As far as I am aware, no-one in model theory has attempted to create a general notion of a “quotient structure”, as seen in algebra. This strikes me as odd. Perhaps it is too algebraic a concept for logicians to dirty their hands with. Anyway, here is a pretty simple definition.
Fix a language \mathcal{L}, and let \mathcal{S} be an \mathcal{L}-structure with support X. Let \sim be an equivalence relation on X. We say that \sim is a congruence relation on \mathcal{S} if it is compatible with the additional logical structure given to X by \mathcal{S}. That is, if x_1 \sim y_1, x_2 \sim y_2, \ldots, x_n \sim y_n are elements of X, then:
- For every n-ary function symbol f \in \mathcal{L} , we have \displaystyle f^{\mathcal{S}} ( x_1, \ldots, x_n ) \sim f^{\mathcal{S}} ( y_1, \ldots, y_n )
- For every n-ary relation symbol R \in \mathcal{L} , we have \displaystyle R^{\mathcal{S}} ( x_1, \ldots, x_n ) \iff R^{\mathcal{S}} ( y_1, \ldots, y_n )
If \sim is such a congruence relation on \mathcal{S}, we define the quotient \mathcal{L}-structure \mathcal{S} / {\sim} as follows:
- The support of \mathcal{S} / {\sim} is the set X/ {\sim} = \big\{ [x]_\sim : x \in X \big\};
- For every n-ary function symbol f \in \mathcal{L} , let \displaystyle f^{ \mathcal{S} / {\sim} } \big( [x_1]_\sim , \ldots, [x_n]_\sim \big) = \big[ f^\mathcal{S}(x_1,\ldots,x_n) \big]_\sim
- For every n-ary relation symbol R \in \mathcal{L} , \displaystyle R^{ \mathcal{S} / {\sim} } \big( [x_1]_\sim , \ldots, [x_n]_\sim \big) \iff R^\mathcal{S}(x_1, \ldots, x_n)
- For every constant symbol c \in \mathcal{L} , let c^{ \mathcal{S} / {\sim} } = \left[ c^\mathcal{S} \right]_\sim
The above interpretations are well-defined, and hence, \mathcal{S} / {\sim} is indeed an \mathcal{L}-structure.
Recall that a (strong) homomorphism of \mathcal{L}-structures \mathcal{M} and \mathcal{N} is a map \pi: \mathcal{M} \to \mathcal{N} such that for every x_1, \ldots, x_n \in \mathcal{M} :
- For every n-ary function symbol f \in \mathcal{L} , we have \displaystyle \pi \big(f^\mathcal{M}( x_1, \ldots, x_n ) \big) = f^\mathcal{N} \big( \pi(x_1), \ldots, \pi(x_n) \big)
- For every n-ary relation symbol R \in \mathcal{L} , we have \displaystyle R^\mathcal{M}( x_1, \ldots, x_n ) \iff R^\mathcal{N} \big( \pi(x_1), \ldots, \pi(x_n) \big)
- For every constant symbol c \in \mathcal{L} , we have \pi( c^\mathcal{M} ) = c^\mathcal{N}.
Given such a map, we define a congruence relation \sim_\pi on \mathcal{M} by \displaystyle x \sim_\pi y \iff \pi(x) = \pi(y)
Note also that \pi(\mathcal{M}) contains the interpretations of all constant symbols in \mathcal{L}, and is closed under all the function symbols in \mathcal{L}. Therefore, \pi(\mathcal{M}) is an induced substructure of \mathcal{N}. The following holds:
First isomorphism theorem: for any (strong) homomorphism \pi: \mathcal{M} \to \mathcal{N}, the quotient map \tilde{\pi}: \mathcal{M} / {\sim_\pi} \to \pi(\mathcal{M}) defined by \displaystyle \tilde{\pi} \big( [x]_\pi \big) = \pi(x) is an isomorphism of \mathcal{L}-structures.
Here’s an application of the above. Fix a language \mathcal{L}, and let \mathcal{S} be an \mathcal{L}-structure with support X. Let \beta X be the set of all (principal and free) ultrafilters on X. We can create a \mathcal{L}-structure \beta \mathcal{S} on \beta X as follows:
- For every n-ary function symbol f \in \mathcal{L} , let \displaystyle f^{\beta\mathcal{S}} \left( \mathcal{U}^{(1)}, \ldots, \mathcal{U}^{(n)} \right) = \left\{ A \subseteq X :\ \mathcal{U}^{(1)} x_1\, \cdots\ \mathcal{U}^{(n)} x_n\ f^\mathcal{S}(x_1, \ldots, x_n) \in A \right\}
- For every n-ary relation symbol R \in \mathcal{L} , \displaystyle R^{\beta\mathcal{S}} \left( \mathcal{U}^{(1)}, \ldots, \mathcal{U}^{(n)} \right) \iff \mathcal{U}^{(1)} x_1\, \cdots\ \mathcal{U}^{(n)} x_n\ R^\mathcal{S}(x_1, \ldots, x_n)
- For every constant symbol c \in \mathcal{L} , let c^{\beta\mathcal{S}} = \mathcal{U}_{c^\mathcal{S}} , i.e. the principal ultrafilter on c^\mathcal{S} .
Here, \mathcal{U}x denotes the ultrafilter quantifier: for any \mathcal{L}-formula \varphi(x) with one free variable, \mathcal{U}x\ \varphi(x) \iff \{ x : \varphi(x) \} \in \mathcal{U} . One can verify that the definitions above are well-posed, and therefore that \beta \mathcal{S} forms an \mathcal{L}-structure.
We can view \mathcal{S} as a substructure of \beta \mathcal{S}, via the embedding \pi(x) = \mathcal{U}_x mapping every x \in X to the principal ultrafilter on x. Furthermore, using the inductive definition of \mathcal{L}-formulae, we can prove that for any \mathcal{L}-formula \varphi and s_1, \ldots, s_n \in \mathcal{S} , we have \displaystyle \mathcal{S} \vDash \varphi(s_1, \ldots, s_n) \iff \beta\mathcal{S} \vDash \varphi \big( \mathcal{U}_{s_1}, \ldots, \mathcal{U}_{s_n} \big)
i.e. that \pi is an elementary embedding, and that \mathcal{S} \equiv \beta \mathcal{S}.
Consider now the nonstandard extension {}^* \mathcal{S} of \mathcal{S}, as constructed by an ultrapower. By Łoś’ theorem, we have \mathcal{S} \equiv {}^* \mathcal{S}.
If X is finite, then we have \mathcal{S} \cong \beta \mathcal{S} \cong {}^* \mathcal{S}. However, for infinite X, none of the structures \mathcal{S}, \beta \mathcal{S}, or {}^* \mathcal{S} are isomorphic.
Nonetheless, we can also construct \beta \mathcal{S} as a suitable quotient of {}^* \mathcal{S}. Consider the mapping \pi : {}^* \mathcal{S} \to \beta \mathcal{S} defined by \displaystyle \pi: \alpha \mapsto \mathcal{U}_\alpha = \{ A \subseteq X: \alpha \in {}^* A \}
\pi is a strong homomorphism from {}^* \mathcal{S} to \beta \mathcal{S}, hence by the first isomorphism theorem, the corresponding quotient {}^* \mathcal{S} / {\sim_\pi} is isomorphic to \pi ({}^* \mathcal{S}) = \beta \mathcal{S} since \pi is surjective.
NB1: Okay, to prove \pi is surjective, we need to assume our model of {}^* \mathcal{S} has the \left( 2^{\vert X \vert} \right)^+-enlarging property. This is a pretty weak assumption – see here for a definition (Defn 2.74) and proof of surjectivity (Prop 3.6).
NB2: Valentino Vito asked a great question: can \mathcal{S} be realised as a quotient of \beta \mathcal{S} or {}^* \mathcal{S} ? In general, the answer is no. Take any infinite set X, and let \mathcal{L} be the language which has a unary relation symbol R_x for every x \in X. Let \mathcal{S} be the \mathcal{L}-structure on X where R_x(y) \iff x = y.
Then, we cannot have a (strong) homomorphism \pi: \beta \mathcal{S} \to \mathcal{S}. Any free ultrafilter \mathcal{U} \in \beta \mathcal{S} does not satisfy any of the relations R_x, but \pi(\mathcal{U}) has to satisfy R_{\pi(\mathcal{U})}, hence \pi is not a h.m.. As any congruence \sim on \beta \mathcal{S} such that there exists an isomorphism \psi: \beta\mathcal{S} / {\sim} \to \mathcal{S} naturally induces a homomorphism \pi_\sim: \beta \mathcal{S} \to \mathcal{S} by \pi_\sim (\mathcal{U}) = \psi \big( [ \mathcal{U} ]_\sim \big) , it is clear that no such congruence relation could exist. The same goes for {}^* \mathcal{S} – there is nothing we can map infinite hyperintegers to.
However, we note that in the case where \mathcal{L} = \varnothing, any function \pi: \beta \mathcal{S} \to \mathcal{S} or {}^* \mathcal{S} \to \mathcal{S} is a strong homomorphism. Hence, \mathcal{S} is trivially a quotient of \beta \mathcal{S} or {}^* \mathcal{S}. It would be interesting to know which languages \mathcal{L}, and \mathcal{L}-structures \mathcal{S}, have the property that \mathcal{S} be realised as a quotient of \beta \mathcal{S} or {}^* \mathcal{S}.