Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-05 (claude-sonnet-5)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The exponential law: for a locally compact metric X and any spaces Z and Y, transposition is a bijection between C(X×Z,Y) and C(Z,C(X,Y)) with the compact-open topology

Statement

Let (X,d) be a locally compact metric space (Locally compact metric space: every point has a compact neighbourhood) carrying its metric topology, and let Z and Y be topological spaces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Give C(X,Y) the compact-open topology (The compact-open topology on C(X,Y) for a metric domain X, with subbasis S(K,V)={f:f[K]⊆V}) and X×Z the product topology (The product set ∏i∈IXi of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Define, for f∈C(X×Z,Y),

Φ(f):Z→C(X,Y),Φ(f)(z)(x):=f(x,z).

Then Φ is a well-defined map C(X×Z,Y)→C(Z,C(X,Y)) and it is a bijection (Injection, surjection, bijection); its inverse sends a continuous F:Z→C(X,Y) to the continuous map (x,z)↦F(z)(x).

Exactly what is and is not claimed

This is an assertion about two sets of continuous maps and a bijection between them. No topology is placed on C(X×Z,Y) or on C(Z,C(X,Y)) anywhere in the statement, and it is not claimed that Φ is a homeomorphism. The homeomorphism form of the exponential law is a genuinely stronger statement, and this library does not have what it needs; the last remark below says exactly what is missing. A reader who wants the categorical slogan "YX×Z≅(YX)Z" should read it here as a bijection of underlying sets, natural in the evident way, and no more.

No choice principle is used.

Facts & Assumptions

Given: A locally compact metric space (X,d) with its metric topology, topological spaces Z and Y, the set C(X,Y) with the compact-open topology, the evaluation map e:C(X,Y)×X→Y (The evaluation map e:C(X,Y)×X→Y, e(f,x)=f(x)), and the assignment Φ of the Statement.

[L1]

If f:X×Z→Y is continuous then Φ(f)(z)∈C(X,Y) for every z, and Φ(f):Z→C(X,Y) is continuous for the compact-open topology (If f:X×Z→Y is continuous then its transpose F:Z→C(X,Y), F(z)(x)=f(x,z), is continuous for the compact-open topology, with no hypothesis on X beyond being metric).

[L6]

A map is a bijection exactly when it is injective and surjective (Injection, surjection, bijection).

Proof

technique · direct
1.1

Let f∈C(X×Z,Y); by [L1] each Φ(f)(z) lies in C(X,Y) and Φ(f) is a continuous map Z→C(X,Y), so Φ(f)∈C(Z,C(X,Y)) and Φ is well defined.

L1
1.2

Let F∈C(Z,C(X,Y)) and define Ψ(F):X×Z→Y by Ψ(F)(x,z):=F(z)(x); this is a function, F(z) being an element of C(X,Y) and hence a function X→Y.

L5construct
2.1

Let h:X×Z→C(X,Y)×X be given by h(x,z):=(F(z),x); its two components are (x,z)↦F(z), which is the composite of the projection onto Z with F, and (x,z)↦x, which is the projection onto X; both are continuous, so h is continuous.

step 1.2L3L4
2.2

Φ is injective: if Φ(f)=Φ(f′) then for all x∈X and z∈Z we get f(x,z)=Φ(f)(z)(x)=Φ(f′)(z)(x)=f′(x,z), so f=f′.

step 1.1L5
3.1

Ψ(F)=e∘h, since (e∘h)(x,z)=e(F(z),x)=F(z)(x)=Ψ(F)(x,z) for every (x,z); hence Ψ(F) is continuous, that is Ψ(F)∈C(X×Z,Y).

step 1.2step 2.1L2L4L5
4.1

Φ is surjective: given F∈C(Z,C(X,Y)), step 3.1 puts Ψ(F) in C(X×Z,Y), and for all z∈Z and x∈X we have Φ(Ψ(F))(z)(x)=Ψ(F)(x,z)=F(z)(x), so Φ(Ψ(F))(z)=F(z) for every z and hence Φ(Ψ(F))=F.

step 1.1step 3.1L5
5.1

By steps 2.2 and 4.1 the map Φ is a bijection from C(X×Z,Y) onto C(Z,C(X,Y)), and step 4.1 identifies its inverse as Ψ, that is F↦((x,z)↦F(z)(x)).

step 2.2step 4.1L6∎

Remarks

Depends on

Used by

Dependency tree · two levels

61 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