Extension of funcion

Question

I think its right but Im not sure. I have topological space (exactly manifold - second countable, Hausdorff, local Euclidean topological space) M, dim M=m. Let $A \subset M$ is closed set, dim A=n, $n<m$. Let $f:A\mapsto R$ is smooth / differentiable function. I argue, there exists smooth / differentiable extension to whole M. Is it correct? Thank you.

Answer 1

This is true, provided you are using correct definition of a smooth function on a closed subset $A$; dimension of $A$ is irrelevant. Smoothness of $f$ means that there exists a smooth extension $h$ of $f$ to an open neighborhood $U$ of $A$ in $M$. Let $V, W$ be another pair of open neighborhoods of $A$, such that $$ \bar{W}\subset V, \bar{V}\subset U. $$ Then, using a partition of unity, one constructs a smooth function $g: M\to {\mathbb R}$ which equals $1$ on $V$ and $0$ on $M\setminus W$. Consider now the product $hg$ and extend it by zero on $M\setminus U$. This is your extension.