Exists open subset and one-to-one $C^1$ mapping such that mapping of intersection is open subset Let $M$ be a smooth $k$-manifold in $\mathbb{R}^n$. Given ${\bf p} \in M$, how would I go about showing there exists an open subsets $W$ of $\mathbb{R}^n$ with ${\bf p} \in W$, and a one-to-one $C^1$ mapping $f: W \to \mathbb{R}^n$, such that $f(W \cap M)$ is an open subset of $\mathbb{R}^k \subset \mathbb{R}^n$?
 A: Let ${\bf p} \in M$. There exists an open subset $V$ of $\mathbb{R}^n$ such that ${\bf p} \in V$ and $P = V \cap M$ is a $k$-dimensional patch. Let $$U \subseteq \mathbb{R}^k, \text{ } h(x_{i_1}, \dots, x_{i_k}): U \to \mathbb{R}^{n-k}$$ define the patch. Let $X$ be an open subset of $\mathbb{R}^{n-k}$ such that $X \supseteq h(U) = \text{Im}\,h$. Then $U \times X \subseteq \mathbb{R}^n$ is open and $W \cap M = V \cap M = P$. In particular, ${\bf p} \in W$. Define $$f: W \to \mathbb{R}^n, \text{ }f(x_1, \dots, x_n) = (x_{i_1}, \dots, x_{i_k}, h(x_{i_1}, \dots, x_{i_k})) - (0, \dots, 0, x_{i_{k+1}}, \dots, x_{i_n}).$$Clearly $f$ is a $C^1$ map as $h$ is. Also $f$ is $1$-$1$; to see this, if $f(x_1, \dots, x_n) = f(y_1, \dots, y_n)$, then$$x_{i_1} = y_{i_1},\, \dots,\, x_{i_k} = y_{i_k}$$and $$-(x_{i_{k+1}}, \dots, x_{i_n}) + h(x_{i_1}, \dots, x_{i_k}) = -(y_{i_{k+1}}, \dots, y_{i_n}) + h(y_{i_1}, \dots, y_{i_k}).$$This implies$$(x_{i+1}, \dots, x_{i_n}) = (y_{i_{k+1}}, \dots, y_{i_n}),$$  so$$(x_1, \dots, x_n) = (y_1, \dots, y_n).$$Hence $f$ is $1$-$1$.
If $(x_1, \dots, x_n) \in W \cap M = P$, $$(x_{i+1}, \dots, x_{i_k}) \in U$$ and $$(x_{i_{k+1}}, \dots, x_{i_n}) = h(x_{i_1}, \dots, x_{i_k}),$$ so $$f(x_1, \dots, x_n) = (x_{i_1}, \dots, x_{i_k}, 0, \dots, 0) = U \subseteq \mathbb{R}^k \subseteq \mathbb{R}^n.$$Therefore, we have $f(W \cap M) = U$, as desired.
