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Suppose $\delta$ is an inaccessible cardinal, and $\mathbb{P}$ is the Levy Collapse $\text{Col}(\kappa, \delta)$ which adds a surjection from $\kappa \to \delta$ (for some regular $\kappa < \delta$). It is well-known that there is a poset---namely $\text{Col}(\kappa, < \delta)$---which is absorbed by $\mathbb{P}$, and which has the same collapsing effects as $\mathbb{P}$, except that it doesn't collapse $\delta$.

Is this true in general? For simplicity, assume GCH and, if necessary, that $\delta$ is a very large cardinal. Suppose $\mathbb{P} \subset V_\delta$ is a poset which collapses $\delta$. Must there exist a poset $\mathbb{Q}$ with the following properties?

  1. $\mathbb{Q}$ is absorbed by $\mathbb{P}$ as a subforcing, i.e. there is a regular embedding from $\mathbb{Q} \to \text{r.o.}(\mathbb{P})$;
  2. $\mathbb{Q}$ collapses a tail end of cardinals below $\delta$, but does not collapse $\delta$.

Note: The answer is ``yes" in the special case that the closure of $\mathbb{P}$ matches $|\delta|^{V^{\mathbb{P}}}$, using standard absorption theory for Levy collapses. In particular, it's true if $\mathbb{P}$ makes $\delta$ countable.

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Do you have a counterexample when you drop the $\mathbb{P}\subset V_\delta$ restriction? – Joel David Hamkins Jun 19 '13 at 2:24
@Joel: No, I don't. – Sean Cox Jun 19 '13 at 3:21
Small observation: if we perform an additional small forcing $\mathbb{R}$, then we can find such a $\mathbb{Q}$ inside $\mathbb{P}\ast\mathbb{R}$. The reason is that we can let $\mathbb{R}$ collapse $|\delta|^{V^{\mathbb{P}}}$ to $\omega$, which is small relative to $\delta$ in the ground model, and so the composition will collapse $\delta$ to $\omega$, placing it into the case you mention at the end of the question. – Joel David Hamkins Jun 19 '13 at 5:09

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