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).

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).

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.

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).