indiscernible sequence?
Morley sequence?
Ramsey theorem?
Erdos-Rado theorem?
Ehrenfeucht-Fraïssé games (back-and-forth games)
Hrushovski construction?
generic predicate?
In the category of sets, we have a bijective correspondence (up to isomorphism) between surjective functions on $X$ and equivalence relations on $X$, where a surjective function $\pi \colon X \to Y$ is taken to the equivalence relation $E_{\pi}$ gotten by pulling back the diagonal of $Y$, and an equivalence relation $E$ is taken to the function $f_E$ which projects $X$ onto the set of $E$-classes. If we replace Set with the category of definable sets $\mathbf{Def}(T)$ of a first-order theory $T$, this correspondence generally fails, because when $E$ is a definable equivalence relation, there may not be a corresponding definable $f_E$. That is to say, internal congruences in $\mathbf{Def}(T)$ are not generally effective.
We say that:
$T$ has uniform elimination of imaginaries if the above correspondence does hold, that is, if internal congruences in $\mathbf{Def}(T)$ are effective.
$T$ has elimination of imaginaries if for every definable set $X = \{m \in \mathbb{M} \operatorname{ | } \varphi(m,b)\}$, there exists a formula $\psi(x,y)$ and a tuple $c$ with the same sort as $y$ such that $c$ uniquely satisfies $X = \{m \in M \operatorname{ | } \psi(m,c)\}.$
Inside a saturated model $\mathbb{M} \models T$, elimination of imaginaries is equivalent to the following statement: for every definable set $X$, there exists a tuple $c$ such that for all $\sigma \in \operatorname{Aut}(\mathbb{M})$, $\sigma(X) = X$ setwise if and only if $\sigma(c) = c$ pointwise. This is sometimes called the coding of definable sets.
Poizat noticed that $T$ having elimination of imaginaries allows one to develop a classical Galois theory classifying definably-closed extensions of small parameter sets in terms of the closed subgroups of a profinite automorphism group in any sufficiently saturated model of $T$.
The imaginary elements of a theory $T$ are precisely $E$-classes, where $E$ is a $0$-definable equivalence relation on $T$.
In the process of inventing stability theory, i.e. classifying theories according to the number of isomorphism classes of their models in each cardinality, Shelah found he needed to eliminate imaginaries in a universal way, so he found a way to conservatively interpret each theory $T$ in a theory $T^{\operatorname{eq}}$ which eliminated all the imaginaries of $T$. The process just involves throwing in for each congruence $E$ on $X$ a new sort (type in type theory) for $X/E$ and a projection map $X \twoheadrightarrow X/E$.
$T^{\operatorname{eq}}$ satisfies the following universal property: any theory which interprets $T$ and eliminates imaginaries must uniquely interpret $T^{\operatorname{eq}}$ as theories under $T$ (in the $2$-category of theories and interpretations.)
Makkai realized not long after that the $(-)^{\operatorname{eq}}$ construction corresponded to the pretopos completion of the syntactic category of $T$.
(Indeed, pretoposes are just toposes that might be missing some power objects. According to the next section, inside a Boolean pretopos these power objects will still be there, just generally only formally.)
If a first-order theory eliminates imaginaries, it interprets infinitely many constants. This is because it interprets at least two constants: a code for the diagonal relation and its complement. Taking binary sequences of these two constants in higher and higher Cartesian products of the model with itself produces infinitely many constants.
In particular, when $T$ eliminates imaginaries, its syntactic category has finite coproducts. In particular, $\mathbf{2}$ exists, and all definable subsets have definable classifying maps.
The following characterization is due to Moshe Kamensky, and can be found here.
Let (T) be a first-order theory which interprets two constants. Then $T$ eliminates imaginaries if and only if for all $X$ and $Y$ in $\mathbf{Def}(T)$, the presheaf $Z \mapsto \operatorname{Hom}_{\mathbf{Def}(T)}(Z \times X, Y)$ is ind-representable.
Suppose first that the presheaves $Y^X \overset{\operatorname{df}}{=}Z \mapsto \operatorname{Hom}_{\mathbf{Def}(T)}(Z \times X,Y)$ are ind-representable. Let $E \overset{s}{\underset{t}{\rightrightarrows}} X$ be a definable equivalence relation on $X$. Let $\phi_E : X \times X \to 2$ classify $E \overset{(s,t)}{\hookrightarrow} X \times X$. Let $\mathbf{y} : \mathbf{Def}(T) \hookrightarrow \widehat{\mathbf{Def}(T)}$ be the Yoneda embedding. Then $\mathbf{y} \phi_E$ has a product-exponential transpose $\overline{\mathbf{y} \phi_E} : \mathbf{y}X \to 2^X$. The Yoneda lemma says that $\operatorname{Hom}_{\widehat{\mathbf{Def}(T)}}(\mathbf{y}X, 2^X) \simeq 2^X(X)$, and by assumption $2^X(X) \simeq \underset{\longrightarrow}{\lim}\operatorname{Hom}(X,J(i))$ for some $J : I \to \mathbf{Def}(T)$ a filtered diagram of definable sets. Since this is filtered, the colimit as a set can be computed as
where the equivalence relation $\sim$ identifies elements that are eventually sent to the same element by transition maps in the diagram. Hence, we can identify $\overline{\mathbf{y}\phi_E}$ with its equivalence class of maps $\{\phi_i : X \to J(i)\}_{i \in I}$. Now, $\mathbf{y}E$ is the kernel pair of $\overline{\mathbf{y} \phi_E}$ in $\widehat{\mathbf{Def}(T)}$, so $E = \bigvee_{i \in I} \ker(\phi_i)$. By compactness, we can replace $I$ with a finite subset $I_0 \subseteq I$. Since $I$ is filtered, there exists a weak coproduct $j$ to $I_0$. Since the $\{\phi_i\}_{i \in I}$ turn $X$ into a cone to $J$, for each $i \in I_0$ there is a transition map $h_{ij} : J(i) \to J(j)$, so $\ker(\phi_j) = \ker(h_{ij} \circ \phi_i) \supseteq \ker(\phi_i)$. Replacing each $\ker(\phi_i)$ with $\ker(\phi_j)$, we see $E = \ker(\phi_j)$, and $\phi_j$ is definable.
On the other hand, suppose we have elimination of imaginaries. Fix $X$ and $Y$. For each $Z$ and $f : Z \times X \to Y$, we may form the equivalence relation $E_f$ on $Z$ which says $z_1 \sim_{E_f} z_2$ if and only if they define the same fiber of $\Gamma(f)$. The projection map induces a map
which is also a map $f \to f/E_f$ of objects over $Y$.
Now, the category of elements $\operatorname{Pt}(1, Y^X)$ of the presheaf $\operatorname{Hom}(- \times X, Y)$ has objects maps $f : Z \times X \to Y$ for $Z$ ranging over $\mathbf{Def}(T)$, and morphisms from $f_1 : Z_1 \times X \to Y$ to $f_2 : Z_2 \times X \to Y$ the maps $g \times X : Z_1 \times X \to Z_2 \times X$ for $g : Z_1 \to Z_2$ such that $f_1 = f_2 \circ g$. For each object $f \in \operatorname{Pt}(1, Y^X)$, $-/ E_f \times X$ as above coequalizes all pairs of maps in $\operatorname{Pt}(1, Y^X)$ into $f$. Furthermore, the natural projection $p$ from $\operatorname{Pt}(1, Y^X) \to \mathbf{Def}(T)$ by sending $f : Z \times X \to Y \mapsto Z$ creates colimits, in particular finite coproducts, which by assumption exist in $\mathbf{Def}(T)$. So $\operatorname{Pt}(1, Y^X)$ is filtered, and since for any presheaf $P$, we have that
so $Y^X$ is ind-representable.
In the above proof, complementation is required to characterize compactness as “every covering of a definable set by an infinite family of definable sets is finitely supported.” The same proof works if we replace inner homs with power objects.
Let $M \models T$ be a model. If a formula $\varphi(x_1, x_2)$ is a $0$-definable equivalence relation $E$ (i.e. an internal congruence in $\mathbf{Def}(T)$) on a definable set $X$, any automorphism $\sigma$ of $M$ must preserve and reflect $E$-equivalence, so that $\sigma$ extends along $X \twoheadrightarrow X/E$ to a permutation on $X/E$. So automorphisms of models “already see” imaginaries. This can be stated rigorously as: the expansion of a monster model $\mathbb{M}$ of $T$ to a monster model $\mathbb{M}^{\operatorname{eq}}$ of $T^{\operatorname{eq}}$ will have the same automorphism group as $\mathbb{M}$.
To make things concrete, let’s consider the structure $(\mathbb{Z}, +)$. Since there is an automorphism which switches signs, the only constant is $0$. So the theory of this structure cannot eliminate imaginaries, since such theories must interpret infinitely many distinct constants.
However, we can tell that this structure has many imaginaries by its rigidity: the above automorphism is the only nontrivial one. This is because any such automorphism of $(\mathbb{Z},+)$ must also act on the internal congruences $E_n$ which say
that is, must extend along the abelian group epimorphisms $\mathbb{Z} \twoheadrightarrow \mathbb{Z}/n\mathbb{Z}$ for each $n$, so that $k \mapsto -k$ is the only one that works.
(It’s essentially for this reason that the automorphism group of the forgetful functor $\mathbf{Grp} \to \mathbf{Set}$ is also $\mathbb{Z}/2 \mathbb{Z}$.)