# Every smooth manifold with boundary is a smooth manifold with corners.

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.