Proving $\gamma \sim \delta (\gamma'(0)=\delta'(0))\iff (\varphi \circ \gamma)'(0) = (\varphi \circ \delta)'(0) \in \Bbb R^m $

Question

Let $M$ be a smooth manifold of dimension $m$, and $p \in M$. A smooth curve through $p$ is one smooth map $\gamma : J \rightarrow M$, with $J \subset \Bbb R$ an open interval containing $0$, and $\gamma(0) = p$. For such a curve we define $\gamma '(0) \in T_pM$ by $\gamma'(0)(f) = (f \circ \gamma)'(0)$: Let $C_p$ be the set of all smooth curves through $p$. For $\gamma; \delta \in C_p$ we define $\gamma \sim \delta$ as $\gamma'(0) = \delta'(0)$.

(a) Prove that this is an equivalence relation.

Let $(U; \varphi )$ be a card around $p$. Suppose that $\gamma; \delta \in Cp$.

(b) Suppose $\gamma \sim \delta$. Prove that $$(\varphi \circ \gamma)'(0) = (\varphi \circ \delta)'(0) \in \Bbb R^m: \tag{*}$$

(c) Suppose $(*)$ holds. Prove that $\gamma \sim \delta$

My solution:

(a) It is obvious that $\sim$ is an equivalente relation because so is $=$.

$\gamma'(0) = \gamma'(0) \iff \gamma \sim \gamma$

$\gamma \sim \delta \iff \gamma'(0) = \delta'(0) \iff \delta'(0) = \gamma'(0) \implies \delta \sim \gamma$

$\gamma_1 \sim \gamma_2 $ and $\gamma_2 \sim \gamma_3 \iff \gamma_1'(0) = \gamma_2'(0)$ and $\gamma_2'(0) = \gamma_3'(0) \implies \gamma_1'(0)=\gamma_3'(0) \iff \gamma_1 \sim \gamma_3$

(b) $\gamma \sim \delta$. Take $f: M \rightarrow \Bbb R$

$(\varphi \circ \gamma)'(0)(f)=(f \circ (\varphi \circ \gamma))'(0) = ((f \circ \varphi) \circ \gamma))'(0) = D_{\gamma (0)}(f \circ \varphi)\gamma'(0)= D_{p}(f \circ \varphi)\delta'(0)=D_{\delta (0)}(f \circ \varphi)\delta'(0)= (\varphi \circ \delta)'(0)(f)$

(c)Let (*) hold take Take $f: M \rightarrow \Bbb R$

$(\varphi \circ \gamma)'(0)(f) = (\varphi \circ \delta)'(0) (f) $

$\iff ( f \circ (\varphi \circ \gamma) )'(0) = ( f \circ (\varphi \circ \delta) )'(0)$

$\iff (D_{(\varphi \circ \gamma )(0)}f)(\varphi \circ \gamma)'(0)= (D_{(\varphi \circ \delta )(0)}f)(\varphi \circ \delta)'(0)$

$\iff (D_{(\varphi(p)}f)(\varphi \circ \gamma)'(0)= (D_{(\varphi(p)}f) (\varphi \circ \delta)'(0)$

Using ( *): $(D_{\varphi(p)}f)=(D_{\varphi(p)}f)$

but I should be getting $\gamma'(0)=\delta'(0)$

1) What am I doing wrong in (c) and how can I fix it?

2) I am not sure my solutions are correct. They look suspiciously simpleIs the rest of the solution correct?

Answer 1

For (c) I don't think it'll come about just by writing out the definitions. You should pick some special functions f that will give you the result you want. Since you have a local chart, use the coordinate projection functions $f_i(x_1,\dots,x_n) = x_i$, to show that the individual components of $\gamma'(0)$ and $\delta'(0)$ are equal. If they are equal in this coordinate chart, then they are equal in all coordinate charts.