \$$ as F is contravariant. But we cannot find the \$$h_1 \$$ and \$$h_2 \$$ which belongs to \$$F((h'_1,h'_2)) \$$. So I guess that there is no right adjoints. But I'm not sure...

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