# Graph representation of a function with a non-zero determinant

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:

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