Two non-equivalent atlas on $\mathbb S^1$ In the same way that I can find two non-equivalent atlas for ${\mathbb{R}}$ (in fact I found an infinite countable number of them) I'm trying to find two non-equivalent atlas for ${\mathbb{S^1}}$.
I tried the atlas obtained by stereographic projection and the canonical projections(the projections onto the x-axis and the y-axis) but I think they are equivalent. Any idea? Maybe my calculus is wrong and they are not equivalent? Anyway, could you spot an idea or heuristic to deal with this problem? 
Thanks in advance. This is my first approach to smooth manifolds.
Notes:
Definition: Two atlas are equivalent if their charts are compatible.
Definition: Two charts (U,$\phi$), (V,$\psi$) are compatible if $U \cap V=\emptyset  $   or    $ \psi \phi^{-1}:\phi(U) \rightarrow \psi(V)$  and  $\phi \psi^{-1}:\psi(V) \rightarrow \psi(U)$ are both $C^{\infty}$.
 A: You are correct that the stereographic projection atlas and the canonical projections atlas are equivalent, and this can be seen by explicitly writing down a formula for each overlap map from a chart in the stereographic projection atlas to a chart in the canonical projection atlas, and checking that this formula defines a $C^\infty$ function.
Here is a very easy way to construct inequivalent atlases on the same differentiable manifold $X$, e.g. $X=\mathbb{R}$ or $X=\mathbb{S}^1$. Pick any homeomorphism $f : X \to X$ which is not a diffeomorphism (one always exists). For each chart in the given atlas $(U,\phi)$, define a chart $(f^{-1}(U),\phi \circ f)$ in the new atlas. The overlap condition holds between charts in this new atlas because the $f$'s cancel out. But an overlap between a chart in the new atlas and one in the old is not smooth, because the $f$ does not cancel out and it would follow that $f$ is smooth which it isn't.
So for an incompatible atlas on $X=\mathbb{R}^1$ you could use, for example, the nondiffeomorphic homeomorphism $f(x)=x^3$. For an incompatible atlas on $X = \mathbb{S}^1$ you could use $f(e^{2 \pi i \theta}) = e^{2 \pi i \theta^2}$ for each $\theta \in [0,1]$.
