I could contribute how I understand **Puzzle TR1**, though I'm not sure how terse it is.

We know that \\(g(b)\\) is an upper bound for the set \\(A_b = \\{a \in A : f(a) \leq_B b\\}\\). We want to show that \\(g(b)\\) is a *least* upper bound for \\(A_b\\).

It is enough to show that \\(g(b) \in A_b\\), from which it follows that any upper bound for \\(A_b\\) must be at least \\(g(b)\\).

But \\(g(b) \in A_b\\) iff \\(f(g(b)) \leq_B b\\) [by the definition \\(A_b\\)], iff \\(g(b) \leq_A g(b)\\) [by the definition of adjoints], which is true.

We know that \\(g(b)\\) is an upper bound for the set \\(A_b = \\{a \in A : f(a) \leq_B b\\}\\). We want to show that \\(g(b)\\) is a *least* upper bound for \\(A_b\\).

It is enough to show that \\(g(b) \in A_b\\), from which it follows that any upper bound for \\(A_b\\) must be at least \\(g(b)\\).

But \\(g(b) \in A_b\\) iff \\(f(g(b)) \leq_B b\\) [by the definition \\(A_b\\)], iff \\(g(b) \leq_A g(b)\\) [by the definition of adjoints], which is true.