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Let $T$ be a theory and let $I$ be an imaginary sort of $T$. We'll let $T_I$ denote the induced theory on $I$, by which I mean specifically the theory of all $\varnothing$-definable relations on $I$.

Suppose that

  1. for all models $\mathfrak{I} \models T_I$ there is an $\mathfrak{M} \models T$ such that $\mathfrak{I} \cong I(\mathfrak{M})$ and furthermore $\mathfrak{M}$ is unique in the sense that for any two $\mathfrak{M}_0$ and $\mathfrak{M}_1$ with isomorphisms $f_i : I(\mathfrak{M}_i) \cong \mathfrak{I}$ there is an isomorphism $g:\mathfrak{M}_0 \cong \mathfrak{M}_1$ such that $$f_1^{-1} \circ I(g) \circ f_0 = \mathrm{id}_\mathfrak{I},$$ where $I(g): I(\mathfrak{M}_0)\cong I(\mathfrak{M}_1)$ is the induced isomorphism and $\mathrm{id}_{\mathfrak{I}}$ is the identity on $\mathfrak{I}$, and

  2. the home sort of $T$ is 'extensional' over $I$ in the sense that for any $a,b \in \mathfrak{M} \models T$, we have that $a=b$ if and only if $a\equiv_{I(\mathfrak{M})} b$.

Does it follow that the interpretation of $T_I$ in $T$ extends to a bi-interpretation between $T$ and $T_I$?

The conclusion does not follow from either 1 or 2 alone:

  • Let $T$ be the theory of an equivalence relation with infinitely many equivalence classes of size $2$, and let $I$ be the imaginary that results from modding by the equivalence relation. $T$ and $I$ satisfy 1 but not 2 and fail the conclusion.
  • Let $T$ be the theory of the collection of subsets of $\omega$ with $\subseteq$. The set of singletons is definable, so we can define an equivalence relation where $xEy$ if either $x$ and $y$ are both singletons and $x=y$ or $x$ and $y$ are both not singletons. If we let $I$ be the imaginary resulting from modding by this equivalence relation then $T$ and $I$ satisfy 2 but not 1 and fail the conclusion.
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  • $\begingroup$ I'm not sure I understand the meaning of "the home sort of $T$ is an imaginary of $T_I$". Do you just mean that the interpretation of $T_I$ in $T$ extends to a bi-interpretation between $T$ and $T_I$? $\endgroup$ Nov 12, 2019 at 20:55
  • $\begingroup$ Yeah, that's a better way to put it. $\endgroup$ Nov 13, 2019 at 14:34

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Your question lies in the realm of relative categoricity (over a predicate), which is related to the theory of stability over a predicate. Here the predicate in question is the imaginary sort $I$.

You can read about this in Section 12.5 of Hodges's (longer) Model Theory. Using Hodges's terminology, your condition 1 is a bit stronger than saying that $T$ is relatively categorical over $I$, and conditions 1 and 2 together imply that $T$ is rigidly relatively categorical over $I$.

The answer to your question is yes, at least when the language is countable. This is a consequence of Gaifman's coordinatization theorem (Theorem 12.5.8 on p. 645 of Hodges), augmented by Exercises 11 and 13 on p. 649.

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  • $\begingroup$ Not that I need the uncountable case, but I always get curious: Do you think it fails for uncountable languages? $\endgroup$ Nov 13, 2019 at 21:03
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    $\begingroup$ Well, the proof of Gaifman's theorem seems to use omitting types in a crucial way... I don't have nearly enough intuition about this stuff to make a guess about whether there's a counterexample. $\endgroup$ Nov 13, 2019 at 22:09

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