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Show that every smooth k-manifold with boundary is a smooth manifold with corners

Definitions:

1.If $M \subseteq R^n$, M is a smooth k-manifold with boundary if for every point $p\in M$ there exists an open neighbourhood $V\subseteq M$ of $p$, and an open set $U$ of $\mathbb{H}^k$ such that $\phi: U\to V$ is a regular embedding.

2.A set $M \subseteq R^n$, M is a smooth k-manifold with corners if for every $p \in M$ there exist open sets $V\subseteq M$, $U\subseteq \overline {\mathbb{R}_+^k}$ and a regular embedding $\phi: U\to V$ such that $p \in V$

So suppose $M$ is a smooth manifold with boundary and assume that $\phi:U\to V$ is a regular embedding which covers the interior points of $M$ and $\Pi:N\to M$ is a regular embedding which covers the boundary.

Then for any $p\in M$ either $p=\phi(x)$ or $p=\Pi(y)$ for some $x,y$ in the interrior or boundary of $\mathbb{H}^k$

Let $p\in int(M)$, then the function $\psi:U\to K$, by $\psi(x_1,...,x_{k-1},x_k)=(e^{x_1},...,e^{x_{k-1}},e^{x_k})$, is a homeomorphism from ${\mathbb{R}_+^k}$ to the interior of $\mathbb{H}^k$ and thus $\phi\circ\psi^{-1}:K\to V$ is a regular embedding from an open subset of $\mathbb{R}_+^k$ to $M$.

I believe this works for dealing with the interior points. What I am not sure how to deal with is the boundary points.

The only thing I could think of would be to take an orthogonal projection of the points in $\partial\mathbb{H}^k$ which are not in $\partial\mathbb{R}^k_+$ say for example in $\mathbb{R}^2$, take $(x,0)\to(0,\vert x\vert)$. But I don't think this works in higher dimension.

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