I have been stuck with the following problem for a few days:
Let $M$ be a manifold with boundary of dimension $n$ (not necessarily compact), then there exists a smooth embedding $h:M\to \mathbb{H}^{2n+1}= \{(x_1,\ldots,x_{2n+1}\mid x_{2n+1}\geq 0\}$, such that $h$ maps the boundary of $M$ into $\partial \mathbb{H}^{2n+1}=\{(x_1,\ldots,x_{2n+1})\in \mathbb R^{2n+1} \mid x_{2n+1}=0\}$ and maps the interior of $M$ into $\text{Int }\mathbb{H}^{2n+1}=\{(x_1,\ldots,x_{2n+1})\in \mathbb R^{2n+1} \mid x_{2n+1}>0\}$, i.e., $$h(\partial M)\subseteq \partial \mathbb H^{2n+1}, \quad h(\text{Int }M)\subseteq \text{Int }\mathbb H^{2n+1}.$$
My attempt.
By the collar neighborhood theorem, $\partial M$ has a neighborhood $U$ in $M$, and there is a diffeomorphism $f:U\to \partial M\times [0,1) $ such that $f(x)=(x,0)$, $\forall x\in \partial M$, and $f^{-1}\left(\partial M\times [0,1/2]\right)$ is a closed subset of $M$. By the Whitney embedding theorem for manifolds without boundary, we can embed $\partial M\times [0,1)$ into $\mathbb R^{2n}\times [0,1)$, namely, $i:\partial M\times [0,1)\hookrightarrow \mathbb R^{2n}\times [0,1)$. Thus we get an embedding $g=i\circ f:U\to \mathbb R^{2n}\times [0,1)\hookrightarrow \mathbb H^{2n+1}$, which maps $\partial M$ into $\partial \mathbb H^{2n+1}$.
But how can I extend $g$ to an embedding defined on the whole $M$? I cannot see it clearly, could you please tell me some ideas? Would be appreciated.