Suppose $ \mathbb{P} := (P, {\leq\_P}, 1\_P) $ is a separative partial order. Let $ \mathbb{B} := \operatorname{RO}(\mathbb{P}) $ denote the Boolean completion.


Fix some dense embedding $ i \colon P \to B^+ $. Then $ i $ is an order-isomorphism (one-to-one, order-preserving) to its image.


Assume that $ \mathbb{P} $ is isomorphic to each of its cones $ \mathbb{P} \mathord{\upharpoonright} p := \{ q \in P : q \leq\_P p \} $ (where $ p \in P $ is some condition).


**Question.** Is $ \mathbb{B} $ isomorphic to each of its cones? If not, is $ \mathbb{B} $ isomorphic to each cone of the form $ \mathbb{B} \mathord{\upharpoonright} i(p) $ (where $ p \in P $) at least?


*Attempt to proof the second statement.* Fix $ p \in P $. Let $ f \colon \mathbb{P} \to \mathbb{P} \mathord{\upharpoonright} p $ denote an isomorphism. Then $ i \circ f \circ i^{-1} \colon \mathbb{B} \to \mathbb{B} \mathord{\upharpoonright} i(p) $ is a canonical candidate for the isomorphism since $ i \mathord{\upharpoonright} (\mathbb{P} \mathord{\upharpoonright} p) $ should be a dense embedding of $ \mathbb{P} \mathord{\upharpoonright} p $ into $ \mathbb{B} \mathord{\upharpoonright} i(p) $, right?



Boolean completion (of a forcing notion) isomorphic to each of its cones
Your second statement is correct, simply because boolean completions are unique up to isomorphism.


For the stronger statement, let $b \in \mathbb{B}^+$. Let $\{ p\_\alpha : \alpha < \kappa \}$ be a maximal antichain of elements of $i[\mathbb{P}]$ below $b$. Pick some maximal antichain $\{ q\_\alpha : \alpha < \delta \}$ in $\mathbb{P}$ where $\delta \geq \kappa$. If $\delta > \kappa$, then using the fact that $\mathbb{P} \cong \mathbb{P} \restriction p\_0$, enlarge the first antichain to one of size $\delta$; denote it $\{ p'\_\alpha : \alpha < \delta \}$. So $\mathbb{B} \restriction b$ is isomorphic to the complete boolean sum of $\delta$ copies of $\mathbb{B}$, since $\mathbb{B} \cong \mathbb{B} \restriction p'\_\alpha$ for each $\alpha$. But so is $\mathbb{B}$ itself.


