Very often in mathematical physics literature I've heard a fact that probably (if I've understood it correctly) translates as follows.
Let $M$ be a differentiable manifold, and denote by $C^\infty(M)$ the ring of smooth real functions $M\to\Bbb{R}$. Let now $E$ and $F$ be vector bundles on $M$, and notice that the spaces of smooth sections $\Gamma(E)$ and $\Gamma(F)$ are modules over $C^\infty(M)$. Let now $f:\Gamma(E)\to\Gamma(F)$ be a smooth, $\Bbb{R}$-linear function. The following are equivalent.
- $f$ is $C^\infty(M)$-linear;
- $f$ is "local", meaning that given sections $\phi,\psi$ which agree at the point $p\in M$, then $f(\phi)$ and $f(\psi)$ agree at $p$ too.
First of all, is the above correct? If not, what is the correct statement? Also, where can I find a reference and a proof?
Let $\{e_i\}$ be a local frame defined in an open set $U\subset M$ over $\Gamma(E)$ and $\{f_j\}$ be the local frame over $\Gamma(F)$ generated by the image of the $e_i$'s under the action of $f$. That is,
$$f_i = f(e_i).$$
Then, any sections $\phi,\psi\in \Gamma(E)$ can be expressed locally in terms of the frame $\{e_i\}$,
$$\phi = \phi^i e_i,$$ $$\psi = \psi^i e_i,$$
where $\phi^i$ and $\psi^i$ are scalar functions on $M$. The two sections $\phi$ and $\psi$ are the same on $U$ if and only if $\psi^i = \phi^i$. We can then we can use $C^\infty(M)$ linearity to show that:
$$f(\phi) = \phi^i f(e_i) = \phi^i f_i,$$ $$f(\psi) = \psi^i f(e_i) = \psi^i f_i.$$
Again, we have that the sections $f(\phi)$ and $f(\psi)$ are the same on $U$ if and only if $\psi^i = \phi^i$.
This means that $\phi = \psi \Leftrightarrow f(\phi) = f(\psi)$ on $U$. Of course the result is valid for every point $p\in U$, so that your conclusion follows.