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Let $U \subseteq \mathbb{R}^n$ be open and $f \in C^1(U, \mathbb{R}^{n+k})$. Furthermore, let $a \in U$ such that $\text{rank}(Df(a)) = n$. The goal of this task is to represent $f$ locally as a graph.

(a) Show that there exists a permutation $\pi \in S_{n+k}$ such that for the canonical unit vectors $e_1, \ldots, e_{n+k}$ in $\mathbb{R}^{n+k}$, the matrix (or induced linear mapping) $P_\pi = (e_{\pi(1)} | \ldots | e_{\pi(n+k)}) \in \mathbb{R}^{(n+k) \times (n+k)}$ and $h = P_\pi \circ f = (h_1, \ldots, h_{n+k})^T$ satisfy $\det D(h_1, \ldots, h_{n})^T(a) \neq 0$.

(b) Show that there exists an open neighborhood $U' \subseteq U$ of $a$, an open set $V' \subseteq \mathbb{R}^n$, a $C^1$-diffeomorphism $\varphi: V' \to U'$, and $g \in C^1(V', \mathbb{R}^k)$ such that $h(\varphi(y)) = (y, g(y))^T$ for all $y \in V'$.

My solution to part $A$

: (a) Show that there exists a permutation $\pi \in S^{n+k}$ such that for the canonical unit vectors $e_1, \ldots, e_{n+k}$ in $\mathbb{R}^{n+k}$, the matrix (or induced linear mapping) $P_{\pi} = (e_{\pi(1)} | \ldots | e_{\pi(n+k)}) \in \mathbb{R}^{(n+k) \times (n+k)}$ and $h = P_{\pi} \circ f = (h_1, \ldots, h_{n+k})^T$, it holds that $\det D(h_1, \ldots, h_{n+k})^T(a) \neq 0$.

To find a local representation of $f$ as a graph, we want to show that there exists a suitable permutation $\pi$ such that the matrix $P_{\pi}$ and the induced linear mapping $h = P_{\pi} \circ f$ satisfy the desired property.

Since $\text{Rang}(Df(a)) = n$, we can assume that the first $n$ columns of $Df(a)$ are linearly independent. We can interpret these columns as $e_1, \ldots, e_n$, corresponding to the canonical unit vectors in $\mathbb{R}^{n+k}$.

Since $f \in C^1(U, \mathbb{R}^{n+k})$, we can apply the inverse function theorem to obtain a local inverse $g$ around $a$. This means that there exists an open neighborhood $V$ of $a$, on which $g: V \to U$ is a continuously differentiable mapping with $g(a) = 0$.

We define $h = g \circ f$, that is, $h(x) = g(f(x))$. Note that $h: U \to V$ is a continuously differentiable mapping since $g$ and $f$ are both continuously differentiable.

Now consider the derivative matrix $D(h_1, \ldots, h_{n+k})(a)$. Since $h = g \circ f$, by the chain rule, we have:

$D(h_1, \ldots, h_{n+k})(a) = Dg(f(a)) \cdot Df(a)$.

Since $g$ is a local inverse around $a$, we have $Dg(f(a)) \cdot Df(a) = I$, where $I$ is the identity matrix.

We can write the matrix $Df(a)$ in the form:

$Df(a) = (e_1 | \ldots | e_n | w_1 | \ldots | w_{n+k-n})$,

where the columns $w_1, \ldots, w_{n+k-n}$ are the remaining columns of $Df(a)$.

Substituting this into the equation, we have:

$Dg(f(a)) \cdot Df(a) = I$,

which gives:

$I = (e_1 | \ldots | e_n | w_1 | \ldots | w_{n+k-n})$.

Since the first $n$ columns of $Df(a)$ are linearly independent, we can find a permutation $\pi \in S^{n+k}$ such that the first $n$ columns of $Df(a)$ correspond to the columns $e_{\pi(1)}, \ldots, e_{\pi(n)}$.

We now set:

$P_{\pi} = (e_{\pi(1)} | \ldots | e_{\pi(n)} | e_{\pi(n+1)} | \ldots | e_{\pi(n+k)})$.

Since the matrix $P_{\pi}$ permutes the columns of $Df(a)$, we have:

$P_{\pi} \cdot Df(a) = (e_1 | \ldots | e_n | w'_1 | \ldots | w'_{n+k-n})$,

where $w'_1, \ldots, w'_{n+k-n}$ are the columns $w_1, \ldots, w_{n+k-n}$ in a new order.

Now consider the mapping $h = P_{\pi} \circ f$:

$h(x) = P_{\pi} \cdot f(x) = (e_{\pi(1)} | \ldots | e_{\pi(n)} | e_{\pi(n+1)} | \ldots | e_{\pi(n+k)}) \cdot f(x) = (h_1(x) | \ldots | h_n(x) | h_{n+1}(x) | \ldots | h_{n+k}(x))$.

The determinant of $D(h_1, \ldots, h_n)(a)$ is given by:

$\det D(h_1, \ldots, h_n)(a) = \det (e_{\pi(1)} | \ldots | e_{\pi(n)} | w'_1 | \ldots | w'_{n+k-n})$.

Since the first $n$ columns of $P_{\pi} \cdot Df(a)$ are linearly independent, $\det D(h_1, \ldots, h_n)(a)$ is nonzero.

Thus, we have shown that there exists a permutation $\pi$ such that the matrix $P_{\pi}$ and the induced linear mapping $h = P_{\pi} \circ f$ satisfy the property $\det D(h_1, \ldots, h_n)(a) \neq 0$. This means that we can locally represent $f$ as a graph.

My solution to part $B$:

To show that there exists an open neighborhood $U' \subseteq U$ of $a$, an open set $V' \subseteq \mathbb{R}^n$, a $C^1$ diffeomorphism $\varphi: V' \rightarrow U'$, and $g \in C^1(V', \mathbb{R}^k)$ such that $h(\varphi(y)) = (y, g(y))^T$ for all $y \in V'$, we can use the inverse function theorem.

From part (a), we know that there exists a permutation $\pi \in S^{n+k}$ such that for the canonical unit vectors $e_1, \ldots, e_{n+k}$ in $\mathbb{R}^{n+k}$, the matrix $P_{\pi} = (e_{\pi(1)} | \ldots | e_{\pi(n+k)}) \in \mathbb{R}^{(n+k) \times (n+k)}$ and $h = P_{\pi} \circ f = (h_1, \ldots, h_{n+k})^T$, it holds that $\det D(h_1, \ldots, h_{n+k})^T(a) \neq 0$.

Consider the mapping $\tilde{h}: U \rightarrow \mathbb{R}^{n+k}$ defined as $\tilde{h}(x) = (h_1(x), \ldots, h_{n+k}(x))^T$. Since $\tilde{h} = (h_1, \ldots, h_{n+k})^T$, we have $\det D\tilde{h}(a) \neq 0$.

Now, we can apply the inverse function theorem to $\tilde{h}$ at $a$. This guarantees the existence of an open neighborhood $U' \subseteq U$ of $a$ and an open set $V' \subseteq \mathbb{R}^{n+k}$ such that $\tilde{h}: U' \rightarrow V'$ has a $C^1$ inverse $\tilde{g}: V' \rightarrow U'$.

Define $\varphi: V' \rightarrow U'$ as $\varphi(x) = \tilde{g}(x)$. Since $\tilde{g}$ is a $C^1$ diffeomorphism, its restriction $\varphi$ to $V'$ is also a $C^1$ diffeomorphism.

Now, let $g: V' \rightarrow \mathbb{R}^k$ be defined as $g(x) = (\tilde{h}{n+1}(x), \ldots, \tilde{h}{n+k}(x))^T$. Since $\tilde{h}(x) = (h_1(x), \ldots, h_{n+k}(x))^T$, we have $g(x) = (h_{n+1}(x), \ldots, h_{n+k}(x))^T$.

For $y \in V'$, we have:

$h(\varphi(y)) = \tilde{h}(\varphi(y)) = \tilde{h}(\tilde{g}(y)) = (h_1(\tilde{g}(y)), \ldots, h_{n+k}(\tilde{g}(y)))^T = (y, g(y))^T$.

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