# An example of a frame homomorphism which does not preserve Heyting implication

A frame is a complete lattice $\langle L,\mathord{\leqslant}\rangle$ which satisfies the following distributivity law: $$a\wedge\bigvee_{i\in I}b_i=\bigvee_{i\in I}a\wedge b_i\,.$$

A frame homomorphism is a mapping $h\colon L_1\rightarrow L_2$ which preserves finite operations and such that $h(\bigvee_{i\in I}b_i)=\bigvee_{i\in I}h(b_i)$.

Every frame is a complete Heyting algebra, yet frame homomorphisms do not generally preserve Heyting implication. Could you please give me an example of such homomorphism?

Let $U^{*}$ denote the pseudocomplement of an element $U$. Then since Heyting algebra homomorphisms preserve pseudocomplements, we shall produce a counterexample by constructing a frame homomorphism which does not preserve pseudocomplements (these frame homomorphisms are quite common).

Let $U\subseteq\mathbb{R}$ be an open set where $U^{**}\neq\mathbb{R}$. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function where $U^{c}=f^{-1}(0)$. Let $\mathcal{T}$ be the collection of all open sets on $\mathbb{R}$. Then $\Omega(f):\mathcal{T}\rightarrow\mathcal{T}$ be the mapping where $\Omega(f)(V)=f^{-1}[V]$ for each open set $V$. Then $\Omega(f)$ is a frame homomorphism. Let $O=\mathbb{R}\setminus\{0\}$. Then $\Omega(f)(O^{**})=\Omega(f)(\mathbb{R})=\mathbb{R}$. However, $\Omega(f)(O)^{**}=U^{**}\neq\mathbb{R}$. Therefore, $\Omega(f(O^{**}))\neq\Omega(f)(O)^{**}$.

It turns out that the complete Heyting homomorphisms between frames are precisely the open mappings between frames. Therefore, every open mapping between topological spaces induces a complete Heyting homomorphism between frames.

• Joseph, I am trying to figure out what $U^c$ is. Shouldn't it say instead that $U^\ast=f^{-1}[\mathbb{R}\setminus\{0\}]$ (or $(U^\ast)^c=f^{-1}[\{0\}]$, with $c$ the standard set-theoretical complement in $\mathbb{R}$)? And $f$ has to be continuous in order to perform the construction, doesn't it? I also assume that by $\mathcal{T}$ you mean the standard order topology on the reals, right? Commented Aug 31, 2016 at 21:07
• Yes. the mapping $f$ is continuous and $U^{c}$ stands for the complement. The topology $\mathcal{T}$ is the standard order topology. I made a couple corrections to the answer. Commented Aug 31, 2016 at 21:45
• One more thing, the pseudocomplement of $O$ is empty. Thus $U$ whose pseudocomplement is non-empty should be chosen at the outset (rather than such which is different from the reals). Commented Aug 31, 2016 at 21:56
• $U^{**}\neq\mathbb{R}$ is equivalent to $U^{*}\neq\emptyset$. Commented Sep 1, 2016 at 15:11

Sort of a more abstract answer. Let $L$ be a frame, and let $a\in L$ be any element of $L$. Then $a\lor\_:L\to\{a'\in L\mid a\leqslant a'\}$ is a (surjective) frame homomorphism. Suppose it preserves implication, i. e. $a\lor(x\to y)=(a\lor x)\to(a\lor y)$ for any $x,y\in L$. Then in particular taking $x=a$ and $y=0$ we obtain $$a\lor\neg a=a\lor(a\to0)=(a\lor a)\to(a\lor 0)=a\to a=1,$$ so that $a$ is complemented.

Thus any non-complemented element $a\in L$ provides an example of a frame homomorphism which does not preserve implication.