homomorphisms of $C^{\infty}(\mathbb R^{n})$ Let $F: \mathbb R^{n} \to \mathbb R^{m} $ be a smooth map, then we have homomorphism of algebras $F^{*}: C^{\infty}(\mathbb R^{m}) \to C^{\infty}(\mathbb R^{n})$. Is it true that any homomorphism of these algebras $f$ is equal to $F^{*}$ for some smooth map $F$?
I think that shouldn't be true, because $ \mathbb R^{n}$ isn't compact. But I can't find the example of this homomorphism.
 A: Let $M$ be a smooth manifold. Observe that an $\mathbb{R}$-algebra homomorphism $C^\infty (M) \to \mathbb{R}$ is necessarily surjective, and that the map
$$p \mapsto \mathrm{ev}_p \text{ where } \mathrm{ev}_p (f) = f (p)$$
defines a bijection between points of $M$ and $\mathbb{R}$-algebra homomorphisms $C^\infty (M) \to \mathbb{R}$. In the case where $M$ has a global chart this is obvious, and in general one can use bump functions to construct a family of functions that determines what point a homomorphism $\phi : C^\infty (M) \to \mathbb{R}$ corresponds to. (Use Hadamard's lemma: given local coordinate functions $(x^i)$ that vanish at $p$ and $f \in C^\infty (M)$, there exist smooth functions $g_i$ such that $f = f(p) + \sum_i x^i g_i$ locally.)
Thus, given any smooth manifold $N$ and any $\mathbb{R}$-algebra homomorphism $\phi : C^\infty (N) \to C^\infty (M)$, there is a unique map $F : M \to N$ such that $\mathrm{ev}_p \circ \phi = \mathrm{ev}_{F (p)}$ for all points $p$ of $M$. Since functions are determined by their values, this implies $\phi = F^*$. It remains to be shown that $F$ is a smooth map, but this is straightforward: just take local coordinate functions on $N$ and apply $\phi$ to get a local expression for $F$ in terms of smooth functions on $M$.
A: Yes. More generally, $M \mapsto C^{\infty}(M)$ provides a contravariant fully faithful functor from the category of smooth manifolds into the category of $\mathbb{R}$-algebras. A reference is:

Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824

