[proofplan] We construct the boundary-straightening diffeomorphism $\Phi$ in two stages. First, the [Local Graph Representation Theorem](/theorems/6) provides rotated coordinates in which $\partial U$ is the graph of a $C^k$ function $\gamma$. Second, we define $\Phi$ by subtracting $\gamma$ from the $n$-th coordinate, which flattens the boundary to $\{y_n = 0\}$ and maps the interior to $\{y_n > 0\}$. We then verify explicitly that $\Phi$ is a $C^k$ diffeomorphism by computing its inverse and checking regularity. [/proofplan] [step:Obtain the local graph representation of $\partial U$ near $z$] By the [Local Graph Representation Theorem](/theorems/6), there exist a radius $r > 0$, a rotation matrix $R \in O(n)$, and a $C^k$ function \begin{align*} \gamma: \mathbb{R}^{n-1} &\to \mathbb{R} \\ (\tilde{x}_1, \ldots, \tilde{x}_{n-1}) &\mapsto \gamma(\tilde{x}_1, \ldots, \tilde{x}_{n-1}) \end{align*} such that in the rotated coordinates $\tilde{x} := R(x - z) \in \mathbb{R}^n$, the [boundary](/page/Boundary) and interior of $U$ within the ball $V := B(z, r)$ are characterised by: \begin{align*} x \in \partial U \cap V &\iff \tilde{x}_n = \gamma(\tilde{x}_1, \ldots, \tilde{x}_{n-1}), \\ x \in U \cap V &\iff \tilde{x}_n > \gamma(\tilde{x}_1, \ldots, \tilde{x}_{n-1}). \end{align*} Here $\tilde{x}' := (\tilde{x}_1, \ldots, \tilde{x}_{n-1})$ denotes the first $n-1$ rotated coordinates. [/step] [step:Define the flattening map $\Phi$ by subtracting the graph function] Define the map \begin{align*} \Phi: V &\to \mathbb{R}^n \\ x &\mapsto y = (y_1, \ldots, y_n) \end{align*} by setting $\tilde{x} := R(x - z)$ and \begin{align*} y_i &:= \tilde{x}_i \quad \text{for } i = 1, \ldots, n-1, \\ y_n &:= \tilde{x}_n - \gamma(\tilde{x}_1, \ldots, \tilde{x}_{n-1}). \end{align*} The map $\Phi$ is the composition of the rigid motion $x \mapsto R(x - z)$ (which is $C^\infty$) with the shear $({\tilde{x}_1, \ldots, \tilde{x}_n}) \mapsto (\tilde{x}_1, \ldots, \tilde{x}_{n-1}, \tilde{x}_n - \gamma(\tilde{x}'))$ (which is $C^k$ since $\gamma$ is $C^k$). Therefore $\Phi$ is $C^k$. [guided] The idea behind the construction is geometrically simple: the boundary $\partial U \cap V$ is the graph $\tilde{x}_n = \gamma(\tilde{x}')$ in rotated coordinates, and we want to flatten it to the hyperplane $\{y_n = 0\}$. The natural choice is to subtract the graph height: define $y_n = \tilde{x}_n - \gamma(\tilde{x}')$, which sends points on the boundary (where $\tilde{x}_n = \gamma(\tilde{x}')$) to $y_n = 0$, and points in the interior (where $\tilde{x}_n > \gamma(\tilde{x}')$) to $y_n > 0$. The first $n-1$ coordinates are left unchanged: $y_i = \tilde{x}_i$ for $i = 1, \ldots, n-1$. This makes the map a "vertical shear" that only adjusts the height coordinate. Explicitly, $\Phi = S \circ L$ where the rigid motion \begin{align*} L: V &\to \mathbb{R}^n, \quad x \mapsto R(x - z) \end{align*} is $C^\infty$ (linear plus translation), and the shear \begin{align*} S: \mathbb{R}^n &\to \mathbb{R}^n, \quad (\tilde{x}_1, \ldots, \tilde{x}_n) \mapsto (\tilde{x}_1, \ldots, \tilde{x}_{n-1}, \tilde{x}_n - \gamma(\tilde{x}')) \end{align*} is $C^k$ because each component is either the identity or involves the $C^k$ function $\gamma$. As a composition of $C^\infty$ and $C^k$ maps, $\Phi$ is $C^k$. [/guided] [/step] [step:Verify that $\Phi$ maps boundary to $\{y_n = 0\}$ and interior to $\{y_n > 0\}$] Let $W := \Phi(V)$, which is an [open set](/page/Open%20Set) in $\mathbb{R}^n$ (shown below once we establish $\Phi$ is a diffeomorphism). **Boundary:** If $x \in \partial U \cap V$, then $\tilde{x}_n = \gamma(\tilde{x}')$ by the graph representation. Therefore \begin{align*} y_n = \tilde{x}_n - \gamma(\tilde{x}') = 0. \end{align*} Hence $\Phi(\partial U \cap V) \subset \{y \in W : y_n = 0\}$. **Interior:** If $x \in U \cap V$, then $\tilde{x}_n > \gamma(\tilde{x}')$, so \begin{align*} y_n = \tilde{x}_n - \gamma(\tilde{x}') > 0. \end{align*} Hence $\Phi(U \cap V) \subset \{y \in W : y_n > 0\}$. The reverse inclusions follow from the bijectivity of $\Phi$ (established in the next step), which gives equality: $\Phi(\partial U \cap V) = \{y \in W : y_n = 0\}$ and $\Phi(U \cap V) = \{y \in W : y_n > 0\}$. [/step] [step:Construct the inverse $\Phi^{-1}$ and verify it is $C^k$] Given $y = (y_1, \ldots, y_n) \in W$, we recover the rotated coordinates $\tilde{x}$ by inverting the defining relations: \begin{align*} \tilde{x}_i &= y_i \quad \text{for } i = 1, \ldots, n-1, \\ \tilde{x}_n &= y_n + \gamma(y_1, \ldots, y_{n-1}). \end{align*} Then $x = R^\top \tilde{x} + z$ (since $R \in O(n)$ implies $R^{-1} = R^\top$), giving the explicit formula \begin{align*} \Phi^{-1}(y) = z + R^\top \bigl(y_1, \ldots, y_{n-1}, \, y_n + \gamma(y_1, \ldots, y_{n-1})\bigr)^\top. \end{align*} Each component of $\Phi^{-1}$ is a $C^k$ function of $y$: the map $y \mapsto (y_1, \ldots, y_{n-1}, y_n + \gamma(y'))$ is $C^k$ because $\gamma$ is $C^k$, and $R^\top$ is a constant linear map. Therefore $\Phi^{-1}$ is $C^k$, and $\Phi: V \to W$ is a $C^k$ [diffeomorphism](/page/Implicit%20Function%20Theorem). [guided] To verify bijectivity, we must show that the system defining $\Phi$ can be uniquely inverted. The first $n-1$ equations $y_i = \tilde{x}_i$ immediately give $\tilde{x}_i = y_i$. The $n$-th equation $y_n = \tilde{x}_n - \gamma(\tilde{x}')$ gives $\tilde{x}_n = y_n + \gamma(y_1, \ldots, y_{n-1})$, since $\tilde{x}' = y' := (y_1, \ldots, y_{n-1})$ is already determined. This shows the map $y \mapsto \tilde{x}$ is uniquely defined and explicit. To recover $x$ from $\tilde{x}$: the rotation $\tilde{x} = R(x - z)$ inverts to $x = R^\top \tilde{x} + z$ (using $R^\top R = I$ since $R \in O(n)$). Combining: \begin{align*} \Phi^{-1}(y) = z + R^\top \bigl(y_1, \ldots, y_{n-1}, \, y_n + \gamma(y_1, \ldots, y_{n-1})\bigr)^\top. \end{align*} For regularity: the function $y \mapsto y_n + \gamma(y')$ is $C^k$ because $\gamma \in C^k$ and addition is smooth. The remaining components are either the identity or application of the constant matrix $R^\top$ and translation by $z$, all of which are $C^\infty$. Therefore $\Phi^{-1}$ is $C^k$. Since both $\Phi$ and $\Phi^{-1}$ are $C^k$, the map $\Phi: V \to W$ is a $C^k$ diffeomorphism satisfying $\Phi(U \cap V) = \{y \in W : y_n > 0\}$ and $\Phi(\partial U \cap V) = \{y \in W : y_n = 0\}$, as claimed. [/guided] [/step]