# Every chart is smoothly compatible, what is wrong with my argument?

Let $A_1$ and $A_2$ be two distinct smooth structures on the manifold $M$. For any two charts $(U,\phi)\subset A_1$ and $(V,\psi)\subset A_2$, both $\phi$ and $\psi$ are diffeomorphisms. If we assume that $U$ and $V$ are not disjoint, then $\psi\circ\phi^{-1}:\phi(U\cap V)\to \psi(U\cap V)$ is a diffeomorphism as the composition of two diffeomorphisms is a diffeomorphism. But this is a contradiction, what is wrong with my argument?

• Why is it contradiction ? – HK Lee Dec 1 '13 at 11:52
• Because this is true for every element of $A_1$ and $A_2$. But these atlases were supposed to be distinct. – gedeon Dec 1 '13 at 11:57
• $\phi$ is a diffeomorphism with respect to $A_1$, not to $A_2$. – user99914 Dec 1 '13 at 11:59
• @John Why does that matter? – gedeon Dec 1 '13 at 12:08

## 1 Answer

If you choose the smooth structure $A_1$, then $\phi^{-1}(U\cap V) \to U\cap V$ is differentiable with respect to $A_1$. But then $\psi : U\cap V \to \psi(U\cap V)$ is not differentiable with respect to $A_1$ and the composition is not differentiable. You may want to use $A_2$ in the second map, but then you have to consider

$$\psi \circ I \circ \phi^{-1},$$

where $I$ is the identity map $I: (U\cap V, A_1) \to (U\cap V, A_2)$. But this map is not differentiable.