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Open-set and closed-set functors on $\mathbf{Top}^{\mathrm{op}}$ 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 $\mathbf{Top}$ , Posets and monotone maps form the large locally small category $\mathbf{Poset}$ , 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 $\mathcal O(X)$ be the open subsets of $X$ ordered by inclusion, and let $\mathcal C(X)$ 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 $(gf)^{-1}=f^{-1}g^{-1}$ . Thus $\mathcal O,\mathcal C:\mathbf{Top}^{\mathrm{op}}\to\mathbf{Poset}$ are functors.

step 1.1L1L2

2.2

Complementation $c_X:\mathcal O(X)\to\mathcal C(X)$ is monotone because $U\subseteq V$ implies $X\setminus U\supseteq X\setminus V$ . 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 $X\setminus f^{-1}(U)=f^{-1}(Y\setminus U)$ 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 $\mathcal O\cong\mathcal C$ as functors $\mathbf{Top}^{\mathrm{op}}\to\mathbf{Poset}$ .

step 2.1step 2.2step 2.3L3∎

Depends on

Used by

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Sources

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

Open-set and closed-set functors on Topop\mathbf{Top}^{\mathrm{op}}Topop are naturally isomorphic by complements