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Let $V \rightarrow M$ be a smooth vector bundle over a smooth compact manifold $M$ (without boundary) and $X \subset M$ a smooth submanifold of $M$, that is not necessarily closed. Suppose $s: X \rightarrow V$ is a smooth section, that extends continuously to $\bar{X}$, the closure of $X$ inside $M$. Does there always exist a smooth section $\tilde{s}: M \rightarrow V$, that restricted to $\bar{X}$ is equal to $s$?

Note that, if I replaced the word smooth with continuous the answer is yes. We can construct such a section using a continuous partition of unity and using the compactness of $\bar{X}$. And similarly, if $X$ was compact then there exists a smooth extension of $s$, this time using a smooth partition of unity and also using the compactness of $X$.

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In this formulation, the answer in general is not. For instance, if you have a smooth function on $(0, + \infty)$ that extends continuously at $0$, then it fails to be smooth in a neighborhood of $0$ in general (this is the case of a trivial $\Bbb R$-bundle). You must assume in advance smooth extension. –  Daniele Zuddas Jun 14 '13 at 7:03
What is $M$ in your example? I said $M$ must be compact. And I also meant $M$ has no boundary (although I didn't say that clearly). –  Ritwik Jun 14 '13 at 7:06
@Ritwik: To flesh out Daniele Zuddas' example, you may take $M=S^1$ consisting of the unit vectors in $\mathbb{R}^2$, and $X$ to be the subset of $M$ consisting of vectors with positive first coordinate. –  Ricardo Andrade Jun 14 '13 at 9:05
Take $M$ to be the circle realized as the 1-point compactification of the real line, $X=(0,\infty)$ and the function $s(x)=\sqrt{x}e^{-x}$. –  Misha Jun 14 '13 at 9:50
Thank you for the examples. I think I understand the argument now and realize that the answer to my question can not possibly be yes as stated. –  Ritwik Jun 14 '13 at 10:18

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