Let $M$ and $N$ be smooth manifolds and $T: M \to N$ be a smooth map. Let $ \mathcal{F}(M,\mathbb{R})$ (resp.$ \mathcal{F}(N,\mathbb{R})$) denote the space of smooth functions from $M$ (resp. $N$) to $\mathbb{R}$ and let $F\_T$ denote the pullback of $T$, i.e. the map $F\_T: \mathcal{F}(N,\mathbb{R}) \to \mathcal{F}(M,\mathbb{R})$ defined by $F\_T(f)= f \circ T$. 


Given an arbitrary map $F: \mathcal{F}(N,\mathbb{R}) \to \mathcal{F}(M,\mathbb{R})$, is there a way to determine if it is of the form $F\_T$ for some $T$, or to associate with it a closest map $F\_T$ of this form (i.e. with respect to some metric) and to determine the corresponding map $T$? Also, could there be some tweaking to the problem statement here that would make this possible? 


Thank you very much.



Inverse Problem for Pullback
Concerning (i):


If $F\colon\mathcal{F}(N,\mathbb R)\to \mathcal{F}(M,\mathbb R)$ is of the form $F\_T$, $F$ is a homomorphism of algebras, this is an obvious necessary condition. It turns out, it is also a sufficient one.


There is a contravariant functor $\mathcal A$ from the category of smooth manifolds and smooth maps to the category of algebras, which send every manifold $M$ to the algebra of smooth functions on it:


$$
\mathcal A\colon M\mapsto \mathcal{F}(M,\mathbb R)
$$
and smooth maps to the corresponding algebra homomorphisms:


$$
\mathcal A\colon \{T\colon M\to N\}\mapsto\{F\_T\}.
$$


Your question is what is the image of this functor on the level of $\hom$'s. The answer is: 


$\mathcal A$ is full and faithful, in particular, any homomorphism of algebras $F\colon\mathcal{F}(N,\mathbb R)\to \mathcal{F}(M,\mathbb R)$ is of the form $F\_T$. 


The proof can be found in <http://www.emis.de/monographs/KSM/kmsbookh.pdf> - see 35.9 and 35.10 therein.


