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.)

> 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.)