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Lets have smooth $n-$ manifold $M\subset R_N$ and define its tangent space at $p\in M$ to be set of equivalence classes of smooth curves with $\gamma: I\to M$, $\gamma(0)=p$ with relation $\gamma_1\sim\gamma_2$ if $\frac{d\gamma_1(0)}{dt}=\frac{d\gamma_2(0)}{dt}$. How to prove that $T_p(M)$ is $n$ dimensional vector space?

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How are you defining $\frac{d \gamma_1(0)}{dt}$ without defining tangent spaces? – Qiaochu Yuan Jun 24 '12 at 10:10
You first have to establish the operations of addition and scalar multiplication. Do you know how to define these? – Olivier Bégassat Jun 24 '12 at 10:11
I think that $\frac{d\gamma}{d t}(0)=(\varphi\circ\gamma)'(0)$ for a chart $(U,\varphi)$. – JBC Jun 24 '12 at 10:12
up vote 5 down vote accepted

Let $(U,\varphi)$ a chart in $p$. Let $\theta_\varphi:T_pM\rightarrow\mathbb R^n$ defined by $\theta_\varphi([\gamma])=(\varphi\circ\gamma)'(0)$.
Then $\theta_\varphi$ is clearly injective (by the definition of your equivalence relation). We check that it's surjective : let $v\in\mathbb R^n$, let $\gamma:t\mapsto\varphi^{-1}(tv+\varphi(p))$ then $v=\theta_\varphi([\gamma])$.
So, $\theta_\varphi$ is a bijection and we can use it to transfer the vector space structure from $\mathbb R^n$ to $T_pM$ :

If $\xi,\eta\in T_pM$, we define $\xi+\eta=\theta_\varphi^{-1}(\theta_\varphi(\xi)+\theta_\varphi(\eta))$ and if $\lambda\in\mathbb R$, we define $\lambda\xi=\theta_\varphi^{-1}(\lambda\theta_\varphi(\xi))$.
You can easily check that you get a space vector and that $\theta_\varphi$ is a linear isomorphism.

This construction is independant of the choice of $(U,\varphi)$ : if $(V,\psi)$ is another chart containing $p$, then $\theta_\varphi\circ\theta_\psi^{-1}(v)=d_{\psi(m)}(\varphi\circ\psi^{-1})(v)$.

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