Galois correspondence induced by a binary relation

Definition
Let $$A$$ and $$B$$ be sets. Suppose $$R \subset A \times B$$ i.e. $$R$$ is a binary relation between $$A$$ and $$B$$. $$R$$ induces a Galois correspondence between subsets of $$A$$ and subsets of $$B$$, as follows. We get two maps, $$F:2^A \to 2^B$$ and $$G:2^B \to 2^A$$:


 * $$F(S) = \{ b \in B | a R b \ \forall \ a \in S \}$$
 * $$G(T) = \{ a \in A | a R b \ \forall \ b \in T \}$$

Key facts

 * $$F$$ and $$G$$ are both reverse-monotone. In category-theoretic jargon, $$F$$ and $$G$$ are contravariant functors between the categories of subsets of $$A$$ and $$B$$, with morphisms being inclusions.

In symbols:

$$S \subset S' \implies F(S') \subset F(S)$$

and:

$$T \subset T' \implies G(T') \subset G(T)$$


 * $$F \circ G$$ and $$G \circ F$$ are both monotone, and ascendant. In other words:

$$S \subset G(F(S))$$

and:

$$S \subset S' \implies G(F(S)) \subset G(F(S'))$$

Similarly for $$F \circ G$$


 * $$G \circ F \circ G = G$$ and $$F \circ G \circ F = F$$. In other words, going back and forth thrice has the same effect as going once. This follows easily from the last two facts.


 * The operator $$F \circ G$$ defines a closure operator on $$B$$, and the operator $$G \circ F$$ defines a closure operator on $$A$$. By closure operator is meant a monotone ascendant operator. In particular, the subsets which are fixed under this operator, which are called the closed sets, include the whole set, and are closed under taking arbitrary intersections


 * In general, the closed sets on either side under a Galois correspondence, need not give a topology. The problem is that the empty set may not be closed, and a finite union of closed sets need not be closed. There are special circumstances under which we do get a topology, for instance the spectrum of a commutative unital ring or the max-spectrum of a commutative unital ring.

Related notions

 * Effect on Galois correspondence of restricting binary relation to a subset