precomposition for the compact open topology
1. Proposition
Let \((X, \mathcal{T}_X)\), \((Y, \mathcal{T}_Y)\) and \((Z, \mathcal{T}_Z)\) be topological space and \(f: Z \rightarrow X\) a continuous map Given the mapping space \(\mathcal{C}(X,Y)\) with the compact open topology, precomposition with \(f\) is a continuous map
\begin{align*} f^{ \rightarrow}: \mathcal{C}(X,Y) \rightarrow& \mathcal{C}(Z,Y) \\ g \mapsto& g \circ f \end{align*}2. Proof
Let \(\mathcal{O}_{K,U} \subseteq \mathcal{C}(X,Y)\) be open. Then the preimage is
\begin{align*} (f^{ \rightarrow})^{-1}[\mathcal{O}_{K,U}] =& (f^{ \rightarrow})^{-1}[\{g \in \mathrm{Hom}_{\mathrm{Top}}(Z,Y) \vert g[K] \subseteq U\}] \\ =& \{g' \in \mathrm{Hom}_{\mathrm{Top}}(X,Y) \vert g \circ f[K] \subseteq U\} \end{align*}where \(f[K]\) is compact as continuous image of a compact space, hence
\begin{align*} \{g' \in \mathrm{Hom}_{\mathrm{Top}}(X,Y) \vert g \circ f[K] \subseteq U\} = \mathcal{O}_{f[K], U} \end{align*}