Looking for an example of a continuous function such that the inverse image is not a Heyting Algebras morphism

Question

Let $f\colon X \longrightarrow Y$ a continuous function between topological spaces. This induces a frame morphism $f^{-1}:\tau_Y\longrightarrow \tau_X$. I'm looking for an example when $f^{-1}$ is not a Heyting Algebra morphism.

Answer 1

Let us try constructing a proof that $f^{-1}$ would be a Heyting algebra morphism, even though we know that in the end it will fail. However, the exact failure mode of the proof will give information which will hint at how to find a counterexample.

So, we have for $U, V$ open in $Y$ that $$ f^{-1}(U \rightarrow V) = f^{-1}(\operatorname{int}((Y \setminus U) \cup V)) \subseteq \operatorname{int}(f^{-1}((Y \setminus U) \cup V))) = \\ \operatorname{int}((X \setminus f^{-1}(U)) \cup f^{-1}(V)) = (f^{-1}(U) \rightarrow f^{-1}(V)). $$

Here, from the continuity of $f$, we can only conclude $f^{-1}(\operatorname{int}(S)) \subseteq \operatorname{int}(f^{-1}(S))$. So, suppose we can find an example of $f$ and $S$ where this is not an equality, and furthermore we have that $S$ is a union of a closed set and an open set. Then we will be able to work backwards from this to find a counterexample to $f^{-1}$ being a morphism of Heyting algebras.

Now, such a counterexample is straightforward to search for. In particular, consider $f : \mathbb{R} \to \mathbb{R}$, $f(x) \equiv 0$. Then $f^{-1}(\{ 0 \}) = \mathbb{R}$, so $\operatorname{int}(f^{-1}(\{ 0 \})) = \mathbb{R}$, while $f^{-1}(\operatorname{int}(\{ 0 \})) = f^{-1}(\emptyset) = \emptyset$. Furthermore, $\{ 0 \}$ is closed. Tracing back, we get this counterexample giving an answer to the original question:

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Consider $f : \mathbb{R} \to \mathbb{R}$, $f(x) \equiv 0$, and let $U := \mathbb{R} \setminus \{ 0 \}$. Then $f^{-1}(U \rightarrow \emptyset) = f^{-1}(\emptyset) = \emptyset$, whereas $(f^{-1}(U) \rightarrow f^{-1}(\emptyset)) = (\emptyset \rightarrow \emptyset) = \mathbb{R}$.

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Just for fun, I'll give another proof that $f^{-1}(U \rightarrow V) \subseteq (f^{-1}(U) \rightarrow f^{-1}(V))$: By the adjunction between $x \rightarrow -$ and $x \wedge -$, it suffices to prove that $f^{-1}(U) \cap f^{-1}(U \rightarrow V) \subseteq f^{-1}(V)$. However, this is easy: $f^{-1}(U) \cap f^{-1}(U \rightarrow V) = f^{-1}(U \cap (U \rightarrow V))$; and since $U \cap (U \rightarrow V) \subseteq V$, the desired result follows. (However, this proof isn't as easy to examine to get to a contradiction to $f^{-1}$ being a morphism of Heyting algebras. On the other hand, it's easy to see that this proof generalizes to show for any lattice morphism $\phi : X \to Y$ between Heyting algebras, we have $\phi(x \rightarrow y) \le (\phi(x) \rightarrow \phi(y))$.)