I'm currently working on tangent spaces on manifolds $M$. Especially the tangent vector space. The book points out that tangent vectors at a point $p\in M$ are a class on curves in which two arbitrary elements of it satisfies:
$$ \phi_{1}(t=0)=\phi_{2}(t=0)$$ So they intersect, and:
$$ \frac{dx^{u}}{dt}(\phi_{1}(t=0))_{t=0} =\frac{dx^{u}}{dt}(\phi_{2}(t=0))_{t=0}$$
I'm trying to prove that the definition of the sum of two tangent vectors vectors as the equivalence class:
$$ u + v := [\psi^{-}(\psi^°\phi_{11}+\psi^°\phi_{21})] $$
Where $\psi$ is a map from the curves on $M$ to the $R^{n}$ and $\phi$ the maps whose equivalence class gives you $u$ and $v$ separately.
Is not dependent on the representation of the class, that is, choose another $\phi_{12}$ for $u$ and $\phi_{22}$ for $v$ and you have the same equivalence class $u+v$.
So it's clear that no matter the representation of the class, the sum vector passes by $p$. Given another representation I have:
$$ \frac{dx^{u}}{dt}(\phi_{11}(t=0))_{t=0} =\frac{dx^{u}}{dt}(\phi_{12}(t=0))_{t=0}$$
$$ \frac{dx^{u}}{dt}(\phi_{21}(t=0))_{t=0} =\frac{dx^{u}}{dt}(\phi_{22}(t=0))_{t=0}$$
So I'm leaving with this which I can not prove:
$$ \frac{dx^{u}}{dt}(\psi^{-}(\psi^°\phi_{11}+\psi^°\phi_{21})(t=0))_{t=0} =\frac{dx^{u}}{dt}(\psi^{-}(\psi^°\phi_{12}+\psi^°\phi_{22})(t=0))_{t=0}$$
Thanks.