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Suppose that one has a diffeomorphism [itex] \phi : M\rightarrow M[/itex] from a manifold onto itself, then a point [itex] p\in M[/itex] is mapped to a different point [itex] p'\in M[/itex] by [itex] \phi[/itex] as follows $$p\mapsto\phi (p)=p'$$

I get that in general these will then be two distinct points on the manifold, but I've heard diffeomorphisms are analogous to active coordinate transformations?! Is this because one can choose a coordinate chart [itex] (U, \psi)[/itex] in the neighbourhood of [itex] p[/itex], i.e. [itex] p\in U\subset M[/itex] , such that the point [itex] p[/itex] has coordinate values $$\psi (p)=\lbrace x^{\mu}(p)\rbrace$$ and then map this point to a distinct point [itex] p'[/itex] via the diffeomorphism [itex] \phi[/itex] . Having done so, can one then use a pullback mapping, [itex]\phi^{\ast}[/itex] to pullback the coordinates in the neighbourhood of [itex] p'[/itex] , [itex] \lbrace \tilde{x}^{\mu}(p')\rbrace[/itex] to the coordinates [itex]\lbrace x^{\mu}(p)\rbrace[itex] in the following manner $$ x^{\mu}(p)= (\phi^{\ast}\tilde{x})^{\mu}(p')$$ In this sense, is it that although [itex] p[/itex] and [itex] p'[/itex] are distinct points on the manifold, due to the diffeomorphism [[itex] \phi[/itex] they can be assigned the same coordinate values?