Let \(\mathcal{C}'(X, Y)\) denote the set \(\mathcal{C}(X, Y)\) in some topology \(\mathcal{T}\). Show that if the evaluation map
\[ e: X \times \mathcal{C}'(X, Y) \to Y \]
is continuous, then \(\mathcal{T}\) contains the compact-open topology. [Hint: The induced map \(E: \mathcal{C}'(X, Y) \to \mathcal{C}(X, Y)\) is continuous.]
Solution
1. Subbasis for the compact-open topology:The compact-open topology \(\mathcal{T}_{\text{co}}\) on \(\mathcal{C}(X, Y)\) is generated by the subbasis of sets of the form:\[S(C, U) = \{f \in \mathcal{C}(X, Y) : f(C) \subseteq U\}\]where \(C \subseteq X\) is compact and \(U \subseteq Y\) is open.To show \(\mathcal{T}_{\text{co}} \subseteq \mathcal{T}\), it suffices to show that each subbasis element \(S(C, U)\) is open in \((\mathcal{C}(X, Y), \mathcal{T})\).
2. Preimage under evaluation:Let \(C \subseteq X\) be compact, \(U \subseteq Y\) open, and let \(f_0 \in S(C, U)\).The set \(e^{-1}(U) = \{(x, f) \in X \times \mathcal{C}'(X, Y) : f(x) \in U\}\) is open in \(X \times \mathcal{C}'(X, Y)\) by continuity of \(e\).Proof: For every \(x \in C\), \(e(x, f_0) = f_0(x) \in f_0(C) \subseteq U\). Thus \(C \times \{f_0\} \subseteq e^{-1}(U)\).
3. Application of the Tube Lemma:There exists an open neighborhood \(W \in \mathcal{T}\) of \(f_0\) in \(\mathcal{C}'(X, Y)\) such that \(C \times W \subseteq e^{-1}(U)\).Proof:
3.1. \(C\) is compact in \(X\).
3.2. \(e^{-1}(U)\) is an open subset of the product space \(X \times \mathcal{C}'(X, Y)\) containing the slice \(C \times \{f_0\}\).
3.3. By the Tube Lemma, there exists an open set \(W \subseteq \mathcal{C}'(X, Y)\) (with \(W \in \mathcal{T}\)) such that \(f_0 \in W\) and \(C \times W \subseteq e^{-1}(U)\).
4. Inclusion of the neighborhood \(W\) in \(S(C, U)\):\(W \subseteq S(C, U)\).Proof:
4.1. Let \(f \in W\). For any \(x \in C\), \((x, f) \in C \times W \subseteq e^{-1}(U)\).
4.2. Hence \(e(x, f) = f(x) \in U\) for all \(x \in C\), which means \(f(C) \subseteq U\).
4.3. By definition of \(S(C, U)\), this shows \(f \in S(C, U)\).
4.4. Thus \(f_0 \in W \subseteq S(C, U)\).
5. Q.E.D.By steps 3 and 4, every point \(f_0 \in S(C, U)\) has an open neighborhood \(W \in \mathcal{T}\) contained in \(S(C, U)\), so \(S(C, U)\) is open in \(\mathcal{T}\). By step 1, \(\mathcal{T}\) contains the compact-open topology \(\mathcal{T}_{\text{co}}\).