Situation: Given two Boolean algebras, $A$ and $B$, where $A$ is defined using a set of generators subject to some relations: \begin{align*} A = \operatorname{BA}\langle \Box a, \Diamond a \; (a \in I) \; \mid \text{ some relations }\rangle. \end{align*} We have a map $\eta : A \to B$ defined on the generators of $A$ and well-defined with respect to the relations on $A$.
Question: For injectivity of $\eta$, does it suffice to prove that $\eta$ is injective on the generators of $A$?
I Feel like this should be true, but I cannot easily find an argument for it.
No, it does not suffice to check that $\eta$ is injective on the generators of $A$. For a simple example, suppose $A$ is generated by one element $a$ with no relations, let $B$ be any Boolean algebra, and let $\eta(a)=1$. Then $\eta$ is trivially injective on the generators, but is not injective since $\eta(a)=\eta(1)$ but $a\neq 1$.