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Open-set and closed-set functors on ${T o p}^{o p}$ are naturally isomorphic by complements

Example

Inverse image makes open and closed subsets contravariant in a space. Ordering closed subsets by reverse inclusion makes complementation a natural isomorphism between the resulting poset-valued functors.

Facts & Assumptions

Given: Topological spaces and continuous maps.

[L1]

Continuous inverse images preserve open and closed sets (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Continuity of a map of topological spaces at a point and globally, The image $R [A]$ and the preimage $R^{- 1} [B]$ of a set under a relation).

[L2]

Topological spaces and posets form categories, and contravariant functors are functors on the opposite category (Topological spaces and continuous maps form the large locally small category $T o p$ , Posets and monotone maps form the large locally small category $P o s e t$ , Covariant functor, identity functor, composite functor, and contravariant functor).

[L3]

A natural isomorphism is a natural transformation with a two-sided inverse natural transformation (Natural isomorphism), and this holds exactly when every component is an isomorphism (A natural transformation is a natural isomorphism exactly when every component is an isomorphism).

Verification

technique · direct

1.1

Let $O (X)$ be the open subsets of $X$ ordered by inclusion, and let $C (X)$ be the closed subsets ordered by reverse inclusion. For $f : X \to Y$ , assign to either kind of subset its inverse image under $f$ .

2.1

Inverse image is monotone for inclusion and for reverse inclusion, preserves identity functions, and satisfies $(g f)^{- 1} = f^{- 1} g^{- 1}$ . Thus $O, C : {T o p}^{o p} \to P o s e t$ are functors.

step 1.1L1L2

2.2

Complementation $c_{X} : O (X) \to C (X)$ is monotone because $U \subseteq V$ implies $X ∖ U \supseteq X ∖ V$ . It is its own order-isomorphism inverse.

step 1.1

2.3

For every continuous $f : X \to Y$ and open $U \subseteq Y$ , the identity $X ∖ f^{- 1} (U) = f^{- 1} (Y ∖ U)$ says exactly that the complement square commutes.

step 1.1L1

3.1

The componentwise order isomorphisms of step 2.2 are natural by step 2.3. Hence complementation gives $O ≅ C$ as functors ${T o p}^{o p} \to P o s e t$ .

step 2.1step 2.2step 2.3L3∎

Depends on

Used by

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Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Emily Riehl, Category Theory in Context, Example 1.4.3 (standard reference, not scraped)

Open-set and closed-set functors on TopopTopop are naturally isomorphic by complements