Linear algebra invertible transformation of coordinates

[LCSphysicist]

Homework Statement

Under what conditions is a coordinate transformation invertible in a neighborhood of some point?

Relevant Equations

N.

$A^{x'} = T(A^{x})$, where T is a linear transformation, in such way maybe i could express the transformation as a changing of basis from x to x' matrix:
$A^{x} = T_{mn}(A^{x'})$, in such conditions, i could say det $T_{mn} \neq 0$. But how to deal with, for example, $(x,y) -> (e^x,e^y)$ ?

 

[haruspex]

Think about the Jacobian.

 

[fresh_42]

Homework Statement:: Under what conditions is a coordinate transformation invertible in a neighborhood of some point?
Relevant Equations:: N.

$A^{x'} = T(A^{x})$, where T is a linear transformation, in such way maybe i could express the transformation as a changing of basis from x to x' matrix:
$A^{x} = T_{mn}(A^{x'})$, in such conditions, i could say det $T_{mn} \neq 0$. But how to deal with, for example, $(x,y) -> (e^x,e^y)$ ?

If the transformation is linear, you used the determinant criterion to check bijectivity. If it is not linear, then you obviously cannot use theorems about linear transformations. In that case you will have to define the inverse function and show that it is a bijection.

There is a difference between a bijection and an isomorphism. Isomorphisms belong to a certain category, here the category of vector spaces. This requires linearity. If the category is e.g. the topological spaces, then continuity in both directions is required, in the case of smooth manifolds it is differentiability. A bijection on the other hand is merely an isomorphism on the category of sets, where no further structure is considered, just sets.

By allowing an arbitrary coordinate transformation, you changed from the category of vector spaces to the category of sets. This means you forgot all linear structures. So you have to show that there is a second transformation $T'$ such that $T\circ T'$ and $T'\circ T$ are both identities of the corresponding set of vectors.