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?

- $\mathbb{Q}$ is absorbed by $\mathbb{P}$ as a subforcing, i.e. there is a regular embedding from $\mathbb{Q} \to \text{r.o.}(\mathbb{P})$;
- $\mathbb{Q}$ 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.