I was reading "Introduction to differentiable manifolds" from Serge Lang, but I got immediately stuck. He defined his atlases as follows:
Let X be a Hausdorff topological space. An atlas of class $C^P (p\geq 0)$ on X is a collection of pairs $(U_i; \phi_i)$ ($i$ ranging in some indexing set), satisfying the following conditions:
- Each $U_i$ is an open subset of $X$ and the $U_i$ cover $X$.
- Each $\phi_i$ is a topological isomorphism of $U_i$ onto an open subset $\phi_iU_i$ of some vector space $E_i$ and for any $i,j, \phi_i(U_i\cap U_j)$ is open in $E_i$.
- The map $$ \phi_j\phi_i^{-1}: \phi_i(U_i\cap U_j) \to \phi_j(U_i\cap U_j) $$ is a $C^p$-isomorphism for each pair of indices $i, j$.
Now the equivalence between two vectors is defined as follows:
We consider triples $(U,\phi, v)$ where $(U,\phi)$ is a chart at x and v is an element of the vector space in which $\phi U$ lies. We say that two such triples $(U,\phi, v)$ and $(V,\psi, w)$ are equivalent if the derivative of $\psi \phi^{-1}$ at $\phi x$ maps $v$ on $w$. The formula reads:
$$ (\psi \phi^{-1})'(\phi x) v = w $$
My question is the following how do you know out of this definition that $\phi^{-1}$ is differentiable? Does it follow from the third bullet point?
Thanks in advance
No, it does not make sense to speak of differentiability for the maps $\psi$, $\phi^{-1}$ alone, as we do not have defined at the points what differentiability is for maps from or to a manifold.
The third bullet point states that for each pair $\phi$, $\psi$ of maps $\psi\phi^{-1}$ is an $C^p$-isomorphism, meaning that both $\psi\phi^{-1}$ and its inverse $\phi\psi^{-1}$ is a $C^p$ map, that is $p$-times continuously differentiable. Note that only for the map $\psi\phi^{-1}$ it makes sense to talk about differentiability, not for the maps $\psi$, $\phi^{-1}$ itself.