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?