Now, back to [Keith Anderson's answers](https://forum.azimuthproject.org/discussion/comment/16691/#Comment_16691) to [Matthew Doty's puzzles](https://forum.azimuthproject.org/discussion/comment/16627/#Comment_16627). I'll only talk about this one:

**MD Puzzle 2**: Find the *right adjoint* \\(r : A\times A \to A\\) to the monotone function \\(\Delta : A \to A \times A \\) given by

$$ \Delta(x) = (x,x) .$$

Here is Keith's answer:




Directly taking John's tutorial and dropping in the functions in the appropriate places we get:

If \\(\Delta: A \to A\times A\\) has a right adjoint \\(r : A\times A \to A\\) and \\(A\\) is a poset, this right adjoint is unique and we have a formula for it:

$$ r(x,y) = \bigvee \\{a \in A : \; \Delta(a) \leq_{A\times A} (x,y) \\} . $$




I think this is correct. As I've emphasized, this formula does _not_ require that _all_ subsets of \\(A\\) have least upper bounds: if we assume \\(\Delta\\) has a right adjoint we know that _the set in question_ has a least upper bound.

But while Keith's answer is correct, we can get a simpler answer... which is the answer Matthew undoubtedly wanted. Namely, I claim:

**Theorem.** If \\(A\\) is any poset and \\(\Delta: A \to A\times A\\) has a right adjoint \\(r : A\times A \to A\\) , this right adjoint is unique and

$$ r(x,y) = x \wedge y .$$

In other words, \\(r(x,y)\\) is the greatest lower bound of the set \\( \\{x,y\\} \\).

To ease our burden, let's prove this assuming that this greatest lower bound \\(x \wedge y\\) exists. (We can worry about why that assumption is true later.)

For this, let's use the definition of right adjoint:

$$ a \le_A r(x,y) \textrm{ if and only if } \Delta(a) \le_{A \times A} (x,y) $$

or in other words

$$ a \le_A r(x,y) \textrm{ if and only if } a \le_A x \textrm{ and } a \le_A y. $$

To prove that \\(r(x,y) = x \wedge y\\) it's therefore enough to show

$$ a \le_A x \wedge y \textrm{ if and only if } a \le_A x \textrm{ and } a \le_A y. $$

**MD Puzzle 2'.** Can someone show this?