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This definition is inspired by the book mentioned in this question. As it turns out later it is identical to that in question mentioned in the comment by Paul Frost.

The tangent space at $p \in M$ can be defined as the space of derivations on $C^\infty(M)$. I thought that the space of functions $C^\infty(M)$, which is a very large jungle of smooth functions, is actually redundant, to charaterise these special maps (derivations) by their action on them. So I wanted to reduce it bya quotient. First I have to prove that if the derivative of a smooth function at $p \in U \subset M$ vanishes in a some coordinates, it vanishes in all other compatible coordinates. Indeed, if $p \in U$ and $(U,\phi),(U,\psi)$ are any compatible charts, then for all $f \in C^\infty(M)$ we have, if $$ \hat f = f \circ \phi^{-1}:\phi(U) \to \mathbb R, \quad D\hat f_{\phi(p)} = 0 $$ then $$ \tilde f = f \circ \psi^{-1} = \hat f\circ \phi\circ\psi^{-1}:\psi(U) \to \mathbb R,\quad D\tilde f_{\psi(p)} = (D\hat f_{\phi(p)})\circ (D (\phi\circ\psi^{-1})_{\psi(p)}) = 0 $$ The subset $W_p(M) \subset C^\infty(M)$ of functions with vanishing derivative is well defined and is a subspace, so one may take the quotient $$ T_p^*M := C^\infty(M)/W_p(M) $$ The elements are equivalence classes of functions, each class is a set of functions with common (first) derivative. The derivative itself as a map $D(f\circ \phi^{-1})_{\phi(p)} \in \mathbb R^{m*}$ will depend on the coordinate map $\phi$. Since these classes are distinguished from each other by the first derivative (in some coordinates) then $\dim T_p^*M = \dim \mathbb R^{m*} = m$, as expected, since $\mathbb R^{m*}$ is the set of all such linear maps $D(f\circ \phi^{-1})_{\phi(p)}$. One has still to prove that under coordinate change the classes are the preserved, which should be an easy part.

Choosing some coordinate map, the coordinate function $x^i:U \to \mathbb R$, or rather their classes $[x^i]$, constitute a basis for $T_p^*M$. Then a tangent vector $v_p$ is the map $$ v_p: [f] \mapsto D(f\circ \phi^{-1})_{\phi(p)}(v^\phi), \quad v^\phi \in \mathbb R^m $$

Could anyone check if every thing above is consistent ?


Update

The Leibniz property emerges from this definition of the tangent vector as a consequence of the old product rule and since $$ (fg)\circ \phi^{-1} = (f\circ \phi^{-1})(g\circ \phi^{-1}). $$ Also the independence of the number $v_p([f])$ of the coordinate map $\phi$ gives the transformation rule $$ v^\phi \mapsto v^\psi = (D (\psi\circ\phi^{-1})_{\phi(p)})(v^\phi). $$

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    $\begingroup$ I think this may be relevant. $\endgroup$
    – Paul Frost
    Sep 20, 2021 at 17:01
  • $\begingroup$ Thanks a lot for the reference, almost identical if not the same definition $\endgroup$
    – Physor
    Sep 20, 2021 at 19:53

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