Suppose $T$ is a complete $\mathcal{L}$-theory with infinite models. Enumerate the collection of all $\emptyset$-definable equivalence relations modulo $T$ as $(E_\alpha(\overline{v}_1,\overline{v}_2))_{\alpha\in\lambda}$ for some cardinal $\lambda$, where each $E_\alpha$ is an equivalence relation on $n_\alpha$-tuples; we may assume $E_0$ is equality of one-tuples. Then we define $\mathcal{L}^{eq}$ to be a $\lambda$-sorted language, with sorts $(S_\alpha)_{\alpha\in \lambda}$, and we refer to $S_0$ as the "home sort". $\mathcal{L}^{eq}$ contains a symbol for every symbol $\mathcal{L}$, interpreted as acting on $S_0$. (So, eg, if $\mathcal{L}$ has an $n$-ary relation symbol $R$, then $\mathcal{L}^{eq}$ has an $n$-ary relation symbol $R^{eq}$, defined on the sort $S_0$.) Also, for each $\alpha\in\lambda$, $\mathcal{L}^{eq}$ has a function symbol $\pi_\alpha:S_0^{n_\alpha}\to S_{\alpha}$, considered to be the projection map taking a tuple to its equivalence class modulo $E_\alpha$.
In particular, every $\mathcal{L}$-sentence can be thought of as an $\mathcal{L}^{eq}$-sentence in a natural way, by considering it as a statement about the $S_0$ sort. Then we define $T^{eq}$ to be the union of all these sentences with the additional axioms
\begin{alignat*}{2}
&\bullet\forall (w\in S_{\alpha}) &&\exists (v_1\dots v_{n_\alpha}\in S_0) [w=\pi_\alpha(v_1,\dots,v_{n_\alpha})] \\
&\bullet\forall(v^{(1)}_1,\dots,&&v^{(1)}_{n_\alpha}\in S_0)\forall(v_1^{(2)},\dots,v_{n_\alpha}^{(2)}\in S_0) \\ & &&[\pi_\alpha(v^{(1)}_1,\dots,v^{(1)}_{n_\alpha})=\pi_\alpha(v_1^{(2)},\dots,v_{n_\alpha}^{(2)})]\leftrightarrow E_\alpha\left(\overline{v^{(1)}},\overline{v^{(2)}}\right)
\end{alignat*}
for every $\alpha\in\lambda$. It is easy to see that $T^{eq}$ is consistent; take any $M$ a model of $T$, and interpret each $S_\alpha^M$ as the equivalence classes of $M^{n_\alpha}$ under $E_\alpha^M$, with $\pi_\alpha$ the canonical projection map. Now, the claim is that $T^{eq}$ is complete. This seems obvious to me, but the resource I'm using (Tent and Ziegler) doesn't include a proof, and the only proof I can come up with uses some machinery that seems like it shouldn't be necessary:
For an arbitrary language $\mathcal{L}$, we say that an $\mathcal{L}$-structure $M$ of cardinality $\kappa\geqslant\aleph_0$ is "special" if $M$ is the union of an elementary chain $(M_\lambda)_{\lambda\in\kappa}$, where each $M_\lambda$ is $\lambda^{+}$-saturated.
Fact 1: If two special $\mathcal{L}$-structures of cardinality $\kappa$ are elementarily equivalent, then they are isomorphic. $\square$
Fact 2: Let $M$ be any infinite $\mathcal{L}$-structure. Then, for any cardinal $\lambda$ with $\beth_\lambda>|M|$, there is a special elementary extension of $M$ of size $\beth_\lambda$. In particular, any $\mathcal{L}$-theory with infinite models has arbitrarily large special models. $\square$
Lemma: The $\mathcal{L}^{eq}$-theory $T^{eq}$ is complete.
Proof: Let $T_1\supseteq T^{eq}$ and $T_2\supseteq T^{eq}$ be any complete $\mathcal{L}^{eq}$ theories. By fact 2, there are special $\mathcal{L}^{eq}$ structures $M_1\models T_1$ and $M_2\models T_2$ of the same cardinality. By construction of $T^{eq}$, each $M_i$ can be considered as a model of $T$ in the natural way, by restricting attention to the $S_0$ sort. Reducts of $\mu$-saturated structures are $\mu$-saturated, so each $M_i$ is in fact a special $\mathcal{L}$-structure. Thus, since $T$ is complete and so $M_1$ and $M_2$ are elementarily equivalent $\mathcal{L}$-structures, by fact 1 there is an $\mathcal{L}$-isomorphism $f:M_1\to M_2$.
We claim we can extend $f$ to an $\mathcal{L}^{eq}$ isomorphism $\overline{f}:M_1\to M_2$ by taking $$\overline{f}(\pi^{M_1}_\alpha(m_1,\dots,m_{n_\alpha}))=\pi_\alpha^{M_2}(f(m_1),\dots,f(m_{n_\alpha})).$$ Note that, since the $\pi_\alpha$ are surjective, this defines $\overline{f}$ on all of $S^{M_1}_\alpha$. Also, for all $\overline{m}$ and $\overline{m}'$, we have
\begin{align}
\pi^{M_1}_\alpha(m_1,&\dots,m_{n_\alpha})=\pi^{M_1}_\alpha(m'_1,\dots,m'_{n_\alpha}) \iff E^{M_1}_\alpha\left(\overline{m},\overline{m}'\right) \iff E^{M_2}_\alpha\left(f(\overline{m}),f(\overline{m}')\right) \\
&\iff \pi^{M_2}_\alpha(f(m_1),\dots,f(m_{n_\alpha}))=\pi^{M_2}_\alpha(f(m'_1),\dots,f(m'_{n_\alpha})).
\end{align}
The forward direction of this shows that $\overline{f}$ is well-defined, and the backwards direction shows that $\overline{f}$ is injective. To show that $\overline{f}$ is surjective, let $\pi^{M_2}_\alpha(m_1,\dots,m_{n_\alpha})\in S_\alpha^{M_2}$ be arbitrary. By surjectivity of $f$, there are $m'_1,\dots,m'_{n_\alpha}$ such that $f(\overline{m}')=\overline{m}$, and then $$\pi^{M_2}_\alpha(m_1,\dots,m_{n_\alpha})=\overline{f}(\pi^{M_1}_\alpha(m'_1,\dots,m'_{n_\alpha})),$$ as desired. Finally, $\overline{f}$ commutes with the $\pi_\alpha$ by construction and takes each $S_\alpha^{M_1}$ to $S_\alpha^{M_2}$, so it is in fact an $\mathcal{L}^{eq}$ isomorphism, and – in particular – $M_1$ and $M_2$ are elementarily equivalent as $\mathcal{L}^{eq}$-structures. Thus $T_1=T_2$ and so $T^{eq}$ is itself complete, as desired. $\square$
Now, I have two questions. First, does this proof look right? Second, it seems to me that there should perhaps be a more immediate proof of the desired result. Special structures take a bit of effort to define (though no set-theoretic assumptions! in contrast with saturated structures); is there a more direct way of seeing the result?
Best Answer
Your proof looks fine, but you're right that there's a more elementary approach, not using special or saturated models, or EF-games.
Point 2 is a bit tedious to write out carefully, but it's worth doing: this is the fundamental fact about the eq construction that makes everything else work.