I am reading John M. Lee's book: Introduction to Smooth Manifolds ...(adsbygoogle = window.adsbygoogle || []).push({});

I am focused on Chapter 3: Tangent Vectors ...

I have some further questions concerning Lee's conversation on computations with tangent vectors and pushforwards ...

The relevant conversation in Lee is as follows:

In the above text we read:

" ... ... we see that [itex]\phi_* \ : \ T_p M \longrightarrow T_{ \phi(p) } \mathbb{R}^n[/itex] is an isomorphism ... ... "

and then further ...

" ... ... [itex]T_{ \phi(p) } \mathbb{R}^n[/itex] has a basis consisting of all derivations [itex]\frac{ \partial }{ \partial x^i } |_{\phi(p)} \ , \ i = 1, \ ... \ ... , n[/itex]. Therefore the pushforwards of these vectors under [itex]( \phi^{-1} )_*[/itex] form a basis for [itex]T_p M[/itex] ... ... "

Question 1

Is [itex]( \phi^{-1} )_*[/itex] the inverse of [itex]\phi_*[/itex] and hence the isomorphism from [itex]T_{ \phi(p) } \mathbb{R}^n[/itex] to the tangent space [itex]T_p M[/itex]?

Why isn't the inverse [itex]( \phi_* )^{-1}[/itex] ?

Question 2

Since [itex]( \phi^{-1} )_* \ : \ T_{ \phi(p) } \mathbb{R}^n \longrightarrow T_p M [/itex] and we know that [itex]T_{ \phi(p) } \mathbb{R}^n[/itex] is a vector space ... then since [itex]( \phi^{-1} )_*[/itex] is an isomorphism ... then ... [itex]T_p M[/itex] is a vector space ...... ...is that correct?

Hope someone can help ... ...

Peter

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# Further Questions on Computations in Coordinates - Lee, Ch 3

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