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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