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• Don #167, #169 - the two definitions are equivalent. If the two partitions have the same cardinality then you can find an $$h: P \rightarrow Q$$, but not necessarily satisfying $$f.h=g$$. For example, we can represent the partitions from the left a…
• I think for Puzzle 18 we also need $$f$$ surjective? Since if we take $$y \in Y \setminus f_{\ast}(X)$$, then for a left adjoint $$g_{\ast}$$: $$g_{\ast}(\{y\}) \subseteq X \iff \{y\} \subseteq f_{\ast}(X)$$ where the left hand side is clearly a…
• In other words, $$g(b)$$ must be an upper bound of this set. But you shouldn't choose $$g(b)$$ to be any bigger that it needs to be! After all, you know $$a \le_A g(b)$$ only if $$f(a) \le_B b$$. So, $$g(b)$$ must be a least upper bound of the …
• Jerry #154 - I think the "and..." part is necessary. For example: Let $$P=\{a\}$$, $$Q=\{1,2\}$$. Let $$f(a)=1$$ and $$g(1)=g(2)=a$$. Then $$f$$, $$g$$ are monotone, $$f.g=id_P$$ but $$g.f\neq id_Q$$.
• I think this is a typo in the book - I submitted it to the mistakes list a couple days ago. The preceding paragraph should say "[The inclusion $$\mathbf{Pos}(S)\rightarrow\mathbf{Rel}(S)$$] is actually the right adjoint of a Galois connection. Its…
• I agree with Patrick #1 on Puzzle 11 (or have the same confusion!). A simpler example: Let $$A = \{a, b\}$$ and $$\leq_A$$ be only $$\leq_A = \{(a, a), (b,b)\}$$. Let $$(B, \leq_B)$$ be given by $$B=\{1, 2\}$$ with the usual $$\leq_B$$. Then \(f = …