I was recently asked this interesting question on set-theoretic geology by Iian Smythe, a set-theory post-doc at Rutgers University; the problem arose in the context of one of this current research projects.

**Question.** Assume that two product forcing extensions are the same $$M[G][K]=M[H][K],$$ where $M[G]$ and $M[H]$ are forcing extensions of $M$ by the same forcing notion $\mathbb{P}$, and $K\subset\mathbb{Q}\in M$ is both $M[G]$ and $M[H]$-generic with respect to this further forcing $\mathbb{Q}$. Can we conclude that $$M[G]=M[H]\ ?$$ Can we make this conclusion at least in the special case that $\mathbb{P}$ is adding a Cohen real and $\mathbb{Q}$ is collapsing the continuum?

It seems natural to hope for a positive answer, because we are aware of many such situations that arise with forcing, where indeed $M[G]=M[H]$. Nevertheless, the answer is negative. Indeed, we cannot legitimately make this conclusion even when both steps of forcing are adding merely a Cohen real. And such a counterexample implies that there is a counterexample of the type mentioned in the question, simply by performing further collapse forcing.

**Theorem.** For any countable model $M$ of set theory, there are $M$-generic Cohen reals $c$, $d$ and $e$, such that

- The Cohen reals $c$ and $e$ are mutually generic over $M$.
- The Cohen reals $d$ and $e$ are mutually generic over $M$.
- These two pairs produce the same forcing extension $M[c][e]=M[d][e]$.
- But the intermediate models are different $M[c]\neq M[d]$.

**Proof.** Fix $M$, and let $c$ and $e$ be any two mutually generic Cohen reals over $M$. Let us view them as infinite binary sequences, that is, as elements of Cantor space. In the extension $M[c][e]$, let $d=c+e \mod 2$, in each coordinate. That is, we get $d$ from $c$ by flipping bits, but only on coordinates that are $1$ in $e$. This is the same as applying a bit-flipping automorphism of the forcing, which is available in $M[e]$, but not in $M$. Since $c$ is $M[e]$-generic by reversing the order of forcing, it follows that $d$ also is $M[e]$-generic, since the automorphism is in $M[e]$. Thus, $d$ and $e$ are mutually generic over $M$. Further, $M[c][e]=M[d][e]$, because $M[e][c]=M[e][d]$, as $c$ and $d$ were isomorphic generic filters by an isomorphism in $M[e]$. But finally, $M[c]$ and $M[d]$ are not the same, because from $c$ and $d$ together we can construct $e$, because we can tell exactly which bits were flipped. $\Box$

If one now follows the $e$ forcing with collapse forcing, one achieves a counterexample model of the type mentioned in the question, namely, with $M[c][e*K]=M[d][e*K]$, but $M[c]\neq M[d]$.

I have a feeling that my co-authors on a current paper in progress, Set-theoretic blockchains, on the topic of non-amalgamation in the generic multiverse, will tell me that the argument above is an instance of some of the theorems we prove in the latter part of that paper. (Miha, please tell me in the comments, if you see this, or tell me where I have seen this argument before; I think I made this argument or perhaps seen it before.) The paper is [bibtex key=”HabicHamkinsKlausnerVernerWilliams2018:Set-theoretic-blockchains”].

Hi Joel, and thanks again for your answer to this question. The solution is very clear.

Out of curiosity, could you say a bit about why various known results in set-theoretic geology, e.g., the definability of the ground model in a forcing extension, do not work here?

Thanks, it was a great question.

For the definability of the ground model, indeed, the model $M$ is definable in $M[c]$ and $M[c][e]$ and so on, using the parameter $P(\omega)^M$. And $M[c]$ and $M[d]$ are each definable in the common extension $M[c][e]=M[d][e]$, using the parameters $P(\omega)^{M[c]}$ and $P(\omega)^{M[d]}$, respectively, but those parameters are not the same, and so this is why the ground-model definability theorem does not give you $M[c]=M[d]$.

Joel, it is not clear to me that your argument follows from something in the paper (or at least I cannot see a straight path from one to the other). I think you advocated using bit-flipping automorphisms like this when we were discussing surgery some time ago, but I am not sure anything exactly like this argument came up.

OK, thanks for the comment. I guess there is a family resemblance in all the bit-flipping arguments.