This definition of a bump function is given in "Introduduction to Manifolds" by Loring W. Tu: Given a point $ p $ in a manifold $ M^n$, a bump function in $p$ supported in $V$ is any non-negative function $ \rho: M \rightarrow \mathbb{R} $ which is identically $ \mathbf{1} $ in some neighborhood of $ p $ with $ supp (\rho) \subset V $.
I understand the process of creating a $C^\infty$ bump function in $R$ and $R^n$, but when I move on to manifolds, the following happens:
Take $V$ a neighborhood of $p$ and $(\varphi,U)$ a chart over $p$ such that $V \subset U$. We have a $C^\infty$ bump function $\rho:\mathbb{R}^n \rightarrow \mathbb{R}$ in $\varphi(p)$ which is identically $\mathbf{1}$ in the closed ball $B[\varphi(p),a]$ supported in $B[\varphi(p),b]$ with $a<b<d(\varphi(p),\partial \varphi(V))$. And now, the composition $\rho \circ \varphi:U\rightarrow \mathbb{R}$ have domain $U$, not $M$ as I wish.
I'm forgetting to do something here? Like, consider the null extension of $\varphi$ over $M$… But, if this is the case, what guarantees me that the composition $\rho \circ \varphi:M\rightarrow \mathbb{R}$ will be differentiable? I know that $\varphi:U\rightarrow \varphi(U)$ a diffeomorphism, but I can't solve the domain issue.
Thanks!
Best Answer
You extend $f=\rho\circ\varphi$ by zero to get $F:M\to \Bbb R$.
Why is $F$ smooth? Let $K$ denote the support of $f$ inside $U$. Then $K$ is compact and so closed in $M$ (manifolds are Hausdorff). So $F$ is the patching of two smooth functions on open sets, namely $f$ on $U$, and the zero function on $M-K$.