I want to extend $f$ locally in the intersection of each coordinate patch of $X$ with $Y$, (and set it to $0$ outside of $Y$) and then use a partition of unity to get a differentiable map that agrees with the old $f$ on $Y$, but the problem is that these intersections may not be open in $X$ since $Y$ is closed. Can this be fixed?
Given a closed submanifold $Y$ of $X$ and a $C^{\infty}$ map $f$ on $Y$, can $f$ be extended to $X$?
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0See also http://math.stackexchange.com/questions/192402/equivalence-of-two-definitions-of-differentiablitity-on-regular-surfaces/192512#192512. – 2012-11-01
1 Answers
Since $Y$ is a submanifold, for each $y \in Y$ we can find a neighborhood $U_y$ in $X$ such that in coordinates $Y \cap U_y$ looks like $\mathbb{R}^k \subseteq \mathbb{R}^n$.
Let's extend $f$ first to $\cup_y U_y$. Take a compactly supported partition of unity subordinate to the $U_y$, $\chi_{y'}$.
We can extend each $\chi_{y'} f$ as follows. In good coordinates, we have $\chi_{y'} f$ is a smooth function on $\mathbb{R}^k$ with compact support, to extend onto $\mathbb{R}^n$ we can do something like $f^e(x,y) = f(x) \rho(|y|)$, where $x \in \mathbb{R}^k, y \in \mathbb{R}^{n-k}$ and $\rho$ is say a bump function on $\mathbb{R}$ supported in $[-\epsilon, \epsilon]$ (which equals 1 at zero, to be an extension :p).
Each $\chi_{y'}f$ should extend to all of $X$ (via extension by zero), so just sum them up to get the desired extension.
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0no worries-glad to help – 2012-11-14