How do we prove that the two basic structures on an n-sphere (projection maps and stereography maps)give the same smooth structure?

Question

We know that structures defined by two atlases on the same topological space give the same smooth structure if their transition mappings are all smooth. For the two mentioned structures - the one defined by projections and the one defined by stereographical projections - I found all the transition mappings. Is it enough to say that these are differentiable because their coordinate mappings are all differentiable? Or would you find the Jacobian and somehow show it's not zero? Perhaps by induction? I tried the other way, since the first one didn't seem precise enough, but I haven't been able to find any general rule for that Jacobian

Answer 1

You have two atlases $\mathcal A, \mathcal A'$. You must show that all transition maps are differentiable. This means that for any two charts $\varphi : U \to V$ in $\mathcal A$, $\varphi' : U' \to V'$ in $\mathcal A$, the two transition maps $$\varphi^{-1}(U \cap U') \stackrel{\varphi}{\rightarrow} U \cap U' \stackrel{(\varphi')^{-1}}{\rightarrow} \varphi'(U \cap U')$$ $$(\varphi')^{-1}(U \cap U') \stackrel{\varphi'}{\rightarrow} U \cap U' \stackrel{\varphi^{-1}}{\rightarrow} \varphi(U \cap U')$$ are differentiable. This is equivalent to showing that one of them is a diffeomorphism.

Thus either you prove differentiability directly for both transition maps or you prove it for one and prove additionally that the Jacobian of this map has non-zero determinant in all points.

A simple example is this. On $\mathbb R$ we have two single chart atlases $\mathcal A = \{ id : \mathbb R \to \mathbb R \}$ and $\mathcal A' = \{ f : \mathbb R \to \mathbb R \}$ with $f(x) = x^3$. The transition maps are $f$ and $f^{-1}$. Note that $f^{-1}(x) = \sqrt[3]{x}$. $f$ is differentiable, but $f^{-1}$ is not (it has no derivative in $0$). The latter can either be shown directly or by invoking that $f'(0) = 0$.