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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

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    One speaks *only* of differentiability of the compositions $\psi\circ\phi^{-1}$ as maps on (open subsets of) the fixed vector space. (Your vector space shouldn't be varying with the index, should it?) – 2017-01-07
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    I haven't look at Lang in ages and no longer have the book. Strange, but OK. Thanks. – 2017-01-07
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    @TedShifrin I have checked it in my copy, the vector spaces vary with the index. I think the author notation refers to "an index" not "dimension" of the vector spaces. – 2017-01-08

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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.

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    You contradicted yourself in your answer. The question asked specifically about $\phi^{-1}$. :) – 2017-01-07
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    You say: "Note that only for the map $\psi\phi^{-1}$ it makes sense to talk about differentiability, not for the maps $\psi$, $\phi^{-1}$ alone" (and I agree). But then your first sentence makes no sense ;) – 2017-01-07
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    @TedShifrin Upps. I misread the question. I read $\psi\phi^{-1}$ where $\phi^{-1}$ is written. – 2017-01-07
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    @KonKan Changed my answer accordingly. – 2017-01-07
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    Nice. and quick! +1 – 2017-01-07