It is a (folklore?) fact that if $\kappa$ is a regular cardinal, and $\mathbb{P}$ is a $\kappa$-closed poset such that $\Vdash\_\mathbb{P} |\mathbb{P}| = \kappa$, then $\mathbb{P}$ is equivalent to $Col(\kappa,\mathbb{P})$, the collection of partial functions from $\kappa$ to $\mathbb{P}$ of size $<\kappa$, ordered by inclusion. Hence if $\mathbb{P}$ is any $\kappa$-closed forcing, then $\mathbb{P} \times Col(\kappa,\mathbb{P}) \sim Col(\kappa,\mathbb{P})$, so $\mathbb{P}$ is completely embeddable into $\mathcal{B}(Col(\kappa,\mathbb{P}))$.


If $\mathbb{P}$ is $\kappa$-strategically closed (following the notation from Cummings' chapter of the *Handbook of Set Theory*, not Jech's book), then one can show that $\mathbb{P}$ is completely embeddable into $\mathcal{B}(Col(\kappa,\mathcal{P}(\mathbb{P})))$, since after forcing with this collapse we can use the strategy to build a filter on $\mathbb{P}$ that is generic over the ground model. It is not clear to me whether we can always embed such $\mathbb{P}$ into the smaller forcing $Col(\kappa,\mathbb{P})$.


**Question:** Are there counterexamples to the statement, "If $\kappa$ is regular and $\mathbb{P}$ is $\kappa$-strategically closed, then there is a complete embedding of $\mathbb{P}$ into $\mathcal{B}(Col(\kappa,\mathbb{P}))$?



absorption of strategically closed posets
If $\mathbb{P}$ is $\kappa$-strategically closed then there is a projection from $Col(\kappa,\mathbb{P})$ onto $\mathbb{P}$. 


The proof is very similar to the $\kappa$-closed case: 


Let $\sigma$ be a winning strategy for the game of length $\kappa$. 
For every condition in $Col(\kappa,\mathbb{P})$, $f:\eta \rightarrow \mathbb{P}$ ($\eta < \kappa$), let define a condition in $\mathbb{P}$, $q\_f$, to be the last condition in the game of length $\eta + 1$ that is played according to $\sigma$ and in which the "bad" player plays in step $\alpha$ the condition $f(\alpha)$, if it is a legal move, and otherwise he doesn't move. Since $\sigma$ is a winning strategy, there is condition $p\_\eta$ that was played in the last move of the game. So $q\_f = p\_\eta$. 


It's clear that if $f\leq g$ as conditions in $Col(\kappa,\mathbb{P})$ then $q\_f \leq q\_g$ and it also clear that for every $g\in Col(\kappa, \mathbb{P})$, the image of the function $f\rightarrow q\_f$ for $f\leq g$ is dense below $q\_g$, so it is a projection. 


Now we can verify that if $G$ is a generic filter for $Col(\kappa, \mathbb{P})$ then $\{ p \in \mathbb{P} | \exists f\in G,\,q\_f \leq p\}$ is a generic filter for $\mathbb{P}$: For every $D\subset \mathbb{P}$ open and dense and $g \in Col(\kappa, \mathbb{P})$, there is $f\leq g$ such that $q\_f \in D$ (by the density of the image of $f\rightarrow q\_f$ below $q\_g$).


