Assume a very simple smooth 1-manifold, with a single chart covering, What I'd like to know is, can we use and Fourier transform for functions on this manifold just as we did for the case of $\mathbb{R}$ with all the properties of the Fourier transform applicable ? If so, Is the proof simple ?
EDIT [in response to comment from Pete L. Clark]
Consider the maps $y : (0,1) \to \mathbb{R}$ and a bijective map $z : (0,1) \to (0,1)$ where $z$ is continuous in $(0,1)$. consider $\tau \in (0,1)$ where $\tau = z(t)$ for some $t \in (0,1)$. Let $x(t) = y(\tau)$, I want to represent $y(\tau)$ using Fourier transform (with all the properties of FT) rather than taking FT of $x(t)$. This would ease me in avoiding some derivations.
EDIT 2 [earlier it should be read as](sorry for the error in earlier edit)
Let $M$ be smooth 1-manifold. Consider the maps $y : M \to \mathbb{R}$ and a bijective map $z : (0,1) \to M$ where $z$ is continuous in $(0,1)$ is a co-ordinate chart which covers $M$. Can i take the Fourier Transform of $y$ as though it ($y$) is acting on $\mathbb{R}$.