Injective frame homomorphism

Question

Let $A$ and $B$ be frames with bottom elements $\bot_A$ and $\bot_B$ respectively. Let $\phi : A \to B$ be a frame homomorphism such that $\phi(a) = \bot_B$ implies $a = \bot_A$ (that is, the kernel of $\phi$ is trivial).

Question: Does this imply that $\phi$ is injective?

Answer 1

The simplest counterexample is given by the map from the three element linear order $\bot < a < \top$ to the two element linear order $\bot < \top$ given by $\bot\mapsto \bot$, $a\mapsto \top$, $\top\mapsto\top$.

Note that this is a special case of Stefan Mesken's answer: The three element linear order is the frame of open sets of the Sierpiński space $S = \{0,1\}$, with open sets $\{\emptyset, \{1\}, S\}$, and the two element linear order is the frame of open sets of the singleton space $\{1\}$, with open sets $\{\emptyset, \{1\}\}$. Now $\{1\}$ is a dense open set in $S$, and taking preimages along the continuous inclusion $\{1\}\to S$ gives the map of frames I described above.

Answer 2

Here is a counterexample:

Let $(X; \tau)$ be a topological space with a dense open set $D \in \tau \setminus \{X\}$. Consider $$ \phi \colon \tau \mapsto \tau \restriction D, a \mapsto a \cap D, $$ where $\tau \restriction D$ is the subspace topology of $(X; \tau)$ induced by $D$.