This question is part of class notes from which I am studying manifolds:
Let $f:S^2 \to \mathbb{R}^3$ be defined by $f(x,y,z)=(-y,x,0)$. (a) Prove that $f$ is a smooth map.(b) Use $f$ to define a smooth vector field on $S^2$.
The issue is following :does everytime to verify that map $f$ is smooth I have to use the following definition : We say that $f$ is smooth at $p$ if for any chart $(U,\phi)$ containing $p$ and $(V,\psi)$ containing f(p) , the map $\psi \circ f\circ \phi^{-1}: U_1 \to V_1$ is smooth.
I think going by this definition it will be a tedious task. So, is there any other result/ theorem by which smoothness can be proved easily.
Unfortunately, I need help on (b) also.
Please shed some light on it.
Well, for one thing the codomain is $\mathbb{R}^3$, so you don't actually need a chart on it. That is, you only need to show that $f \circ \psi^{-1}$ is smooth for any chart $\psi: U \subseteq S^2 \to \mathbb{R}^2$.
To be a bit more slick, you can observe that $f$ extends smoothly to a map defined on all of $\mathbb{R}^3$, so in reality $f$ is the composition of two smooth maps $\widetilde{f} \circ \iota$, where $\iota: S^2 \to \mathbb{R}^3$ is the inclusion map and $\widetilde{f}(x,y,z) = (-y,x,0)$ as above. Of course this requires that you already have shown that $S^2$ is a submanifold of $\mathbb{R}^n$ (from which we can conclude that the inclusion is smooth); if you don't have the ability to claim that at this point in the class the work to do so would be just as much as using the original definition in the first place.