Comparing Cohen algebra between models By Cohen algebra I mean $RO(\mathbb{R})$, the regular open algebra on real numbers, or equivalently the completion of (the) countable atomless algebra. Let $M\subseteq N$ be transitive models of set theory. I would like to know if $RO(\mathbb{R})^M$ can be embedded into $RO(\mathbb{R})^N$; I am mainly curious about the case when $M=L^N$ or $N$ is a forcing extension of $M$. There seem to be two options: (1) $i(U):=\text{Int}^N(\text{Cl}^N(U))$; (2) first define $i$ on rational intervals in the obvious way and then extend it to all regular open $U$.
Do these define the same map $i$? Is it actually an embedding? Actually for (2) I don't know how to prove that $i$ is well-defined or $i(U)$ is regular open.
Edit: Thanks for Eric's very detailed answer. As he mentions at the end of his answer, $RO(\mathbb{R})^M$ does embed into $RO(\mathbb{R})^N$ by the simple fact about boolean completion, so my question was stupid......Nevertheless it is nice to know how does the map exactly look like.
 A: The following topological fact is helpful here.

Lemma: Let $X$ be a topological space and let $Y\subseteq X$ be a dense subspace.  Then the map $f:RO(X)\to RO(Y)$ defined by $f(U)=U\cap Y$ is an isomorphism, and its inverse is the map $g:RO(Y)\to RO(X)$ given by $g(U)=\text{Int}^X(\text{Cl}^X(U))$.
Proof: First, let us check that $U\cap Y$ really is regular open in $Y$ if $U$ is regular open in $X$, so that $f$ is well-defined.  Let $V$ be the interior of the closure of $U\cap Y$ in $Y$ and suppose $y\in V$; we wish to show that $y\in U$.  There is some open set $W\subseteq X$ such that $V=W\cap Y$.  Since $Y$ is dense in $X$, $V$ is dense in $W$, so $U$ is dense in $W$.  Since $U$ is regular open, this means $W\subseteq U$, and hence $y\in U$, as desired.
Now it is clear that $f$ and $g$ are both order-preserving, so all that remains is to show they are inverses.  If $U\in RO(X)$, then $f(U)$ is dense in $U$ since $Y$ is dense in $X$, and then it follows that $g(f(U))=U$ since $U$ is regular open.  If $U\in RO(Y)$, let $V\subseteq X$ be an open set such that $U=V\cap Y$.  Since $U$ is dense in $V$, $g(U)$ is the interior of the closure of $V$, which makes it clear that $U\subseteq f(g(U))$.  For the reverse inclusion, note that $f(g(U))$ is an open subset of $Y$ contained in $\text{Cl}^X(U)\cap Y=\text{Cl}^Y(U)$, and thus is contained in $\text{Int}^Y(\text{Cl}^Y(U))=U$.

Let us now apply this fact to the topological spaces $\mathbb{Q}\subseteq\mathbb{R}^M\subseteq\mathbb{R}^N$, where each inclusion is dense.  To be clear, I am talking about these topological spaces externally, not internally to $M$ and $N$, so for instance by $RO(\mathbb{R}^M)$ I mean the algebra of all regular open subsets of $\mathbb{R}^M$ equipped with its order topology in the external universe.  So $RO(\mathbb{R}^M)$ may be larger than what you refer to as $RO(\mathbb{R})^M$, since not every open subset of $\mathbb{R}^M$ is an element of $M$.
The lemma tells us that $RO(\mathbb{Q}),$ $RO(\mathbb{R}^M)$, and $RO(\mathbb{R}^N)$ are all isomorphic, with the isomorphisms given by intersection in one direction and taking interiors of closures in the other direction.  Your question is not about these algebras but about the subalgebras $RO(\mathbb{R})^M\subseteq RO(\mathbb{R}^M)$ and $RO(\mathbb{R})^N\subseteq RO(\mathbb{R}^N)$.  But it is clear our isomorphism from $RO(\mathbb{R}^M)$ to $RO(\mathbb{R}^N)$ maps elements of $RO(\mathbb{R})^M$ to elements of $RO(\mathbb{R})^N$, since $M\subseteq N$.  So, $RO(\mathbb{R})^M$ does embed in $RO(\mathbb{R})^N$ by your proposed map $i(U)=\text{Int}^N(\text{Cl}^N(U))$.  Using the isomorphisms from $RO(\mathbb{R}^M)$ and $RO(\mathbb{R}^N)$ to $RO(\mathbb{Q})$, we also see that if $U\in RO(\mathbb{R})^M$ then $U$ and $i(U)$ contain exactly the same rational intervals, so $i(U)$ can also be described as the union (in $\mathbb{R}^N$) of all the rational intervals that $U$ contains.
Using the isomorphisms to $RO(\mathbb{Q})$ also gives another useful perspective on this embedding.  Applying the Lemma internally to $M$ and $N$, we see that $RO(\mathbb{R})^M\cong RO(\mathbb{Q})^M$ and $RO(\mathbb{R})^N\cong RO(\mathbb{Q})^N$.  So from this perspective, the embedding $RO(\mathbb{R})^M\to RO(\mathbb{R})^N$ is just the inclusion map $RO(\mathbb{Q})^M\to RO(\mathbb{Q})^N$.
Finally, I claim that this embedding is complete, i.e. it preserves all joins that exist in $RO(\mathbb{R})^M$.  In particular, this means that $i(U)$ can also be computed by taking any decomposition in $M$ of $U$ as a union of rational intervals and taking the corresponding union in $\mathbb{R}^N$.  To prove this completeness, let us work with the inclusion $RO(\mathbb{Q})^M\to RO(\mathbb{Q})^N$, so suppose $(U_i)$ is a family of regular open subsets of $\mathbb{Q}$ such that $U_i\in M$ for each $i$, and suppose that this family has a join $U$ in $RO(\mathbb{Q})^M$.  I claim that $U$ is actually the join of $(U_i)$ in $RO(\mathbb{Q})$ (and thus also in $RO(\mathbb{Q})^N$).  To prove this, suppose there is an upper bound $V\in RO(\mathbb{Q})$ of the $U_i$ which is strictly smaller than $U$.  Then $U\wedge \neg V\neq 0$ in the Boolean algebra $RO(\mathbb{Q})$, so there is some rational interval $W\leq U\wedge\neg V$.  Then $U\wedge \neg W<U$ in $RO(\mathbb{Q})^M$ but $U\wedge\neg W\geq V$ so $U\wedge\neg W\geq U_i$ for each $i$.  This contradicts the fact that $U$ is the least upper bound of the $U_i$ in $RO(\mathbb{Q})^M$, so no such $V$ can exist and $U$ is indeed the join of $(U_i)$ in $RO(\mathbb{Q})$.
(More generally, this last argument shows that if $B$ is a Boolean algebra and $A$ is a subalgebra which is dense in the sense that every nonzero element of $B$ contains some nonzero element of $A$, then the inclusion map $A\to B$ is complete.)

Finally, here is a different approach using a bit of machinery of forcing posets.  Recall the notion of a dense embedding $i:P\to Q$ of forcing notions, which is an order-preserving map with dense image which preserves incompatibility.  Any poset $P$ has a unique (up to unique isomorphism) dense embedding into a poset of the form $B\setminus\{0\}$ where $B$ is a complete Boolean algebra, known as the completion of $P$.  When $P$ is already a Boolean algebra with its least element removed, then $B$ is just its usual Boolean algebra completion.  If $i:P\to Q$ is a dense embedding, then it induces an isomorphism between the completions of $P$ and $Q$.
Now, the Cohen algebra can also be viewed as the completion of the poset $P$ of finite partial functions from $\omega$ to $2$.  This $P$ is absolute between $M$ and $N$.  So, in $M$, we have a dense embedding $P\to RO(\mathbb{R})^M\setminus\{0\}$, which $M$ thinks is the completion of $P$.  From $N$'s perspective, the completion of $P$ is actually the dense embedding $P\to RO(\mathbb{R})^N\setminus\{0\}$.  But $N$ also knows about the dense embedding $P\to RO(\mathbb{R})^M\setminus\{0\}$, and so it knows that this induces an isomorphism between the completion of $P$ (i.e., $RO(\mathbb{R})^N\setminus\{0\}$) and the completion of $RO(\mathbb{R})^M\setminus\{0\}$.  Thus in $N$, the completion of the Boolean algebra $RO(\mathbb{R})^M$ is $RO(\mathbb{R})^N$, and this gives the embedding you want.
