TITLE: Exponential rule for Whitney-$\mathcal{C}^{\infty}$-topology
QUESTION [5 upvotes]: Let $M,N,X$ be smooth manifolds. Equip the space of smooth functions between two manifolds with the (strong) Whitney- $\mathcal{C}^\infty$-topology.
The evaluation map $$ev\colon N\times\mathcal{C}^\infty(N,X)\rightarrow X$$ is continuous. (This can e.g. be found in "Margalef-Roig/Dominguez - Differential Topology", Proposition 9.6.7)
Therefore one gets a map of sets $$\mathcal{C}(M,\mathcal{C}^\infty(N,X))\rightarrow \mathcal{C}(M\times N,X).$$
I searched the literature for a reference to the "other direction", but did not find anything. I.e. starting with a smooth map $M\times N\rightarrow X$ I want to know whether the adjoint $M\rightarrow \mathcal{C}^\infty(N,X)$ is continuous or if not, in which cases.

REPLY [4 votes]: The other direction is not true. The map
$$\mathcal{C}^\infty(M,\mathcal{C}^\infty(N,X))\rightarrow \mathcal{C}^\infty(M\times N,X)$$
is not surjective if $N$ is not compact. The image consists of functions $f$ such that

for each compact $K\subset M$ there is a compact $L\subset N$ such that $f(x,y)$ is constant in $x\in K$ for each $y\in N\setminus L$.

This is, because smooth (and continuous) curves in $C^\infty (N,X)$ can move only within a compact of $N$ in finite time. See 4.4.4 (page 34) of

Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp. (pdf)

for the very easy proof. There even the continuous case that you ask for is treated.
Added:
The following might give you feeling for the reason:
Consider $f\in C^\infty(N,\mathbb R)$, equip $C^\infty(N,\mathbb R)$ with the strong Whitney topology. If $t\mapsto t.f$ is continuous $\mathbb R\to C^\infty(N,\mathbb R)$, then $f$ has compact support. Thus $C^\infty_c(N,\mathbb R)$ is the maximal topological vector space in ($C^\infty(N,\mathbb R)$, Whitney topology).