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A Galois connection between posets satisfies $F G F = F$ and $G F G = G$

Statement

Let $A, B$ be posets and let $F : A \to B$ , $G : B \to A$ form a Galois connection. Then

$F G F = F, G F G = G .$

For preorders, the same argument gives pointwise equivalences $F G F ≅ F$ and $G F G ≅ G$ in the associated thin categories, but equality need not follow without antisymmetry.

Facts & Assumptions

Given: Posets $A, B$ and a Galois connection $F ⊣ G$ .

[L1]

A Galois connection satisfies $a \leq G F (a)$ and $F G (b) \leq b$ , and both maps are monotone (Galois connection between preorders).

[F1]

Antisymmetry says that $x \leq y$ and $y \leq x$ imply $x = y$ (Partial order and partially ordered set).

Proof

technique · direct

1.1

Applying $F$ to $a \leq G F (a)$ gives $F (a) \leq F G F (a)$ , while the counit inequality at $F (a)$ gives $F G F (a) \leq F (a)$ .

1.2

Applying $G$ to $F G (b) \leq b$ gives $G F G (b) \leq G (b)$ , while the unit inequality at $G (b)$ gives $G (b) \leq G F G (b)$ .

2.1

Antisymmetry applied to steps 1.1 and 1.2 gives $F G F (a) = F (a)$ and $G F G (b) = G (b)$ for every $a, b$ , hence the two equalities of maps.

step 1.1step 1.2F1

3.1

Without antisymmetry, steps 1.1 and 1.2 still give morphisms in both directions between the corresponding objects of each thin category, which are inverse because parallel morphisms are unique.

step 1.1step 1.2∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 5 results over 4 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Emily Riehl, Category Theory in Context, 2nd ed., Corollary 4.2.10 (standard reference, not scraped)

A Galois connection between posets satisfies FGF=FFGF=F and GFG=GGFG=G