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Alex Chen

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  • @ThomasRead Yes, thanks for pointing this out. Continuing with Puzzle 18, if \(f_{\ast}\) has a left-adjoint \(g_{\ast}\), then \(f\) must be surjective. From before, we know that if \(f_{\ast}\) has a left-adjoint \(g_{\ast}\), then \(f\) must be…
  • I think there is another example of a poset (Puzzle 4) from Owen Biesel's puzzles in the comments of Lecture 17 (OB1 to OB4): the set of galois connections \( \{ (f, g) : f \dashv g \} \) between two posets \(A\) and \(B\) forms a poset. A galois c…
  • 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…
  • Ah, I see where my "counterexamples" went wrong. Thanks a lot for your help, John and Dan!
  • Following Dan Schmidt's post 5, it seems that if \(f_{\ast}\) has a left adjoint \(g_{\ast}\), then \(f\) must be injective. Otherwise, if \(f\) is not injective, there are elements \(a_1, a_2 \in A, a_1 \neq a_2\) such that \(f(a_1) = b = f(a_2)\).…
  • Hi, I'm confused about the proof of proposition 1.88, where they show that if \(Q\) is a preordered set with all meets, then any monotone map \(g: Q \to P\) that preserves meets is a right adjoint. They do this by constructing a left adjoint \(f: P …