Continuity in the Strong(Whitney) Topology

Question

Let $P,M$ and $N$ are smooth manifolds and let $F:P\times M \to N$ be a smooth map. We know its associated map $\tilde F:P \to C^\infty(M,N)$ given by $p \mapsto F_p(m)$ is continuous if and only if $F$ is continuous whenever we equip $C^\infty(M,N)$ with the weak topology. However this is not true in genreal if $C^\infty$ comes with the the strong (Whitney) topology; I feel the counterexample should not be that hard but I am unable to find it. Does anybody know such a function? Thanks in advance.

Answer 1

$M$ must be non-compact or else the topologies are the same, so the answer should exploit non-compactness in some way. Let $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ send $(t, x)$ to $tx$. We wish to show that

\begin{align*} \tilde{F}: \mathbb{R} & \to C^\infty(\mathbb{R}, \mathbb{R}) \\ t & \mapsto F(t, -) \\ \end{align*}

is not continuous in the Whitney topology. Let $U = \{ (x, y) \in \mathbb{R}^2 \mid |y| < 1\}$. Then $M(U) = \{ f\in C^\infty(\mathbb{R}, \mathbb{R}) \mid \text{graph}(F(t,-)) \subset U\}$ is a basic open subset of $C^\infty(\mathbb{R}, \mathbb{R})$ in the Whitney $C^0$ topology and hence sub-basic in the Whitney $C^\infty$ topology. (Here I am using notation and definitions from Golubitsky and Guillemin). This is because $j^0 f (X)$ of $f: X \to Y$ is simply the graph of $f$ in $X\times Y$. But notice that $$\tilde{F}^{-1}(M(U)) = \{ t \in \mathbb{R} \mid \quad |tx| < 1 \quad \forall x\} = \{ 0\}$$

so $\tilde{F}$ is not continuous. This is because for any $t \neq 0$ there exists an $x$ such that $|xy| \geq 1$.