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I can't get a thing from "Advanced calculus" of Lynn H. Loomis, page 374.

Defyning the tangent space, it says that two curves through $x$, named $\varphi$ and $\psi$ are equivalent, i.e. $\varphi \sim \psi$, if for every differentiable real-valued function $f$ we have

$$\left. \frac{d f \circ\varphi}{dt} \right|_{t=0} = \left. \frac{d f \circ\psi}{dt} \right|_{t=0}$$

Then it says that for a chart $(U, \alpha)$ through $x$ we can write $F=f\circ\alpha^{-1}$ and $\Phi=\alpha \circ\varphi$ such that

$$\left. \frac{d f \circ\varphi}{dt} \right|_{t=0} = dF(\Phi'(0))$$

So it concludes that $\varphi \sim \psi$ iff $\Phi'(0) = \Psi'(0)$.

The $\Longleftarrow$ direction is obvious but for me it is not obvious the $\Longrightarrow$. I guess we must prove that if $\Phi'(0) \neq \Psi'(0)$ there is a function $f$ for witch the differential is different.

Any help? Thanks.

The question has already been done here Loomis and Sternberg: Tangent Space to a manifold, using equivalence classes; help justifying one step of an argument but not properly answered for me. I've already studied the difinition of the tangent spaces in other books without this problem but I would like to follow the passages of this book.

Giovanni Barbarani
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    I see that I'm the one who asked that question. The thing I didn't know at the time was the Hahn-Banach theorem from functional analysis (which is a triviality in finite-dimensions but non-trivial in infinite dimensions). Once you know this, all you have to show is that given a Banach space $V$, and a smooth manifold $M$ modelled on $V$, for every continuous linear map $\lambda:V\to \Bbb{R}$, there exists a smooth $f:M\to \Bbb{R}$ (you'll need to use bump functions to make the domain all M) such that $D(f\circ \alpha^{-1})_{\alpha(x)}=\lambda$. – peek-a-boo Sep 20 '21 at 10:02
  • Thanks. It sufficient to consider $f:U \rightarrow \mathbb{R}$ an to choose $f=\lambda \circ \alpha$. In that way $f \circ \alpha^{-1} = \lambda$. And thi concludes the proof. – Giovanni Barbarani Sep 20 '21 at 14:16
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    Yes, this works (ignore my previous comment about bump functions; I just realized the book doesn't require $f$ to be defined on all of $M$) – peek-a-boo Sep 20 '21 at 15:37

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