Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
See also…. – levap Nov 1 '12 at 22:40
up vote 2 down vote accepted

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.

share|cite|improve this answer
thank you very much, sorry for taking so long to accept – roo Nov 6 '12 at 20:40
no worries-glad to help – uncookedfalcon Nov 14 '12 at 0:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.