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Kification, compact tests, and finite constructions
Statement
Kification preserves exactly the continuous maps from compact Hausdorff spaces, is idempotent and functorial, and satisfies: for CG , a function is continuous if and only if it is continuous into . Finite k-products are categorical products of CG spaces. Ordinary quotients, finite disjoint unions and closed subspaces of CG spaces are CG. Products of closed inclusions are closed inclusions in this category, and finite clopen decompositions commute with kification. If is CG, its ordinary product is CG. Kification leaves cubical maps and their relative homotopies unchanged.
Facts & Assumptions
K-closed sets are tested by all compact Hausdorff maps. Compactly generated conventions for based homotopy
The continuity of a map into an ordinary product is equivalent to coordinate continuity. A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice
A fibre-constant continuous map factors continuously through a quotient. For a quotient map , a map out of is continuous iff its composite with is; a continuous map on constant on the fibres of factors uniquely through ; and a composite of quotient maps is a quotient map
Closed subsets of compact spaces are compact. A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact
Finite products of compact spaces are compact. A product of finitely many compact spaces is compact in the product topology
Closed bounded Euclidean subsets, in particular intervals and cubes, are compact. A subset of with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology
Proof
Given: The spaces, maps, and hypotheses in the statement above.
Inverse images commute with arbitrary intersections and finite unions, and send to . Thus the k-closed sets are closed sets of a topology containing all original closed sets. Each original test is continuous into by that definition; the reverse follows by composing with the continuous identity . Since the tests are identical, .
For a finite disjoint union, a k-closed subset restricts to a closed subset of each CG summand by testing the inclusion composed with every test; it is therefore closed. For closed in CG and k-closed in , a test restricts on the compact Hausdorff closed set . Thus is closed there, hence in . Consequently is closed in , proving that the subspace is CG.
If is CG and is continuous, then for every k-closed and test , is closed. Hence is k-closed in , thus closed. This proves continuity into ; composition with proves the converse. Applied to , this also proves functoriality. A compact Hausdorff is CG since its identity is a test.
Let be k-closed in the ordinary and . The vertical test shows closed. Choose a closed interval neighbourhood of in disjoint from , using relative intervals at 0 and 1. Set . For a test , the inverse image of in the compact Hausdorff is closed, hence compact. Its projection is compact and closed in and equals . Thus is k-open and hence open. The rectangle misses , so is ordinary closed. Therefore is CG.
A family of continuous coordinates from CG induces a continuous map into the ordinary product, which lifts to its kification by step 2.1. Conversely projections from the k-product are continuous. Coordinate uniqueness proves the product property, and inverse coordinate rearrangements prove finite associativity and symmetry. For an ordinary quotient with CG, step 2.1 makes continuous. Each k-closed therefore has closed , so quotient finality makes closed in .
The inverse image of a closed factor under a projection from a k-product is closed, hence CG by step 1.2. Its ordinary subspace topology identifies with the corresponding k-product: the continuous coordinate map in one direction comes from step 3.1, while its inverse is continuous into the subspace because its composite into the ambient product is continuous. Iterating handles products of closed inclusions. A compact test into a finite clopen decomposition splits into compact Hausdorff clopen domains. Testing each piece proves that kification commutes with that decomposition.
Cubes and their cylinders are compact Hausdorff; the zero-fold cube is a point. Step 1.1 therefore preserves all maps and homotopies from these domains. The underlying functions do not change, so all specified boundary equalities are preserved as well. Empty spaces and empty coproducts satisfy the same closed-set tests vacuously.
Depends on
- Compactly generated conventions for based homotopy
- A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice
- For a quotient map $q : X \to Y$, a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$; and a composite of quotient maps is a quotient map
- A product of finitely many compact spaces is compact in the product topology
- A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism
- A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact
- In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones
- A subset of $\mathbb{R}^n$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology
Used by
Dependency tree · two levels
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Sources
- N. P. Strickland, The category of CGWH spaces (standard reference, not scraped)