Specifying an arbitrary point on a manifold It is known that any arbitrary point x on the sphere $\mathbb{S}^2$ can be parametrised by the spherical coordinates
$$\bf{x}=r(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\quad 0\le\theta\le\pi,0\le\phi\le 2\pi.$$
Here the arbitrary point x appears to be a vector function on $\mathbb{R}^3$.See for instance Steward calculus (parametrisation of the unit sphere)
Now I know that the sphere is embedded in $\mathbb{R}^3$, I start asking and want to know a very elementary general fact...
Can any arbitrary point on a general manifold be viewed as a vector function on the corresponding Euclidean space? (Given the manifold can be embedded into that Euclidean space) 

To be a manifold, a space $X$ has to be covered in coordinate maps $\varphi: A \subseteq \Bbb R^n \to X$. This is part of the definition. In particular, consider $\varphi$ to be a coordinate map onto a neighborhood of your point $x$. If $X$ is embedded into $R^m$ by some other map $\psi$, then $\psi \circ \varphi$ is a smooth map that could be considered a vector function.
