Is the codomain of a surjective lattice-homomorphism Heyting if the domain is Heyting? Let $L$ and $L'$ be lattices and let $F:L\to L'$ be a surjective lattice-homomorphism in the sense that $F$ respects $\wedge$ and $\vee$.
I managed to prove for several properties of $L$ that they will be inherited by $L'$.
Let me mention some:

*

*if $L$ is bounded then $L'$ is bounded (and $F$ preserves $0$ and $1$).

*if $L$ is distributive then $L'$ is distributive.

*if $L$ is Boolean then $L'$ is Boolean (and $F$ preserves complements).

Now my question:

(1) Do we also have: "if $L$ is Heyting then $L'$ is Heyting"?
(2) And if so then will $F$ preserve $a\to b$ in the sense that $F(a\to b)=F(a)\to F(b)$?

Actually the question can be reformulated as:

Is it true that $F(a)\wedge F(x)\leq F(b)\implies F(x)\leq F(a\to b)$?

It is clear to me that the inverse of this implication is true.
It would not surprise me if the answer is "no", so let me conclude with requesting for a counterexample if that is indeed the case.
Thank you in advance.
 A: This is not a proper answer as the main question eluded me for quite some time, when trying to answer.
However, it is clear that, in the finite case, the answer is yes.  
So there's the question of whether or not $f(a \to b) = f(a) \to f(b)$.
And the answer is that it is not necessarily so.
For a very simple counter-example, take $L$ to be the three-element chain, with $0<e<1$ and $L'$ the two-element one, $0<1$.
The map $f:L\to L'$ given by
$$f(0)=f(e)=0 \quad\text{and}\quad f(1) = 1,$$
is certainly an onto lattice homomorphism.
However,
$$f(e \to 0) = f(0) = 0 \neq 1 = 0 \to 0 = f(e) \to f(0).$$
The equalities above are easy to check, either directly by definition, or taking into account that, as you can see here, second example, in a chain, $a \to b = b$ if $b < a$, and in general, if $a \leq b$, then $a \to b = 1$.
In a tentative to be more insightful, a good reason to come up with the example above is as follows.
As you know for every homomorphism $f:L \to L'$, the kernel of $f$ is a congruence of $L$.
So the smaller question you made is equivalent to ask if $\theta$ is a Heyting algebra congruence, whenever $L$ is a Heyting algebra and $\theta$ is a lattice congruence on $L$.
But Heyting algebras are $1$-regular, that is, a congruence on a Heyting algebra is determined by the congruence class of $1$.
To be more precise,
$$(a,b) \in \theta \Leftrightarrow ((a \to b) \wedge (b \to a), 1) \in \theta.$$
More to the point, a Heyting algebra is subdirectly irreducible iff it has an element $e$ such that, for $x \neq 1$ in that algebra, $x \leq e$, and in that case, 
$$\mu = \{e,1\}^2 \cup \Delta_L$$
is the monolith, meaning that $\mu \leq \theta$, whenever $\theta$ is a non-trivial congruence ($\theta \neq \Delta_L$).
Hence, if $(e,x) \in \theta \neq \Delta$, then $(e,1) \in \theta$, which doesn't happen for $\theta = \ker f$, in the example above.
