An algebra A has the congruence extension property (CEP) if for every B ≤ A and θ ∈ ConB there is a φ ∈ Con A such that θ = φ ∩ (B x B) . A class K of algebras has the CEP if every algebra in the class has the CEP. Show that the class of Heyting algebras has the CEP.
I was thinking of defining φ = θ $\cup \Delta$ where $\Delta$ is the diagonal. Then φ $ \in$ ConA but I'm not sure if this gives θ = φ ∩ (B x B) (or how to show it).
Let us use some preliminary results about congruences and filters on Heyting algebras.
I'll leave their proofs to you.
Let $\mathbf H$ be a Heyting algebra and $\theta$ a congruence of $\mathbf H$.
Then $$F_{\theta} = \{ x \in H : (1,x) \in \theta \}$$ is a filter of $\mathbf H$.
And now the converse:
Let $\mathbf H$ be a Heyting algebra and $F$ be a filter of $\mathbf H$.
Then $$\Theta_F = \{ (x,y) \in H^2 : (x \to y) \wedge (y\to x) \in F \}$$ is a congruence of $\mathbf H$.
Now let $\mathbf A$ and $\mathbf B$ be Heyting algebras with $\mathbf B \leq \mathbf A$, and $\theta$ a congruence of $\mathbf B$.
Define $F$ to be the filter of $\mathbf A$ generated by $F_{\theta}$, that is
$$F = \{ x \in A : \exists y \in F_{\theta} ( y \leq x ) \}.$$
I claim that
$$\Theta_F \cap B^2 = \theta.$$
In order to prove it, we need the following Heyting algebra identities:
$$x\to x=1, \quad (x\to y)\wedge y=y \quad\text{and}\quad x\wedge(x\to y)=x\wedge y.$$
Now, if $(u,v) \in B^2$ and $(u\to v)\wedge(v\to u) \in F$, then
$$(u\to v)\wedge(v\to u) \equiv 1 \pmod{\theta},$$
whence
$$(u\to v) \wedge v \wedge (v\to u) \equiv v \pmod{\theta},$$
that is,
$$v \wedge (v\to u) \equiv v \pmod{\theta},$$
or yet, $v \wedge u \equiv v \pmod{\theta}$.
By symmetry of congruence relations, $u \wedge v \equiv u \pmod{\theta}$, and so $u \theta v$.
Thus, we proved that $\Theta_F \cap B^2 \subseteq \theta$.
Conversely, if $u \theta v$, then $(u \to v, v \to v) \in \theta$, that is, $(u\to v,1) \in \theta$, whence $u \to v \in F_{\theta}$, yielding $u \to v \in F$; by symmetry of congruence relations, $v \to u \in F$, and therefore $$(u\to v) \wedge (v\to u) \in F,$$ or $(u,v) \in \Theta_F$. Hence $\theta \subseteq \Theta_F \cap B^2$.
Therefore, Heyting Algebras have the CEP.
Concerning your approach, it is not true that $\theta \cup \Delta$ is always a congruence of $\mathbf A$, whenever $\theta$ is a congruence of $\mathbf B$ and $\mathbf B \leq \mathbf A$.
For example, if $\mathbf A$ is a Heyting algebra with more than two elements, and $B = \{0,1\}$, then $\mathbf B \leq \mathbf A$ and $\{0,1\}^2 = \nabla_B$ is a congruence of $\mathbf B$, but $\Delta_A \cup \{0,1\}^2$ is not a congruence of $\mathbf A$, since a congruence class on a lattice is always a convex sub-lattice.
