In the book that I am reading (Introduction to Differential Manifolds by Barden and Thomas), they note that, for each chart $(U,\phi)$ in the atlas of some smooth $m$-manifold and each $p \in U,$ we have a bijection $\phi_*(p):T_p(M)\to \mathbb{R}^m$ given by
$$\phi_*(p):[\gamma]_p \mapsto (\phi\circ\gamma)'(0).$$
Transporting the linear structure from $\mathbb{R}^m$ onto $T_p(M)$ makes each $\phi_*(p)$ into a linear isomorphism.
They then say that, by taking the dual map of $(\phi_*(p))^{-1}\!,$ we have, for each $p \in U,$ a linear isomorphism $T_p^*(M)\to(\mathbb{R}^m)^*$ and so taking these maps all together we obtain a bijection $$\phi^*:T^*(U)=\coprod_{p \in U} T_p^*(M) \to \phi(U)\times (\mathbb{R}^m)^*$$ The next step is to view $\phi(U)\times (\mathbb{R}^m)^*$ as an open subspace of $\mathbb{R}^{2m}$ and then transport the topological structure of this onto $T^*(U)$ via the map $\phi^*.$
BUT... Surely this last step surely requires us to choose an identification of $(\mathbb{R}^m)^*$ with $\mathbb{R}^m$ and so, since this cannot be done canonically, I'm a bit confused...
Any explanations or advice is very welcome. Thanks!
This is where differential geometry "cheats": once you've transported things in $\Bbb R^m$, you are allowed to use tools there that are not in the spirit of differential geometry: the identification of $\Bbb R^m$ with $(\Bbb R^m)^*$ (via the usual isomorphism) is just one of them, but you may use many others if necessary (the Euclidean structure, for instance, the existence of a Fourier transform in order to construct Sobolev spaces on manifolds, and others). Everything that is forbidden on manifolds is allowed in $\Bbb R^m$!
(Also, don't forget that when you say "$\Bbb R^m$" you don't just say "a real $m$-dimensional vector space", but rather "a real $m$-dimensional vector space, together with a preferred basis" - and this makes a big difference.)