Lang's "Fundamentals of Differential Geometry" Chapter 2.

Let $X$ be a set. Let $\{U_i\}$ be an Atlas of pairs $(U_i,\alpha_i)$ such that:

i) $U_i \subset X$ for all $i$ and $\cup_i U_i = X$

ii) Each $\alpha_i$ is a bijection of $U_i$ onto an open subset $\alpha_i U_i$ of some Banach space $E_i$, and for any indices$i,j$ we have that $\alpha_j\alpha_i^{-1}: \alpha_i(U_i \cap U_j) \rightarrow \alpha_j(U_i \cap U_j)$ is a $C^p$ isomorphism.

It goes on to say that the vector spaces $E_i$ do not need to be the same for all $I$, yet that by taking the derivative of $\alpha_j\alpha_i^{-1}$ we see that $E_i$ and $E_j$ are toplinearly isomorphic.

Thus I am confused because $d(\alpha_j\alpha_i^{-1})$ is only a linear homeomorphism from $\alpha_i(U_i \cap U_j)$ to $\alpha_j(U_i \cap U_j)$, which are both mearly contained in $E_i$ and $E_j$.... SO where is the linear homeomorphism from $E_i$ to $E_j$ which must exist if they are toplinearly isomorphic!!

(the morphisms in the toplinear category are linear contiuous maps between topological vector spaces)


The derivative $d(\alpha_j\alpha_i^{-1})$ is not a map from $\alpha_i(U_i\cap U_j)$ to $\alpha_j(U_i\cap U_j)$. Rather, it is a map from $\alpha_i(U_i\cap U_j)$ to the space of continuous linear maps $E_i\to E_j$: at each point of $\alpha_i(U_i\cap U_j)$, the derivative is some linear map $E_i\to E_j$. Moreover, since $\alpha_j\alpha_i^{-1}$ is assumed to be a $C^p$ isomorphism, its derivative must be invertible at each point, so the derivative at each point is a toplinear isomorphism $E_i\to E_j$.

Note, though, that you can only conclude that $E_i$ and $E_j$ are toplinearly isomorphic if $U_i\cap U_j$ is nonempty, so that there actually is a point of $\alpha_i(U_i\cap U_j)$ to take the derivative. As a result, the conclusion that the $E_i$ are all toplinearly isomorphic is only valid if $X$ is connected (so any two charts can be connected by a chain of overlapping charts).


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