embedded submanifold in euclidean space derivative Let $y$ be the coordinate function on $\mathbb{R}^2$. Consider $\mathbb{S}^1$ as an embedded submanifold. If $p\in (1,0)$, how do I show that $\frac{\partial}{\partial{y}}|_p\in T_p\mathbb{S}^1$? It clearly is in $T_p\mathbb{R}^2$.
My attempt: Since $\mathbb{S}^1$ is defined by the smooth function $F(x,y)=x^2+y^2-1$,
$T_p\mathbb{S}^1=Ker(F_{*,p})$ for all $p\in \mathbb{S}^1$.
Hence if $h\in C^{\infty}$, $F_{*,p}(\frac{\partial}{\partial{y}}|_p)h=\frac{\partial{h\circ F}}{\partial{y}}|_p=0$
where the last equality holds because $\frac{\partial{F}}{\partial{y}}|_p=0$
Question 2: In Loring Tu's book, does the differentiable structure as constructed in Proposition 9.4, coincide with the differentiable structure as constructed in example 5.16 for the sphere?
 A: Yes, that argument works. Perhaps it wouldn't hurt to make the use of the chain rule explicit. Also note that $\frac{\partial F}{\partial y}\Big\vert_p=0$ holds true precisely because $p=(1,0)$, this would fail at other points (convince yourself of this).
Alternatively, you don't need to be this abstract. Since we're working in $\mathbb{R}^2$, we can identify all tangent spaces with subspaces of $\mathbb{R}^2$. Then $\frac{\partial}{\partial y}\Big\vert_p$ is just the vector $(0,1)$ and, for all $p\in S^1$, the tangent space $T_pS^1$ is precisely the subspace of $\mathbb{R}^2$ orthogonal to $p$. And, of course, $(1,0)$ is orthogonal to $(0,1)$ (this is what's illustrated in Fig. 19.1. in Tu's book).
A: Another way to see this is to actually produce a slice chart about $p=(1,0).$ Note that if $f(x,y)=x^2+y^2-1$ then $\partial f_x(1,0)=2\neq 0$ so if we set $\phi(x,y)=(f,y)$ then, we get a slice chart in a small neighborhood $p\in U\mathbb \subseteq R^2.$ That is, $(\phi|_{S^1},U\cap S^1)$ is a chart about $p$ in $S^1$. But $\phi|_{S^1}(f,y)=(0,y)\cong y$ and so $T_pS^1=\operatorname {span} \{\partial_y\}.$
