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?



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


