Injective frame homomorphism condition Let $A$ and $B$ be frames and $f : A \to B$ a frame homomorphism. That is, $f$ preserves finite meets and arbitrary joins. I want to prove the following:
Claim: if $f^{-1}(\top_B) = \{ \top_A \}$ then $f$ is injective (under some condition).
Heyting case: I know the prove in case $f$ is a Heyting algebra morphism (i.e. there is a Heyting arrow and $f$ preserves it): If $a \not\leq b$ then $a \to b \neq \top_A$. Suppose $f(a) \leq f(b)$, then $f(a \to b) = f(a) \to f(b) = \top_B$ hence by assumption $a \to b = \top_A$, a contradiction. So $f(a) \not\leq f(b)$ and injectivity follows.
Attempt: I know one can define a Heyting arrow on a frame by setting $a \to b := \bigvee \{ x \mid x \wedge a \leq b \}$. If our frame morphism $f$ would preserve this arrow the claim follows.
Remark: Without any extra conditions the claim is false. Maybe it helps to assume that $f$ is injective on the generators.
 A: Corrected me if I'm wrong about the definition of a frame, but I think it is a complete lattice $\mathbf A = \langle A, \vee, \wedge, \top, \bot \rangle$ satisfying
$$\left(\bigvee_{i\in I}x_i\right)\wedge y = \bigvee_{i\in I}(x_i \wedge y).$$
In other words, it's the lattice reduct of a complete Heyting Algebra.
In this case, every finite distributive lattice (in particular, a chain) is a frame.
Let $A=\{\bot_A,a,\top_A\}$ and $B=\{\bot_B,b,\top_B\}$, with $\bot_A<a<\top_A$ and $\bot_B<b<\top_B$, and define $f:A\to B$ by making $f(\bot_A)=f(a)=\bot_B$ and $f(\top_A)=\top_B$.
Then $f$ is a frame homomorphism, $f^{-1}(\top_B)=\{\top_A\}$ but $f$ is not injective.
If you add the condition that $f^{-1}(\bot_B)=\{\bot_A\}$, it's still not enough, since a similar counter-example can be build taking for $\mathbf A$ the four-element chain, and $\mathbf B$ as above.
Notice that, although a frame is the reduct of a Heyting Algebra, it's crucial that the implication is not part of the signature in this case, because we get more homomorphisms.
In the example above, if we were seeing $\mathbf A$ and $\mathbf B$ as Heyting Algebras, $f$ would not be a homomorphism, since
$$f(a)\to f(\bot_A)=\bot_B\to\bot_B=\top_B\color{red}{\neq}\bot_B=f(\bot_A)=f(a\to\bot_A).$$
