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For a definition of differentiable manifold, we require transition maps to be diffeomorphisms. Do we also require that for every chart $(U,V,\phi)$ map $\phi: U \to V$ is diffeomorphism, or just a homeomorphism?

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    @NickAlger Be careful. Nowadays, the most common definition of a differentiable manifold starts saying: _Let $M$ be a topological space_.2018-01-15

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Given a topological manifold you only require that the charts are homeomorphisms and that the transition functions are diffeomorphisms. Saying that the charts are differentiable does not make sense in the topological setting. But if you have a differentiable atlas you can define that any map from or to the manifold is differentiable if its concatenations with the charts are differentiable. So after you have defined such a differentiable structure the charts are automatically diffeomorphisms.

As for $dφ_p$: After you have defined such a differentiable structure on the manifold, the map $φ$ is just another map between two manifolds where one happens to be $V ⊂ ℝ^n$. Then the notion of tangent space is defined on both of them and for $φ: U → V$ when $p\in U, φ(p) = 0 \in V$ we have that $dφ_p$ is the induced map $T_pU → T_0V \simeq ℝ^n$. Depending on how you construct the tangent spaces also the understanding of $dφ$ will vary. If, like in your MIT text book, you define $T_pU$ to be the space of curves in $U$ through $p$ that “point in the same direction”, all you have to do, to calculate $dφ_p$ is take a representative curve $γ$ in $U$ and push it to $V$ like $φ∘γ$ which gives you a curve in $V$ that represents some element of the tangent space $T_0V$.

Of course in this special case for $φ$ being a chart you can also give a basis of $T_pU$ in terms of $φ$ just by taking the coordinate curves $γ_i : t ↦ (0, …, 0, t, 0, …, 0) \in V$ and then defining $X_i := [φ^{-1}∘γ_i]_{T_pM}$ where $[\cdot]$ stands for taking all curves in the same direction.

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    Thank you. As I understood when we have differential structure then any chart function $\phi$ is diffeomorphism as $\bf{1}_{{Id}}=\phi\circ \phi^{-1}$ is diffeomoprhism. How does one make sense out of $d\phi_{p}$? Why I am asking this is that I am stuck on [link](http://math.mit.edu/classes/18.952/spring2012/Chapter4-18.952.pdf) diagram 3, Remark 2.14, page 15. Can someone provide me with clarification please? Thank you again!2012-06-24