TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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If $f : X \times Z \to Y$ is continuous then its transpose $F : Z \to C (X, Y)$ , $F (z) (x) = f (x, z)$ , is continuous for the compact-open topology, with no hypothesis on $X$ beyond being metric

Statement

Let $(X, d)$ be a metric space carrying its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), 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), and let

$f : X \times Z \to Y$

be continuous, the product carrying the product topology (The product set $\prod_{i \in I} X_{i}$ 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). For $z \in Z$ define $F (z) : X \to Y$ by $F (z) (x) : = f (x, z)$ . Then:

$F (z) \in C (X, Y)$ for every $z \in Z$ ;
the transpose $F : Z \to C (X, Y)$ is continuous when $C (X, Y)$ carries the compact-open topology (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ ).

No local compactness and no separation hypothesis is used, on any of the three spaces; $X$ is metric only because the compact-open topology is defined here over the compact subsets of a metric space. No choice principle is used.

Facts & Assumptions

Given: A metric space $(X, d)$ with its metric topology, topological spaces $Z$ and $Y$ , a continuous $f : X \times Z \to Y$ , and $F (z) (x) = f (x, z)$ .

[L1]

A map $h$ into a product is continuous exactly when each of its components is, the components being the composites with the projections; the projections are continuous (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, claims 1 and 2, The product set $\prod_{i \in I} X_{i}$ 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).

[L2]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, claim 1).

[L3]

A constant map into a topological space is continuous, the preimage of an open set being the whole domain or the empty set, both open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b), Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L4]

The identity map of a topological space is continuous, being its own preimage assignment (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b)).

[L5]

Continuity may be checked on a subbasis: $h$ is continuous exactly when $h^{- 1} [S]$ is open for every member $S$ of a subbasis of the target (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (d), Basis and subbasis for a topology, and the topology generated by a family of sets, A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis).

[L6]

The sets $S (K, V) = {g \in C (X, Y) : g [K] \subseteq V}$ , for compact $K \subseteq X$ and open $V \subseteq Y$ , are a subbasis for the compact-open topology on $C (X, Y)$ (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ , Open cover, subcover, compact metric space, and compact subset of a metric space, A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[L7]

Tube lemma: for compact $K \subseteq X$ , a point $z_{0} \in Z$ and an open $N \subseteq X \times Z$ with $K \times {z_{0}} \subseteq N$ there is an open $W \subseteq Z$ with $z_{0} \in W$ and $K \times W \subseteq N$ (Tube lemma: if $K$ is a compact subset of a metric space $X$ , $Z$ is a topological space and $N$ is open in $X \times Z$ with $K \times {z_{0}} \subseteq N$ , then $K \times W \subseteq N$ for some open $W ∋ z_{0}$ ).

[L8]

A subset of a topological space is open exactly when it is a neighbourhood of each of its points, that is when each of its points lies in an open set inside it (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, consequence 4).

[L9]

$f$ is continuous, so $f^{- 1} [V]$ is open in $X \times Z$ for every open $V \subseteq Y$ (Continuity of a map of topological spaces at a point and globally, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b)).

Proof

technique · direct

1.1L1L3L4

Fix $z \in Z$ and let $j_{z} : X \to X \times Z$ be the map $j_{z} (x) : = (x, z)$ ; its components are the identity of $X$ and the constant map at $z$ , both continuous, so $j_{z}$ is continuous.

2.1step 1.1L2

$F (z) = f \circ j_{z}$ , since $(f \circ j_{z}) (x) = f (x, z) = F (z) (x)$ for every $x \in X$ ; hence $F (z)$ is continuous and $F (z) \in C (X, Y)$ , which is claim 1.

3.1step 2.1L5L6suffices: preimages of subbasic sets are open

For claim 2 it suffices, by [L5] and [L6], to show that $F^{- 1} [S (K, V)]$ is open in $Z$ for every compact $K \subseteq X$ and every open $V \subseteq Y$ .

3.2step 2.1L6

Unwinding the definitions, $F^{- 1} [S (K, V)] = {z \in Z : F (z) [K] \subseteq V} = {z \in Z : f (x, z) \in V for every x \in K}$ .

4.1step 3.2L9

Let $z_{0} \in F^{- 1} [S (K, V)]$ and put $N : = f^{- 1} [V]$ , an open subset of $X \times Z$ ; by step 3.2 every $x \in K$ satisfies $f (x, z_{0}) \in V$ , that is $K \times {z_{0}} \subseteq N$ .

5.1step 4.1L7choose

The tube lemma applied to $K$ , $z_{0}$ and $N$ gives an open $W \subseteq Z$ with $z_{0} \in W$ and $K \times W \subseteq N$ .

6.1step 3.2step 4.1step 5.1

Every $z \in W$ then satisfies $f (x, z) \in V$ for every $x \in K$ , that is $z \in F^{- 1} [S (K, V)]$ ; so $W$ is an open set with $z_{0} \in W \subseteq F^{- 1} [S (K, V)]$ .

7.1step 3.1step 6.1L8∎

As $z_{0}$ was an arbitrary point of $F^{- 1} [S (K, V)]$ , that set is open in $Z$ ; by step 3.1 this proves claim 2.

Remarks

The tube lemma is the entire content. The condition defining $F^{- 1} [S (K, V)]$ is "the whole slice $K \times {z}$ lands in $V$ ", and openness of that condition in $z$ is exactly the statement that a neighbourhood of a slice contains a tube. Everything else is unwinding.
This half of the exponential law is the cheap half. It needs no hypothesis on $X$ beyond compactness being available for its subsets, and none at all on $Z$ or $Y$ . The converse half — that every continuous $F : Z \to C (X, Y)$ arises from a continuous $f$ — runs through continuity of the evaluation map and is where local compactness of $X$ is spent.
The map $F$ determines $f$ and conversely, as functions. That the assignment $f \mapsto F$ is injective, and that under the local compactness hypothesis it is onto the continuous maps $Z \to C (X, Y)$ , is the exponential law below; this theorem is the statement that the assignment lands in the right place.

If $f : X \times Z \to Y$ is continuous then its transpose $F : Z \to C (X, Y)$ , $F (z) (x) = f (x, z)$ , is continuous for the compact-open topology, with no hypothesis on $X$ beyond being metric

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If f:X×Z→Yf:X×Z→Y is continuous then its transpose F:Z→C(X,Y)F:Z→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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If $f : X \times Z \to Y$ is continuous then its transpose $F : Z \to C (X, Y)$ , $F (z) (x) = f (x, z)$ , is continuous for the compact-open topology, with no hypothesis on $X$ beyond being metric