Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 XX and any spaces ZZ and YY, transposition is a bijection between C(X×Z,Y)C(X \times Z, Y) and C(Z,C(X,Y))C(Z, C(X,Y)) with the compact-open topology

Statement

Let (X,d)(X,d) be a locally compact metric space (Locally compact metric space: every point has a compact neighbourhood) carrying its metric topology, and let ZZ and YY 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)C(X,Y) the compact-open topology (The compact-open topology on C(X,Y)C(X,Y) for a metric domain XX, with subbasis S(K,V)={f:f[K]V}S(K,V) = \{f : f[K] \subseteq V\}) and X×ZX \times Z the product topology (The product set iIXi\prod_{i \in I} X_i 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 fC(X×Z,Y)f \in C(X \times Z, Y),

Φ(f):ZC(X,Y),Φ(f)(z)(x):=f(x,z).\Phi(f) : Z \to C(X,Y), \qquad \Phi(f)(z)(x) := f(x,z) .

Then Φ\Phi is a well-defined map C(X×Z,Y)C(Z,C(X,Y))C(X \times Z, Y) \to C\big(Z, C(X,Y)\big) and it is a bijection (Injection, surjection, bijection); its inverse sends a continuous F:ZC(X,Y)F : Z \to C(X,Y) to the continuous map (x,z)F(z)(x)(x,z) \mapsto 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)C(X \times Z, Y) or on C(Z,C(X,Y))C(Z, C(X,Y)) anywhere in the statement, and it is not claimed that Φ\Phi 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)ZY^{X \times Z} \cong (Y^{X})^{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)(X,d) with its metric topology, topological spaces ZZ and YY, the set C(X,Y)C(X,Y) with the compact-open topology, the evaluation map e:C(X,Y)×XYe : C(X,Y) \times X \to Y (The evaluation map e:C(X,Y)×XYe : C(X,Y) \times X \to Y, e(f,x)=f(x)e(f,x) = f(x)), and the assignment Φ\Phi of the Statement.

[L1]

If f:X×ZYf : X \times Z \to Y is continuous then Φ(f)(z)C(X,Y)\Phi(f)(z) \in C(X,Y) for every zz, and Φ(f):ZC(X,Y)\Phi(f) : Z \to C(X,Y) is continuous for the compact-open topology (If f:X×ZYf : X \times Z \to Y is continuous then its transpose F:ZC(X,Y)F : Z \to C(X,Y), F(z)(x)=f(x,z)F(z)(x) = f(x,z), is continuous for the compact-open topology, with no hypothesis on XX 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 fC(X×Z,Y)f \in C(X \times Z, Y); by [L1] each Φ(f)(z)\Phi(f)(z) lies in C(X,Y)C(X,Y) and Φ(f)\Phi(f) is a continuous map ZC(X,Y)Z \to C(X,Y), so Φ(f)C(Z,C(X,Y))\Phi(f) \in C(Z, C(X,Y)) and Φ\Phi is well defined.

L1
1.2

Let FC(Z,C(X,Y))F \in C(Z, C(X,Y)) and define Ψ(F):X×ZY\Psi(F) : X \times Z \to Y by Ψ(F)(x,z):=F(z)(x)\Psi(F)(x,z) := F(z)(x); this is a function, F(z)F(z) being an element of C(X,Y)C(X,Y) and hence a function XYX \to Y.

L5construct
2.1

Let h:X×ZC(X,Y)×Xh : X \times Z \to C(X,Y) \times X be given by h(x,z):=(F(z),x)h(x,z) := (F(z), x); its two components are (x,z)F(z)(x,z) \mapsto F(z), which is the composite of the projection onto ZZ with FF, and (x,z)x(x,z) \mapsto x, which is the projection onto XX; both are continuous, so hh is continuous.

step 1.2L3L4
2.2

Φ\Phi is injective: if Φ(f)=Φ(f)\Phi(f) = \Phi(f') then for all xXx \in X and zZz \in Z we get f(x,z)=Φ(f)(z)(x)=Φ(f)(z)(x)=f(x,z)f(x,z) = \Phi(f)(z)(x) = \Phi(f')(z)(x) = f'(x,z), so f=ff = f'.

step 1.1L5
3.1

Ψ(F)=eh\Psi(F) = e \circ h, since (eh)(x,z)=e(F(z),x)=F(z)(x)=Ψ(F)(x,z)(e \circ h)(x,z) = e(F(z), x) = F(z)(x) = \Psi(F)(x,z) for every (x,z)(x,z); hence Ψ(F)\Psi(F) is continuous, that is Ψ(F)C(X×Z,Y)\Psi(F) \in C(X \times Z, Y).

step 1.2step 2.1L2L4L5
4.1

Φ\Phi is surjective: given FC(Z,C(X,Y))F \in C(Z,C(X,Y)), step 3.1 puts Ψ(F)\Psi(F) in C(X×Z,Y)C(X \times Z, Y), and for all zZz \in Z and xXx \in X we have Φ(Ψ(F))(z)(x)=Ψ(F)(x,z)=F(z)(x)\Phi(\Psi(F))(z)(x) = \Psi(F)(x,z) = F(z)(x), so Φ(Ψ(F))(z)=F(z)\Phi(\Psi(F))(z) = F(z) for every zz and hence Φ(Ψ(F))=F\Phi(\Psi(F)) = F.

step 1.1step 3.1L5
5.1

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

step 2.2step 4.1L6

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 129 results over 23 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