I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.

I have always assumed that this fact was part of the classical forcing folklore results, but ~~it doesn’t seem to be mentioned explicitly in the usual forcing literature~~ (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.)

**Theorem 1**. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.

The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.

**Theorem 2**. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.

By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.

**Proof**: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.

I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.

In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.

Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that

- $d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$.
- $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$.
- $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.

Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.

Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.

To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.

Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.**QED**

Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions.