Showing that the image of a regular frame is regular

Question

Suppose $F$ is frame. I am trying to show that the homomorphic image of a regular frame is again regular.

Here by regular I mean: $x = \vee_{y\prec x} y$ for any $x\in F$. The symbol $\prec$ means "rather below" and $x\prec y$ iff there is some $a\in F$ such that $x\wedge a = 0$ while $y\vee a = 1$.

I have managed to show that if $f:L\rightarrow K$ where $f$ is a frame homomorphism and WLOG we can assume it is onto (since we are considering images) then $x\prec y \implies f(x)\prec f(y)$. Hence if $p\in K$ i.e. $p = f(x)$ for some $x\in L$ we get $p\leq \vee_{r\prec p} p$.

I am struggling to show that $p = \vee_{r\prec p} p$. It would easily follows if for example $x\prec y \implies x\leq y$ but I am not sure this is true.

Answer 1

If $x\prec y$, let $a$ be such that $x\wedge a=0$ and $y\vee a=1$. We then have $$x=x\wedge 1=x\wedge (y\vee a)=(x\wedge y)\vee(x\wedge a)=(x\wedge y)\vee 0=x\wedge y.$$ Thus $x\leq y$.