# Week 1 Notes

• EX 1.110:
1. If $$f : P \rightarrow Q$$ has a right adjoint $$g$$, then for any other right adjoint $$g'$$, and for all $$p \in P$$ nd $$q \in Q$$, $$p \leq g(q)$$ iff $$f(p) \leq q$$ iff $$p \leq g'(q)$$. So $$$\begin{split} & &&p \leq g(q) \implies p \leq g'(q) \\ &\implies &&p \leq g(q) \leq g'(q) \\ \end{split}$$$ and $$$\begin{split} & &&p \leq g'(q) \implies p \leq g(q) \\ &\implies &&p \leq g'(q) \leq g(q) \\ \end{split}$$$ Since $$P$$ is a preorder, $$g(q) \cong g'(q)$$.
2. If $$h : P \rightarrow Q$$ has a left adjoint $$g$$, then for any other left adjoint $$g'$$, and for all $$p \in P$$ nd $$q \in Q$$, $$g(q) \leq p$$ iff $$q \leq h(p)$$ iff $$g'(q) \leq p$$. So $$$\begin{split} & &&g(q) \leq p \implies g'(q) \leq p \\ &\implies &&g'(q) \leq g(q) \leq p \\ \end{split}$$$ and $$$\begin{split} & &&g'(q) \leq p \implies g(q) \leq p \\ &\implies &&g(q) \leq g'(q) \leq p \\ \end{split}$$$ Since $$P$$ is a preorder, $$g(q) \cong g'(q)$$.
• EX 1.112:

Let $$f : P \rightarrow Q$$ be left adjoint to $$g : Q \rightarrow P$$, $$A \subseteq P$$ be any subset, and $$j := \bigvee A$$ be its join. $$f(j)$$ is a upper bound for $$f(A)$$, since $$f(a) \leq f(j)$$ for all $$a \in A$$ according to the monotonicity of $$f$$. For any other upper bound $$b$$ for $$f(A)$$, we have $$f(a) \leq b$$ for all $$a \in A$$, then $$a \leq g(f(a)) \leq g(b)$$ (P 1.107). In other words, $$g(b)$$ is an upper bound for $$A$$. But then $$j$$ is the lowest upper bound for $$A$$, so $$j \leq g(b)$$, and so $$f(j) \leq b$$ (D 1.95). Hence $$f(j)$$ is the lowest upper bound for $$f(A)$$.

• EX 1.119:

Let $$f : P \rightarrow Q$$ be left adjoint to $$g : Q \rightarrow P$$.

1. $$p \leq (g \circ f)(p)$$ for all $$p \in P$$ (P 1.107).
2. Since $$(g \circ f)(p) \in P$$, then $$(g \circ f)(p) \leq (g \circ f \circ g \circ f)(p)$$ (P 1.107). Since $$f(p) \in Q$$, then $$f(g(f(p))) \leq f(p)$$ (P 1.107), and since $$g$$ is monotone, $$(g \circ f \circ g \circ f)(p) \leq (g \circ f)(p)$$.
3. Hence $$(g \circ f \circ g \circ f)(p) \cong (g \circ f)(p)$$.

Created: 2020-01-13 Mon 14:36