# Definition of tangent space does not depend on choice of regular embedding

Definition: Let $$M\subseteq \mathbb{R}^n$$ be a smooth k-manifold and fix a point $$p\in M$$. Let $$C_p$$ denote the collection of $$C^1$$ maps of the form $$\gamma:(-1,1)\to M$$ such that $$\gamma(0)=p$$. Fix a regular embedding $$\phi :U\to v$$ such that $$p\in V$$ and let $$\phi^{-1}:V\to U$$ denote its inverse. Endow $$C_p$$ with and equivalence relation such that $$\gamma_1 \sim \gamma_2$$ if $$(\phi^{-1}\circ \gamma_1)^\prime(0)=(\phi^{-1}\circ \gamma_2)^\prime(0)$$.

Show that this definition doesn't depend on the choice of regular embedding.

So previous to this I had to show that for 2 regular embeddings $$\phi_1,\phi_2$$ that $$\phi_2^{-1}\circ\phi_1$$ is a diffeomorphism.

So let $$\phi_1,\phi_2$$ be 2 regular embeddings for $$M$$. Suppose $$\gamma_1\sim\gamma_2$$.

then $$(\phi^{-1}\circ \gamma_1)^\prime(0)=(\phi^{-1}\circ \gamma_2)^\prime(0)$$

then $$(\phi_2^{-1}\circ\gamma_1)^\prime(0)=(\phi_2^{-1}\circ\phi_1\circ\phi_1^{-1}\circ\gamma_1)^\prime(0)$$$$=(\phi_2^{-1}\circ\phi_1)^\prime\circ\phi_1^{-1}\circ\gamma_1)(\phi_1^{-1}\circ\gamma_1)^\prime(0)$$$$=(\phi_2^{-1}\circ\phi_1)^\prime\circ\phi_1^{-1}\circ\gamma_1)(\phi_1^{-1}\circ\gamma_2)^\prime(0)$$

Really not sure where to go from here.

The chain rule states that

$$D(g\circ f)(x)=D(g)(f(x))\cdot D(f)(x)$$

Since $$\gamma_1(0)=\gamma_2(0)=p$$ your computation goes like

$$(\phi_2^{-1}\circ\gamma_1)^\prime(0) =(\phi_2^{-1}\circ\phi_1\circ\phi_1^{-1}\circ\gamma_1)^\prime(0)$$$$=(\phi_2^{-1}\circ\phi_1)^\prime(\phi_1^{-1}(p))\cdot(\phi_1^{-1}\circ\gamma_1)^\prime(0)$$ $$=(\phi_2^{-1}\circ\phi_1)^\prime(\phi_1^{-1}(p))\cdot(\phi_1^{-1}\circ\gamma_2)^\prime(0)$$

$$=(\phi_2^{-1}\circ\phi_1\circ\phi_1^{-1}\circ\gamma_2)^\prime(0)=(\phi_2^{-1}\circ\gamma_2)^\prime(0)$$