# absorption of strategically closed posets

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 $\mathcal{B}(Col(\kappa,\mathbb{P}))$?

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 $q_g$).

• Actually more is true: $\mathbb{P}$ is $\kappa$-strategically closed iff $Col(\kappa,\lambda) \cong \mathbb{P}\ast \mathbb{R}$ for some $\mathbb{R}$ and $\lambda=|\mathbb{P}|$. May 9, 2014 at 11:58

Meanwhile, let's observe that if you weaken strategic $\kappa$-closure to the requirement that the forcing is strategically ${\lt}\kappa$-closed, that is, strategically $\beta$-closed for all $\beta<\kappa$, then the conclusion does not necessarily hold. The difference is between player II being able to continue playing for $\kappa$ many moves in a single game, versus being able to play for $\beta$ many moves in the length $\beta$ games for arbitrarily large $\beta$ below $\kappa$.

Suppose $\kappa$ is a Mahlo cardinal, and let $\mathbb{P}$ be the forcing to kill the Mahloness of $\kappa$. Conditions in $\mathbb{P}$ consist of closed bounded subsets of $\kappa$, containing no regular cardinals, ordered by end-extension. This forcing has size $\kappa$, and it adds a club in $\kappa$ containing no regular cardinals. It is strategically $\lt\kappa$-closed, since for any $\beta<\kappa$, the collection of conditions mentioning an ordinal above $\beta$ is $\leq\beta$-closed and dense in $\mathbb{P}$. So player II can win the games of length $\beta+1$ simply by playing into that dense set at the first opportunity.

Meanwhile, $\mathbb{P}$ does not completely embed into $\text{Coll}(\kappa,\kappa)$, because that forcing is ${\lt}\kappa$ closed and hence preserves every ground model stationary subset of $\kappa$ (one can see this by a pseudo-generic argument, since for any name for a club avoiding a given set, one can produce a club in the ground model avoiding that set), while $\mathbb{P}$ destroys the stationarity of the set of regular cardinals below $\kappa$. So the collapse forcing cannot add a generic filter for $\mathbb{P}$.