x∈A − → R (x) = R∗(A) − → R (x) = u({x}) u(∅) = ∅ ∀B ⊂ Y, v(B) = {x ∈ X|− →R (x) ⊂ B} = − →R ∗(P(B)) ←− R (y) = {Xv({Y {y}) v(Y ) = X Indeed, ∀A ⊂ X,∀B ⊂ F,A ⊂ v(B)⇔ (∀x ∈ A,− →R (x) ⊂ B)⇔ u(A) ⊂ B. The case R = Gr f where f ∈ Y X gives (f∗, f∗) ∈ Gal(P(X),P(Y )). The case R = Gr g where g ∈ X gives (g∗, {X ◦ g∗ ◦ {Y ) ∈ Gal(P(X),P(Y )). Proposition. Let E, F , G three ordered sets, (u, v) ∈ Gal(E,F ), and (u′, v′) ∈ Gal(F,G) (resp. Gal(F,G)). Then (u′ ◦ u, v ◦ v′) ∈ Gal(E,G) (resp. Gal(E,G)) The monotone case is written ∀x ∈ E,∀y ∈ G, x ≤ v(v′(y))⇔ u(x) ≤ v′(y)⇔ u′(u(x)) ≤ y. If E = P(X), F = P(Y ), G = P(Z), if (u, v) is associated to R ⊂ X×Y and (u′, v′) to R′ ⊂ Y ×Z, then (u′ ◦ u, v ◦ v′) is associated to R′′ ⊂ X × Z defined by: — Antitone case: xR′′z ⇔ − →R (x) ⊂ ←− R′(z), i.e. −→ R′′(x) = ⋂ y∈~ R(x) −→ R′(y). — monotone case: xR′′z ⇔ − →R (x) ∩ ←− R′(z) 6= ∅, i.e. −→ R′′(x) = ⋃ y∈~ R(x) −→ R′(y). In both cases, if R = Gr f where f ∈ Y X then −→ R′′ = −→ R′ ◦ f . In the monotone case, if R′ = Gr g where g ∈ Z then −→ R′′ = g∗ ◦ − → R .
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