TITLE: When does a left adjoint between Heyting algebras preserve 1?
QUESTION [9 upvotes]: This is an exercise from Johnstone's book Stone Spaces:

Let $f: A \to B$ and $g: B \to A$  be order-preserving maps between complete Heyting algebras with $f$ left adjoint of $g$. Show that $f$ preserves binary meet iff the equation 
  $$g (f a \to b)= (a \to g b)$$
  holds for all $a,b$.
Can you find a similar condition for $f$ to preserve $1$?

I have already proved the first part, but I cannot find a condition for the second one.
It would be very helpful if you can give a hint.
Thanks!

REPLY [2 votes]: I've been thinking about this question for a bit. It seems to me that we can't be looking for an answer too "similar" to $g(fa\to b)=a\to gb$ because this is an equation in $A$, and such an equation will not (I believe) suffice for $f1=1$. The idea is that requiring $f1=1$ asks that "$1\in B$ must be in the image of $f$", whereas requiring $f$ to be meet-preserving doesn't ask that "every meet in $B$ must be of the form $f(a_1\wedge a_2)$ for some $a_1,a_2\in A$".
So, in my opinion, a reasonable answer to the question would be to come up with an indirect way of writing $1\in B$ in terms of $\to$ and then simply require $f1$ to equal that expression. For example, it turns out that $1=fa\to f1$ for all $a\in A$, and thus we could say that $f$ preserves 1 if and only if
$$f1=fa\to f1$$
for all $a\in A$. I am not totally satisfied with this answer, but I will at least prove it.

Proof: Let's first establish what properties we can use. Since $B$ is a Heyting algebra we have
  $$\tag{1}b\wedge c\leq d\quad\Leftrightarrow\quad b\leq c\to d$$
  for all $b,c,d\in B$. We are also given that $f$ is left adjoint to $g$, which means that
  $$\tag{2}fa\leq b\quad\Leftrightarrow\quad a\leq gb$$
  for all $a\in A$ and all $b\in B$.
Now since $f1\leq f1$ we have $1\leq g(f1)$ by (2), so if $a\in A$ then we have $fa\leq f1$ by (2) again because $a\leq 1\leq g(f1)$. Therefore $1\wedge fa\leq f1$ for all $a\in A$, and as such $1\leq fa\to f1$ for all $a\in A$ by (1). Hence $fa\to f1=1$ for all $a\in A$.