Theorem 3

Let $X, Y, Z$ be three topological spaces and let $f$ be a mapping of $X \times Y$ into $Z$. If $f$ is continuous then $\tilde{f}: x \to f(x, .)$ is a continuous mapping of $X$ into $C_c(Y; Z)$. The converse is true if $Y$ is locally compact.

Suppose that $f$ is continuous. To show that $\tilde{f}$ is continuous we have to prove that, for each compact subset $K$ of $Y$ and each open subset $U$ of $Z$, the inverse image $V$ of $T(K, U)$ under $\tilde{f}$ is open in $X$. Let $x_0 \in V$; for each $y \in K$, we have $f(x_0, y) \in U$, and since $f$ is continuous there is a neighbourhood $V_y$ of $x_0$ in $X$ and a neighbourhood $W_y$ of $y$ in $Y$ such that $f(V_y \times W_y) \subset U$. Since $K$ is compact, there exists a finite number of points $y_i \in K$ such that the sets $W_{y_i} (1 \leq i \leq n)$ cover $K$. Let $V'$ be the intersection of the neighbourhoods $V_{y_i}$ of $x_0$, which is a neighbourhood of $x_0$; if $x \in V'$ and $y \in K$, we have $f(x, y) \in U$, since $y$ is contained in some one of the $W_{y_i}$ and $x$ is contained in each $V_{y_i}$; hence $V' \subset V$, and therefore $V$ is a neighbourhood of each of its points and consequently is open in $X$.

Conversely, suppose that $\tilde{f}$ is continuous and that $Y$ is locally compact, and let us show that $f$ is continuous. Let $x_0 \in X$, let $y_0 \in Y$ and let $U$ be an open neighbourhood of $f(x_0, y_0)$ in $Z$; we shall show that there is a neighbourhood $V$ of $x_0$ in $X$ and a neighbourhood $W$ of $y_0$ in $Y$ such that $f(V \times W) \subset U$. Since $y \to f(x_0, y)$ is continuous, there is a compact neighbourhood $W$ of $y_0$ such that $f(\{x_0\} \times W) \subset U$. On the other hand, since $\tilde{f}$ is continuous, the set $V$ of $x \in X$ such that $f(x, .) \in T(W, U)$ [i.e. such that $f(x, y) \in U$ for all $y \in W$] is an open subset of $X$ and therefore a neighbourhood of $x_0$; and we have $f(V \times W) \subset U$.

Q.E.D.