The Set-Theoretic Multiverse and Joint Embeddings I am curious whether or not the following axiom is independent of Hamkins's axioms for the Set-Theoretic Multiverse. Hamkins's axioms can be found here on pages 1-2 and here on pages 24-26. 
Consider the following axiom: If $(M,\in_M)$ and $(N,\in_N)$ are set-theoretic universes, then there exists a universe $(W,\in_W)$ such that there exists injective functions $j:M\to W$ and $i:N\to W$ where $(W|_{j(M)}, \in_W) \cong (M,\in_M)$ and $(W|_{i(N)},\in_W) \cong (N,\in_N)$.  
Remark: Independence might not be the correct work I am looking for (since the Set-Theoretic Multiverse is a foundational structure). I am looking to see whether this axiom (as well as its negation) is coherent in the same sense that the Multiverse axioms are coherent (i.e. in the same way that the collection of countable computably-saturated models of ZFC satisfies the multiverse axioms). 
 A: Let me add to Noah's remarks that your axiom (and Noah's) is true in the model that Victoria Gitman and I provide in our paper A natural model of the multiverse axioms, which you also had mentioned. In that paper, we prove that the collection of all countable computably saturated models of ZFC satisfies all the other multiverse axioms. It also satisfies your joint embedding property axiom, because of the main theorem of my article, Every countable model of set theory embeds into its own constructible universe, which shows that every countable computably saturated model of ZFC is universal for all countable acyclic binary relations. So in fact all these models are isomorphic to submodels of each other. 
Theorem. In any multiverse consisting of countable models, satisfying the multiverse axioms, the joint embedding property also holds. 
Proof: It follows from the well-founded mirage axiom that the multiverse must have an ill-founded model $W$, and the ordinals of $W$ must therefore contain a copy of $\mathbb{Q}$. It follows by my embeddings theorem that $\langle W,\in^W\rangle$ is universal for all countable acyclic binary relation. In particular, any two countable models of set theory embed into $W$, and so we have the joint embedding property. QED
In fact, it follows from the well-founded mirage axiom that every model $W$ of the multiverse is ill-founded, and hence universal for all countable models of set theory. 
But let me also point out that it follows from results in that paper that this notion of embedding is probably not what you should mean. A mere $\in$-homomorphism need not map ordinals to ordinals, and can map the empty set to a nonempty set and so on. Probably one wants to consider at least $\Delta_0$-elementary embeddings, if not much more than this. 
A: Some time ago, I looked into arithmetical consequences of the set-theoretic multiverse.
I started with the assumption that there are numbers that are standard, which I will denote $\omega$ and that we have some useful conception of subsets of $\omega$, I found that if two multiverses are compatible in the sense that they are both elements of some larger universe then they must have the same standard system. There is also an interesting converse to this that I discuss in my post.
In detail, if $U$ is a universe then elements $n$ of $\omega^U$ code certain subsets of $\omega$. Namely, $n$ codes the set $x_n$ of all standard $i \in \omega$ such that the $i$-th binary digit of $n$ is $1$. The collection of all sets coded by elements of $\omega^U$ in this way is called the standard system of $U$ and is denoted $\operatorname{SSy}(U)$.
If a universe $U$ is an element of a larger universe $V$ then we must have $\operatorname{SSy}(U) = \operatorname{SSy}(V)$. Indeed, because $V$ can recognize that $\omega^U$ is a model of a reasonable arithmetic theory, $V$ realizes that $\omega^V$ is an initial segment of $\omega^U$ since $V$ believes that $\omega^V$ is the standard model of arithmetic. From this, we find that $U$ and $V$ have the same standard system. Indeed, if $p$ is any nonstandard element of $V$ then every $x \in \operatorname{SSy}(V)$ is coded by some $n < 2^p$, and such $n$'s are both in $U$ and $V$.
Note that the assumptions I used are debatable, as discussed by Joel in this post. I never got around to writing the sequel of this post where I weaken the hypotheses somewhat. Perhaps I will find some time now that I am reminded of this...
Also note that compatibility, as defined above, is much stronger than your joint embedding property. However, the underlying reason why I chose to look at compatibility rather than embeddings remains: how would we know that such embeddings exist unless such embeddings exist in some larger universe?
A: The answer seems to be "no": since none of the multiverse axioms involve two universes at once, the union of two multiverses is a multiverse. This lets you easily (EDIT: As Joel points out, this is totally false: it is not at all easy, although it may be true) construct counterexamples to the claim.
Meanwhile, if we add the (in my opinion, desirable) axiom that for every pair of universes $M$, $N$ there is a $W$ in which both $M$ and $N$ are countable sets, then the answer is clearly "yes;" and moreover, these two statements are equivalent (given $W$ interpreting $M$ and $N$, just consider a $V$ in which $W$ is countable.
