I'm trying to solve the following exercise (number 2.59, from M. Abate, F. Tovena). Consider the fact that, as control-system engineer, I'm trying to learn differential geometry as self-taught. So, please, be patient regarding my mistakes. Thank you.
The text is the following.
Be $M$ a manifold, and $p\in M$. We say that two differential curves. $\sigma_{1},\sigma_{2}:\mathbb{R}\rightarrow M$ with $\sigma_{1}(0)=\sigma_{2}(0)=p$ are equivalent if $(\phi\circ\sigma_{1})'(0)=(\phi\circ\sigma_{2})'(0)$ for some local chart $(\phi,U)$ centerd in $x_{0}$. Show that this is an equivalence relation and that the set of equivalence class is isomorph to $T_{x_{0}}M$.
Here there is my attempt.
Proof
Denoting the above relation between curves as $\sim$, to guarantee that it is an equivalence realation, we need to prove that $\sigma_{1} \sim \sigma_{1}$ (refelxivity), $\sigma_{1} \sim \sigma_{2}$ iff $\sigma_{2} \sim \sigma_{1}$ (simmetry), $\sigma_{1} \sim \sigma_{2}$ and $\sigma_{2} \sim \sigma_{3}$ implies $\sigma_{1} \sim \sigma_{2}$ (transitivity). We have that $\phi\circ\sigma_{1}= (\sigma^{1}_1,\dots,\sigma^{n}_{1})$ hence we have that $$(\phi\circ\sigma_{1})^{'}(0)=\frac{d (\phi\circ\sigma_{1})}{dt}=\sum^{n}_{j=1}(\sigma^{i}_{1})^{'}(0)$$
where $t$ is the variable of the curve.
For the above expression, we have trivially that the reflexivity property holds. We have also that if $$(\phi\circ\sigma_{1})^{'}(0)=\frac{d (\phi\circ\sigma_{1})}{dt}=\sum^{n}_{j=1}(\sigma^{i}_{1})^{'}(0) =\sum^{n}_{j=1}(\sigma^{i}_{2})^{'}(0)=\frac{d (\phi\circ\sigma_{2})}{dt}=(\phi\circ\sigma_{2})^{'}(0)$$ it holds $(\phi\circ\sigma_{2})^{'}(0)= (\phi\circ\sigma_{1})^{'}(0)$. On the other hand we have that $$(\phi\circ\sigma_{2})^{'}(0)=\frac{d (\phi\circ\sigma_{2})}{dt}=\sum^{n}_{j=1}(\sigma^{i}_{2})^{'}(0) =\sum^{n}_{j=1}(\sigma^{i}_{1})^{'}(0)=\frac{d (\phi\circ\sigma_{1})}{dt}=(\phi\circ\sigma_{1})^{'}(0)$$ implies $(\phi\circ\sigma_{1})^{'}(0) = (\phi\circ\sigma_{2})^{'}(0)$. Hence we have proven also that the simmetry property holds. It remains to prove the transitivity one. Assuming to have $$(\phi\circ\sigma_{1})^{'}(0)=\frac{d (\phi\circ\sigma_{1})}{dt}=\sum^{n}_{i=1}(\sigma^{i}_{1})^{'}(0) =\sum^{n}_{i=1}(\sigma^{i}_{2})^{'}(0)=\frac{d (\phi\circ\sigma_{2})}{dt}=(\phi\circ\sigma_{2})^{'}(0)$$ and $$(\phi\circ\sigma_{2})^{'}(0)=\frac{d (\phi\circ\sigma_{2})}{dt}=\sum^{n}_{i=1}(\sigma^{i}_{2})^{'}(0) =\sum^{n}_{i=1}(\sigma^{i}_{3})^{'}(0)=\frac{d (\phi\circ\sigma_{3})}{dt}=(\phi\circ\sigma_{3})^{'}(0)$$ thanks to the previously proven simmetry property, we have that $(\phi\circ\sigma_{1})^{'}(0) = (\phi\circ\sigma_{3})^{'}(0)$.
In order to solve the exercise, it remains to prove that the set of the equivalence classes define a natural isomorphism with the tangent space to $T_{x_{0}}M$.
We need to prove that there is a linear bijection between the set of equivalence classes and $T_{x_{0}}M$.
The equivalence class for the curve $\sigma_{j}$ is defined as $$[\sigma_{j}]=\{\sigma_{i} : (\phi\circ\sigma_{i})^{'}(0)=(\phi\circ\sigma_{j})^{'}(0),\hspace{2mm} \sigma_{i}, \sigma_{j}:\mathbb{R}\rightarrow M \hspace{2mm}\}$$ for some chart $(\phi,U)$ centerd in $x_{0}$ .
Let we consider the map $\iota : M\rightarrow TM_{x_{0}} $, $\rho : [\sigma_{j}]\rightarrow [\sigma_{j}]^{'}(0) $, where
$$\rho([\sigma_{j}])=[\sigma_{j}]^{'}(0) = \sum^{n}_{i=1}([\sigma^{i}_{j}])^{'}(0)\frac{\partial}{\partial x^{i}}\big{|}_{x_{0}}$$
where $[\sigma^{i}_{j}]$ is the $i$-th component of $[\sigma^{}_{j}]$
We have to prove that that $\rho$ is an isomrphism. Clearly $\rho$ it is linear. We prove that it is injective. If $[\sigma]_{j}\neq O$ we will have $[\sigma^{h}_{j}] \neq 0$. In fact, considering the germs which represents the $j$-h coordinates, ${x}^{h}\in C^{\infty}(x_{0})$ represented by $(\mathbb{R}^{n}, x^{h})$ we have $$\rho([\sigma_{j}])(x^{h}) = \sum^{n}_{i=1}([\sigma^{i}_{j}])^{'}(0)\frac{\partial x^{h}}{\partial x^{i}}\big{|}_{x_{0}}=[\sigma^{i}_{j}] \neq 0$$.
In order to prove that it is surijective, we have to prove that $X =\rho([\sigma])$, where $X\in TM_{p}$.
We will use the following lemma
Lemma
Be $x_{0}= (x_{0}^{1},\dots, x_{0}^{n} )\in \mathbb{R}^{n}$. Then there are germs $g_{1},\dots, g_{n} \in C^{\infty}(x_{0})$ such that $g_{j}(x_{0})= \frac{\partial f}{\partial x^{j}}(x_{0})$ and $$f = f(x_{0}) + \sum_{j= 1}^{n}(x^{j}- x_{0}^{j})g_{j}$$
where $x^{j} \in C^{\infty}(x_{0})$ is the germ represented by the j-th coordinate function.
We set $[\sigma_{h}] = X(x^{h})$ and $[\sigma_{h}] = ([\sigma^{1}_{h}],\dots, [\sigma^{n}_{h}])$. Be $f\in C^{\infty}(x_{0})$. By applying the previous lemma, we get $$ X(f) = X(f(x_{0})) + \sum_{i= 1}^{n}X((x^{i}- x_{0}^{i})g_{i})$$. Now, since $X$ it's a derivation, thanks to the Leibniz property and the the fact that $X(c) = 0$ where $c\in \mathbb{R}$, we get $$ X(f) = \sum_{i= 1}^{n}((X(x^{i})- X(x_{0}^{i}))g_{j}) + (((x^{i})- (x_{0}^{i}))X(g_{i})$$
which gives $$X(f) =\sum_{i= 1}^{n} [\sigma^{i}_{j}]\frac{\partial f}{\partial x^{i}}) = \rho([\sigma_{j}])f$$ which proves the fact that $\rho$ is an isomophism.
Let me know what you think about it.
Thank you for your support.