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Let $ M $ be a countable, transitive model for $ \mathsf{ZFC}^* $, i.e. for a sufficiently large finite fragment of $ \mathsf{ZFC} $. Suppose that $ \mathbb{P} := (P, {\leq_P}, \mathbb{1}_P) \in M $ and $ \mathbb{Q} := (Q, {\leq_Q}, \mathbb{1}_Q) \in M $ are forcing notions.


Reminder

Definition ([Kun80, VII.7.1]). A mapping $ i \colon P \to Q $ is a complete embedding of $ \mathbb{P} $ into $ \mathbb{Q} $ iff

(i) $ \forall r, s \in P \ (r \leq_P s \implies i(r) \leq_Q i(s)) $,

(ii) $ \forall r, s \in P \ (r \perp_P s \implies i(r) \perp_Q i(s)) $, and

(iii) $ \forall q \in Q \ \exists p \in P \ \forall r \in P \ (r \leq_P p \implies i(r) \parallel_Q q) $.

A condition $ p \in P $ as in (iii) is called a reduction of $ q $ to $ \mathbb{P} $.

Remark. Note that in clause (ii) equivalence holds because of (i).

Definition ([Kun80, VII.7.7]). A mapping $ i \colon P \to Q $ is a dense embedding of $ \mathbb{P} $ into $ \mathbb{Q} $ iff

(i) $ \forall r, s \in P \ (r \leq_P s \implies i(r) \leq_Q i(s)) $,

(ii) $ \forall r, s \in P \ (r \perp_P s \implies i(r) \perp_Q i(s)) $, and

(iii) $ i[P] $ is dense in $ \mathbb{Q} $.

Remark. Every dense embedding is a complete embedding.

Theorem ([Kun80, II.3.3]). There exists a dense embedding of $ \mathbb{P} $ into $ \operatorname{RO}(\mathbb{P}) \setminus \{ \mathbb{0} \} $.

Lemma ([Kun80, VII.Ex.C2]). If $ \mathbb{P} $ and $ \mathbb{Q} $ are separative and $ i \colon P \to Q $ is a complete embedding of $ \mathbb{P} $ into $ \mathbb{Q} $, then $ i $ is one-to-one, $ i(\mathbb{1}_P) = \mathbb{1}_Q $, and $ (r \leq_P s \iff i(r) \leq_Q i(s)) $ holds for all $ r, s \in P $.


Now, consider the following statements:

(C1) For each $ \mathbb{Q} $-generic $ H $, there exists a $ G \in M[H] $ such that $ G $ is $ \mathbb{P} $-generic over $ M $. (Then $ M[G] \subseteq M[H] $.)

(C2) For each $ \mathbb{P} $-generic $ G $, there exists a $ \mathbb{Q} $-generic $ H $ such that $ G \in M[H] $. (Then $ M[G] \subseteq M[H] $.)

(C3) There exists a complete embedding $ i \in M $ of $ \mathbb{P} $ into $ \mathbb{Q} $.

(C4) There exists a complete embedding $ i \in M $ of $ \mathbb{P} $ into $ \operatorname{RO}(\mathbb{Q}) \setminus \{ \mathbb{0} \} $.

(D1) For each $ \mathbb{Q} $-generic $ H $, there exists a $ G \in M[H] $ such that $ G $ is $ \mathbb{P} $-generic over $ M $ and $ H \in M[G] $. (Then $ M[G] = M[H] $.)

(D2) For each $ \mathbb{P} $-generic $ G $, there exists a $ \mathbb{Q} $-generic $ H $ such that $ G \in M[H] $ and $ H \in M[G] $. (Then $ M[G] = M[H] $.)

(D3) There exists a dense embedding $ i \in M $ of $ \mathbb{P} $ into $ \mathbb{Q} $.

(D4) There exists a dense embedding $ i \in M $ of $ \mathbb{P} $ into $ \operatorname{RO}(\mathbb{Q}) \setminus \{ \mathbb{0} \} $.

(Pr) If $ \mathbb{Q} $ is proper, then $ \mathbb{P} $ is also proper.


Question

What implications between the above statements are provable in $ \mathsf{ZFC} $? Which are not?

If it is helpful, you may assume that $ \mathbb{P} $ and $ \mathbb{Q} $ are separative partial orders (in the strict sense).


Main problem

Suppose that (C1) holds. What additional assumptions do we need to show (Pr)?

(Note that (C2) implies (Pr). So what additional assumptions does one need to show (C2) from (C1)?)


Bibliography

[Kun80] Kenneth Kunen: Set Theory: An Introduction to Independence Proofs. North Holland, 1980

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  • 2
    $\begingroup$ In order to show that C1 doesn't imply C3 you can take $\mathbb{Q}$ to be Cohen forcing and take $\mathbb{P}$ to be the lottery sum of $\mathbb{Q}$ and something which is not c.c.c. (for example the collapse $Col(\omega, \omega_1)$). Now clearly $\mathbb{Q}$ adds a generic for $\mathbb{P}$ but there is no complete embedding from $\mathbb{P}$ to $\mathbb{Q}$ since by condition (ii) it would give you an uncountable antichain in $\mathbb{Q}$. $\endgroup$ – Yair Hayut Jul 17 '14 at 14:19
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    $\begingroup$ @Yair: Why are you online? $\endgroup$ – Asaf Karagila Jul 17 '14 at 14:56
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    $\begingroup$ To answer the first question in Justus87's second comment: Yes. If a dense subset of a poset is ccc then so is the whole poset. $\endgroup$ – Andreas Blass Jul 17 '14 at 15:31
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    $\begingroup$ A fact that I heard mentioned is that if $\mathbb Q$ adds a generic for $\mathbb P$, then there is some condition $p\in\mathbb P$ such that $\mathbb P$ below $p$ embeds into $\text{ro}(\mathbb Q)$. (Note that this avoids Yair's counterexample) Is there a simple proof of this? $\endgroup$ – Victoria Gitman Jul 17 '14 at 18:14
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    $\begingroup$ In general in my eyes all these type of arguments are better understood looking not at complete embeddings but at complete homomorphisms (possibly non injective - see the comment of Victoria Gitman above) of the boolean completions. May be you can find some other informations on some of these questions in arxiv.org/abs/1402.1714 $\endgroup$ – matteo viale Jul 17 '14 at 22:43
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The following implications are either trivial or well known:

(D$ n $) implies (C$ n $): Trivial.

(C3) implies (C1): Use $ G := i^{-1}[H] $. (See [Kun80, VII.7.5].)

(C3) implies (C4): The composition of two complete embeddings is a complete embedding.

(D3) implies (D1): Use $ G := i^{-1}[H] $. Then $ H = \{ q \in Q : \exists p \in G \ i(p) \leq q \} $. (See [Kun80, VII.7.11].)

(D3) implies (D2): Use $ H := \{ q \in Q : \exists p \in G \ i(p) \leq q \} $. Then $ G = i^{-1}[H] $. (See [Kun80, VII.7.11].)

(D3) implies (D4): The composition of two dense embeddings is a dense embedding.


Regarding (C1) does not imply (C3) and (C1) does not imply (C4):

Let $ \mathbb{Q} $ be the Cohen forcing and let $ \mathbb{P} $ be the lottery sum of $ \mathbb{Q} $ and $ \operatorname{Col}(\omega, \omega_1) $. Then (C1) clearly holds (use $ G := H $), but (C3) and (C4) do not hold since $ \mathbb{P} $ is not c.c.c. whereas $ \mathbb{Q} $ and hence $ \operatorname{RO}(\mathbb{Q}) $ satisfy the countable chain condition.

(See the comments of Yair Hayut and Andreas Blass.)

Also see A common forcing extension obtained via different forcing notions by J.D. Hamkins or Lemma 25.5 in T. Jech's book Set Theory (1978 edition).


Regarding (C3) implies (C2):

Each $ H \subseteq Q $ which is $ \mathbb{Q} / G $-generic over $ M[G] $ is also $ \mathbb{Q} $-generic over $ M $, and $$ G \in M[G][H]_{\mathbb{Q} / G} = M[H]_{\mathbb{Q}}. $$

(See [Kun80, VII.Ex.D3] and [Kun80, VII.Ex.D4].)

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