# Failure of Cantor-Bernstein for the Levy Collapse

Related to this question, is it possible to give an example of the failure of Cantor-Bernstein for complete embeddings of forcing notions involving the Levy Collapse $Col(\omega,<\kappa)$? Suppose $\kappa$ is a large cardinal. Does there exist some forcing $\mathbb{P}$ such that $\mathbb{P} \triangleleft Col(\omega,<\kappa)$, and $Col(\omega,<\kappa) \triangleleft \mathbb{P}$, but $\mathbb{P}$ is not equivalent to $Col(\omega,<\kappa)$? The analogous question for $Col(\mu,<\kappa)$, $\mu$ regular, is also of interest.

Clarification: By $\mathbb{P} \triangleleft \mathbb{Q}$, I mean there exists a complete embedding (aka regular embedding) $e : \mathbb{P} \to \mathcal{B}(\mathbb{Q})$, or equivalently a projection map $\pi : \mathcal{B}(\mathbb{Q}) \to \mathcal{B}(\mathbb{P})$.

If you mean as in the other question that $\mathbb{P}\triangleleft\mathbb{Q}$ just in case every generic filter for $\mathbb{Q}$ adds a generic filter for $\mathbb{P}$, then the answer is yes, because Jonas Reitz's idea on the other question is also applicable here. Namely, let $\mathbb{P}$ be the lottery sum $\text{Col}(\omega,\lt\kappa)\oplus\mathbb{Q}$, where $\mathbb{Q}$ is some much larger poset above, such as $\text{Col}(\omega,\lt\theta)$, which definitely adds a generic for $\text{Col}(\omega,\lt\kappa)$. The point now is that forcing with $\mathbb{P}$ will definitely add a generic for $\text{Col}(\omega,\lt\kappa)$, regardless of which choice is made in the lottery, and conversely every generic for $\text{Col}(\omega,\lt\kappa)$ is also generic for $\mathbb{P}$, by opting for the first term in the lottery. But they are not forcing equivalent. The same idea works with $\text{Col}(\mu,\lt\kappa)$, and indeed, with essentially any forcing notion at all.
But I find it more likely that you intend $\mathbb{P}\triangleleft\mathbb{Q}$ just in case there is a complete embedding of $\mathbb{P}$ into the (Boolean completion) of $\mathbb{Q}$, which in many ways is the natural relation to consider with forcing. In this case, your question is related to the question whether the Lévy collapse $\text{Col}(\omega,\lt\kappa)$ up to an inaccessible cardinal $\kappa$ is characterized up to forcing equivalence as the unique $\kappa$-c.c. forcing notion of size $\kappa$ that collapses all smaller cardinals to $\omega$. Namely, if this characterization is correct, then whenever $\mathbb{P}\triangleleft\text{Col}(\omega,\lt\kappa)\triangleleft\mathbb{P}$, we would see that $\mathbb{P}$ would be $\kappa$-c.c. and have a size $\kappa$ dense set by the first relation, and it would collapse all smaller cardinals by the second relation, and so by the assumed characterization it would be forcing equivalent to $\text{Col}(\omega,\lt\kappa)$. Thus, if this characterization of the Lévy collapse is correct, then there are no counterexamples of the kind you seek.
The analogous characterization is true in the case of $\text{Col}(\omega,\theta)$, which is in fact the unique poset of size $|\theta|$ necessarily collapsing $\theta$ to $\omega$, and a proof of this can be found for example as (folklore) lemma 18 of my paper Structural connections between a forcing class and its modal logic. I have always expected that this characterization extends to the Levy collapse in the way I have suggested, but I do not have a proof of this.
• Yes, I meant there is a complete embedding, sorry for the ambiguity. Actually your conjecture about $Col(\omega,<\kappa)$ is wrong: see Mohammad's answer here mathoverflow.net/questions/121352/resembling-the-levy-collapse/…. However Mohammad's example does not provide a failure of Cantor-Bernstein. Mar 6, 2014 at 15:49