Identification of the tangent space of a manifold and the tangent vectors to curves

I'm studying the different definitions of the tangent space for abstract manifolds, and I'm struggling to prove that these abstract concepts reduce to the classical ones when dealing with submanifolds of $$\mathbb{R}^n$$. What I know:

• The tangent space $$T_pM$$ can be defined as the space of all equivalence classes of curves $$\gamma:(-\epsilon, \epsilon)\to M$$ with $$\gamma(0)=p$$ under the equivalence relation $$\gamma \sim \beta \quad\iff\quad (\mathbf{x}^{-1}\circ\gamma)'(0) = (\mathbf{x}^{-1}\circ\beta)'(0),$$ where $$\mathbf{x}$$ is a local chart of $$M$$ at $$p$$. This is easily shown to be independent of the chosen chart.

• The tangent space $$T_pM$$ can also be defined as the space or all derivations $$D$$ which are linear functionals obeying the Leibniz rule, which act on the space of smooth functions defined locally around $$p$$.

• I know how to prove that the latter two are equivalent (in the sense that the spaces are isomorphic), and related by $$[\gamma]\mapsto \left(f \mapsto (f\circ\gamma)'(0)\right),$$ where the latter map can be thought of as the directional derivative with respect to $$[\gamma]$$

• I also know that they are both isomorphic to $$\mathbb{R}^n$$.

• For the the differential of a smooth map $$\varphi:M\to N$$, I use the definition using derivations derivations, i.e. $$\mathrm d\varphi_p(X)(f) = X(f\circ \varphi).$$

• I know how to prove the chain rule and I also know that the differential $$\mathrm d\varphi_p$$ is an isomorphism whenever the map $$\varphi$$ is a diffeomorphism.

What I don't know how to prove: If $$M$$ is a submanifold of $$\mathbb{R}^n$$, the tangent space $$T_pM$$ is isomorphic to the space of all tangents $$\dot\gamma(0)\in \mathbb{R}^n$$ of curves $$\gamma$$ passing through $$p$$?

I can intuitively see why this should be the case given the information that I have, but I don't know how to write a proper proof of it.

One of the ideas is to show that two curves belong to the same equivalence class if and only if their tangent vectors at $$p$$ coincide, in which case I could do something as shown in this post. However, with my definition of the differential I do not directly see that $$(\mathbf{x}^{-1}\circ\gamma)'(0) = \mathrm d\mathbf{x}^{-1}_p(\dot\gamma(0)).$$ That being said, I would guess that one can use the aforementioned isomorphisms to identify the differential with a map that operates in this way.

Another idea would be to prove that the directional derivative with respect to $$[\gamma]$$ actually coincides with the tangent vector $$\dot\gamma(0)$$, but I don't see how one can do this.

Could you help me with this?

EDIT: I have added a bounty. Let me know if my question is unclear or missing any of the details.

EDIT 2: After thinking about this a bit more, I don't think it is possible to show what I'm trying to show without first defining the tangent space and the differential for submanifolds of $$\mathbb{R}^n$$ in the natural way (i.e. using tangents to curves) and showing that the differential of a chart is an isomorphism, and then using that fact to show that all of the new notions reduce to the classical ones. I would love to be wrong regarding this though.

• I agree. You should show that two curves are equivalent if they have the same tangent vector. I mean, that is essentially the definition of the equivalence relation. The relation you are worried about, $(\mathbf{x}^{-1} \circ \gamma)'(0) = \mathrm{d} \mathbf{x}^{-1}_p (\gamma'(0))$ is just the "chain rule". – Nick Feb 13 at 21:39