Extension of a sequence of complete embeddings

Question

Suppose for $\langle P_i : i < \omega \rangle$ is a sequence of countably closed partial orders. Suppose for each $n$, there is a complete embedding $$e_n : \prod_{i<n} P_i \to B(Q),$$ where $Q$ is countably closed, $B(Q)$ is its Boolean completion, the quotient $B(Q)/\mathrm{ran}(e_n)$ is forced to be equivalent to a countably closed partial order, and for $m<n$, $e_n \restriction \prod_{i<m} P_i = e_m$.

Does there exist a complete embedding $e_\omega : \prod_{i<\omega} P_i \to B(Q)$ extending each $e_n$? Can we ensure that $Q/\mathrm{ran}(e_\omega)$ is forced to be equivalent to a countably closed poset? If it helps, also assume each $P_i$ and $Q$ has the property that any descending $\omega$-sequence has an infimum, and the $e_n$’s are continuous.

Answer 1

I think I have a counterexample.

Let $Q$ be the partial order of functions $q$ with domain a countable ordinal $\alpha$, such that for each $\beta<\alpha$, $q(\beta)$ is an ordinal below $\omega^\omega$. Obviously this is isomorphic to adding a Cohen subset of $\omega_1$. Let $g$ be a name for the generic function.

For each $n$, let $R_n$ be the subforcing that decides the coefficient of $g(\beta)$ on $\omega^n$ for each $\beta$. Let each $P_n$ be the poset of functions with domain a countable ordinal into the natural numbers.

For each $n$, $\prod_{i<n} P_i$ embeds into $Q$ by determining the coefficients on $\omega^i$ for $i<n$. The quotient is the countably closed poset that determines the rest. We can write the quotient as $Q_n$, the poset of functions from countable ordinals where the output is an ordinal $<\omega^\omega$ with zero coefficients on $\omega^i$ for $i<n$. For each $n$, there is a dense subset of the product $\prod_{i<n} P_i \times Q_n$ that is isomorphic to $Q$. For $(p_0,\dots,p_{n-1},q)$ with each coordinate having domain $\alpha$, map this to the function $r$ with for all $\beta<\alpha$, $$r(\beta) = q(\beta) + \omega^{n-1}p_{n-1}(\beta)+\dots+\omega p_1(\beta)+p_0(\beta).$$

Notice that there is no extension of these embeddings as desired, since for any sequence with infinite support at some point common to the domains, this would map to an ordinal with an infinite sum as its Cantor normal form, which is absurd.