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# Exercise 90 - Chapter 1

edited May 2018

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 $$f \Big( \bigvee A \Big) = \bigvee f(A)$$.

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)\$