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…
Marcello #25: Presumably there are many pages with no links, and many which are never linked to, so you'd definitely get some sort of non-trivial structure. I'd love to see what it actually looks like though.
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 = …