Options

Exercise 90 - Chapter 1

edited May 2018 in Exercises

Complete the proof of Proposition 1.84 by showing that left adjoints preserve joins.

Proposition 1.89 (Left adjoints preserve joins).

Let \( f: P \rightarrow Q \) have as its right adjoint \( g: Q \rightarrow P \).

Suppose \( A \subseteq P \) any subset, and let \( f(A) = { f(a) = a \in A } \) be its image.

Then if \( A \subseteq P \) is any subset that has a join \( \bigvee f(A) \in P \) and we have $$ f \Big( \bigvee A \Big) = \bigvee f(A) $$.

Comments

  • 1.
    edited July 2018

    This proof is dual to the one given in the first half of the proof of the proposition.

    By definition, an element \(t\in P\) is the join of the elements of \(A\) if(f) for every p\(\in A\), p≤t, and t is the least such element of \(P\). By monotonicity of f, f(p)≤f(t), so f(t) is an upper bound for f(\(A\)) in \(Q\). Let u\(\in Q\) be any upper bound for f(\(A\)), so, for every p\(\in A\), f(p)≤u. By adjointness of f,g, p≤g(u) must hold for every p\(\in A\), so g(u) is an upper bound for \(A\) in \(P\). By definition, the join t is the least upper bound, so t≤g(u). By adjointness of f,g one has that f(t)≤u, so, f(t) is the least upper bound in \(Q\) for f(\(A\)). In symbols \[f\left(\bigvee A\right)=\bigvee f(A)\].

    Note that, in general, the last equation is only true up to isomorphism, i.e.: \[f\left(\bigvee A\right)\cong\bigvee f(A)\]

    Comment Source:This proof is dual to the one given in the first half of the proof of the proposition. By definition, an element \\(t\in P\\) is the join of the elements of \\(A\\) if(f) for every p\\(\in A\\), p≤t, and t is the least such element of \\(P\\). By monotonicity of f, f(p)≤f(t), so f(t) is an upper bound for f(\\(A\\)) in \\(Q\\). Let u\\(\in Q\\) be any upper bound for f(\\(A\\)), so, for every p\\(\in A\\), f(p)≤u. By adjointness of f,g, p≤g(u) must hold for every p\\(\in A\\), so g(u) is an upper bound for \\(A\\) in \\(P\\). By definition, the join t is the least upper bound, so t≤g(u). By adjointness of f,g one has that f(t)≤u, so, f(t) is the least upper bound in \\(Q\\) for f(\\(A\\)). In symbols \\[f\left(\bigvee A\right)=\bigvee f(A)\\]. Note that, in general, the last equation is only true up to isomorphism, i.e.: \\[f\left(\bigvee A\right)\cong\bigvee f(A)\\]
Sign In or Register to comment.