How to show that $T_{(1,0)}\mathbb S^1 \cong \operatorname{span}(\{e_2\})$? I want to show that $T_{(1,0)}\mathbb S^1 \cong \operatorname{span}(\{e_2\})$ using the stereographic chart and using the definition that $T_xM$ is the set of velocity vectors $v$ where each vector $v$ is the equivalence class of curves that goes through point $x$ and tangent to each other.
I got so far the following:

*

*Since $\varphi:U\to\mathbb{R}$ is given by $\varphi(x,y)=\frac{x}{1-y}$ and $v=\frac{d}{dt}(\varphi\circ \gamma)(t)\Big|_{t=0}$ for some $\gamma:I\to \mathbb S^1$ with $\gamma(0)=x=(1,0)$, we can compute that
\begin{align}
v& =\frac{d}{dt}(\varphi\circ \gamma)(t)\Big|_{t=0}\\
&=\frac{d}{dt}\Big(\frac{x(t)}{1-y(t)}\Big)\Big|_{t=0}\\
&=\frac{x^{\prime}(t)(1-y(t))-x(t)(-y^{\prime}(t))}{(1-y(t))^2}\Big|_{t=0}\\
&=\frac{x^{\prime}(0)(1-y(0))+x(0)y^{\prime}(0))}{(1-y(0))^2}\\
&=x^{\prime}(0)+y^{\prime}(0).
\end{align}
I don't know how to interprete that and how to actually show that $T_{(0,0)}\mathbb S^1$ should be a span of $e_2$.


*I know that if $i:\mathbb S^1\to\mathbb{R}^2$ is an inclusion, then $$di_x:T_x \mathbb S^1\to T_{i(x)}\mathbb{R}^2\text{ is 
 injective}.$$ So, we need somehow show $di_x(v)=\operatorname{span}(\{e_2\})$.
What should I do?

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $f((x,y))=x^2+y^2$ then $1$ is a regular value of $f$ and hence $f^{-1}(1)$ is a submanifold of $\mathbb{R}^2$ but $S^1=f^{-1}(1)$ and furthermore given $p \in S^1$, $T_p (S^1)=Ker(df_p)$
and calculating we obtain that $df_p=2(p_1,p_2)$ then is easy to see that $$T_p(S^1)=\{x \in \mathbb{R}^2 ; \langle p,x \rangle =0\}=span(p) ^{\perp}$$
But $\mathbb{R}^{2}=span(p) \oplus span(p)^{\perp}$ and $dim(span(p))=1$ then $dim(span(p)^{\perp})=1$ then it is only necessary to find one element of $span(p)^{\perp}$ to characterize it, it is easy to see that if $p=(p_1,p_2)$ then $(-p_2,p_1) \in span(p)^{\perp}$ it follows that
$$T_p(S^1)= span((-p_2,p_1))$$
And in particular in your problem taking $p=(1,0)$ then $T_{(1,0)}(S^1)=span((0,1))$
