# Inverse Problem for Pullback

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.

• What is $C(M)$? Continuous functions or smooth? If the latter, then you want $T$ to be smooth; if the former, why are you demanding that $M$ and $N$ be smooth? Nov 10 '14 at 1:20
• @JoséFigueroa-O'Farrill I meant to indicate smooth throughout, and I have edited the question now to correct this, although if any of the listed questions have answers for continuous replacing smooth, that would also be of equal interest. Thank you very much. Nov 10 '14 at 1:30

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.