Zeroes of $dx_1$ on $\mathbb{R}^2$ vs. zeroes of $dx_1|_{S^1}$ on $S^1$ Let us consider $S^1$ as a manifold embedded in $\mathbb{R}^2$. Let $dx_1\in\Omega^1(\mathbb{R^2})$. 
$$
Z_{\mathbb{R}^2}:=\{p\in\mathbb{R}^2:(dx_1)_p=0\}\\
Z_{S^1}:=\{p\in\mathbb{R}^2:(dx_1|_{S^1})_p=0\}
$$
I have to prove that $Z_{\mathbb{R}^2}\cap S^1\ne Z_{S^1}$.
I have checked that $Z_{\mathbb{R}^2}\cap S^1\subset Z_{S^1}$. Now I have to prove the inequality. Any help?
 A: For any $p \in \mathbb{R}^2$, 
$$\dfrac{\partial}{\partial x_1}\bigg|_p \in T_p\mathbb{R}^2$$ 
and 
$$(dx_1)\left(\dfrac{\partial}{\partial x_1}\bigg|_p\right) = 1 \neq 0$$ 
so $p \not\in Z_{\mathbb{R}^2}$. Therefore, $Z_{\mathbb{R}^2} = \emptyset$.
Now let $i : S^1 \to \mathbb{R}^2$ be the inclusion map, then $dx_1|_{S^1} = i^*dx_1$. Note that $(i_*)_p : T_pS^1 \to T_p\mathbb{R}^2$ is just the inclusion of $T_pS^1$ as a subspace of $T_p\mathbb{R}^2$, so if $v \in T_pS^1$, 
$$(dx_1|_{S^1})_p(v) = (i^*dx_1)_p(v) = (dx_1)_p((i_*)_pv) = (dx_1)_p(v).$$
If $p = (a, b) \in S^1$, 
$$T_pS^1 = \operatorname{span}\left\{b\frac{\partial}{\partial x_1}\bigg|_p - a\frac{\partial}{\partial x_2}\bigg|_p\right\} \subset \operatorname{span}\left\{\frac{\partial}{\partial x_1}\bigg|_p, \frac{\partial}{\partial x_2}\bigg|_p\right\} = T_p\mathbb{R}^2.$$ 
Note that
$$(dx_1|_{S^1})_p\left(b\frac{\partial}{\partial x_1}\bigg|_p - a\frac{\partial}{\partial x_2}\bigg|_p\right) = (dx_1)_p\left(b\frac{\partial}{\partial x_1}\bigg|_p - a\frac{\partial}{\partial x_2}\bigg|_p\right) = b,$$
so $p \in Z_{S^1}$ if and only if $b = 0$. Therefore, $Z_{S^1} = \{(1, 0), (-1, 0)\}$.
So we see that $Z_{\mathbb{R}^2}\cap S^1 = \emptyset\cap S^1 = \emptyset \neq \{(1, 0), (-1, 0)\} = Z_{S^1}$.
