Let $E,F\rightarrow X$ be two smooth vector bundles over a manifold $X$ and suppose we have a (linear) map $T:\Gamma(E)\rightarrow \Gamma(F)$ which is invertible and local. I would like to know if the inverse $T^{-1}:\Gamma(F)\rightarrow \Gamma(E)$ is also local.
Some definitions: $\Gamma(E)$ denotes the space of smooth section over $E$ (and similarly for $F$). A linear map $T:\Gamma(E)\rightarrow \Gamma(F)$ is called local if for every section $s\in\Gamma(E)$ and open set $U\subset X$ such that $s|_U=0$, we have $(Ts)|_U=0$. Equivalently, $$\text{supp}(Ts)\subset \text{supp}(s).$$
Disclaimer: I don't know if this is true, so a counter-example is also welcome. I tried proving it in different ways, but I couldn't understand how to use the locality property. At best, what I got was as follows: let $s\in\Gamma(F)$ and $U\subset X$ open such that $s|_U=0$. Since $T$ is surjective, let $w\in\Gamma(E)$ be such that $Tw=s$. If we prove that $w|_U=0$, we are done. From here on, anything I tried doesn't work (e.g. using bump functions; using the above mentioned support property of local maps).
Ideally, I would like to say something like "injective maps which are local are "locally injective"", so I could conclude from $Tw|_U=v|_U=0$ that $w|_U=0$ (using this "local injectivity"), but I don't know if that is the case, nor how to prove it. Any help is appreciated!
This is false. One reason is that local operators include differential operators, and "inverting" a differential operator (i.e. solving an inhomogeneous PDE) is seldom a local procedure.
As a counterexample, take both $E$ and $F$ to be the trivial bundle $S^1\times\mathbb{R}\to S^1$. We can identify the sections $\Gamma(S^1\times\mathbb{R})$ with the smooth functions $\mathbb{R}\to\mathbb{R}$ with period $1$. With this identification in place, consider the operator $L$ defined by $$ Lf(x)=f(x)-f''(x) $$ Using Green's functions, we can find an inverse $$ L^{-1}f(x)=\frac{1}{2e-2}\int_0^1(e^t+e^{1-t})f(x+t)dt $$ which is clearly not a local operator.