# A function from a smooth manifold with boundary to $[0,\infty)$

Suppose $M$ is a smooth manifold with boundary, show that there exists a smooth function $f: M \rightarrow [0, \infty)$ such that $\partial M = f^{-1}(0)$.

My attempt is that given a chart $(U_\alpha, \varphi _\alpha)$ of $M$, $U_\alpha$ is diffeomorphic to some open subset of $\mathbb H ^m = \mathbb R^{m-1} \times [0,\infty)$. Then the m-th coordinate function is a smooth function from $U_ \alpha$ to $[0, \infty)$. But I don't know how to do this globally.

• Have you seen partitions of unity? Nov 7 '14 at 2:00
• I tried but still don't know how partition of unity could work here. Would you please elaborate a bit more? Thanks! @Jose27
– lyx
Nov 7 '14 at 2:20
• If $f_n$ is the function you obtained in a coordinate patch $U_n$, and $\psi_n$ is a partition of unity subordinate to this cover by closed sets then consider $\sum_n f_n\psi_n$ (Notice that you still have to do something about points in $M\setminus \cup U_n$, but this is not a problem because...) Nov 7 '14 at 3:08
• Do you already know that $\partial M$ has a neighborhood diffeomorphic to $\partial M \times [0,1)$? Nov 7 '14 at 3:53
• @Jose27, thank you so much! The explanation is pretty clear and straightforward
– lyx
Nov 7 '14 at 14:50

Take a collar neighborhood, ie a diffeomorphism $\partial M \times [0,1) \to U \subset M$ where $U$ is a neighborhood of $\partial M$ and this diffeomorphism takes $\partial M \times 0$ identical to the boundary of $M$. Hence after collapsing $M/M-U$ you get actually the cone $C$ over $\partial M$, which has the projection and you define $M \to M/M-U \to [0,1]$ which is onto.