Daniel Fava wrote:

> It seems like the existence of a Galois connection between \$$(A,\leq_A) \$$ and \$$(B, \leq_B ) \$$ requires least upper bounds and greatest lower bounds to be defined. Does a Galois connection between \$$(A,\leq_A) \$$ and \$$(B, \leq_B ) \$$ imply that \$$(A,\leq_A) \$$ and \$$(B, \leq_B ) \$$ are complete lattices?

I guess you answered that question in the negative in comment #23, since neither poset in your picture has all least upper bounds - nor does either have all greatest lower bounds.

But you answered your question so un-verbosely that I can't resist stating the answer in words. If \$$f : A \to B\$$ has a right adjoint \$$g : B \to A\$$ and \$$A\$$ is a poset, we have

$$g(b) = \bigvee \\{a \in A : \; f(a) \le_B b \\}$$

so the least upper bound _in this formula_ must exist in \$$A\$$ for any \$$b\$$. Similarly, if \$$g : B \to A\$$ has a left adjoint \$$f : A \to B\$$ and \$$B\$$ is a poset, we have

$$f(a) = \bigwedge \\{b \in B : \; a \le_A g(b) \\}$$

so the greatest lower bound _in this formula_ must exist in \$$B\$$ for any \$$a\$$. But no further greatest lower bounds or least upper upper bounds need exist!

Here's an extreme example: for any poset \$$A\$$ whatsoever, the identity function \$$1_A : A \to A\$$ has a left and right adjoint, namely itself. This is easy to check straight from the definition:

$$a \le a \textrm{ if and only if } a \le a .$$

If you compute these adjoints using the formulas above, you see that only sets of the form

$$\\{ a \in A : \; a \le b \\}$$

need have greatest lower bounds - and they do, namely the element \$$b\$$. Similarly, only sets of the form

$$\\{a \in A: \; a \ge b \\}$$

need have least upper bounds - and they do, namely the element \$$b\$$.

(Trivial examples can be very enlightening.)