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Hello everyone, I was wondering if anyone knew how to prove that the map from $C^{\infty}(M)$ to $\xi (p)$, that is, from the infinitely differentiable functions on a manifold M to the space of (once)-differentiable function germs, where the map is associating to each f in $C^{\infty}(M)$ its class in $\xi (p)$ is onto. By the way, since you ask, the reason I'm interested in this is because its a question that WAS on my final for differential topology, I've tried to work it out since then but no luck so far, this is not homework it's just curiosity now, hope its ok ill have to check the post regultaions, sorry, if not just tell me and i'll delete the question...

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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Willie Wong Nov 28 '13 at 12:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo
If this question can be reworded to fit the rules in the help center, please edit the question.

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1. Your question's well-formed, but nearly impossible to parse. Edit? 2. I don't know any differential geometry, but is this homework? It reads kind of like it might be, and it's gotten downvoted. If it's homework, don't bother editing. –  Harrison Brown Dec 6 '09 at 0:37
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Assuming you mean $C^{\infty}$ germs, then this is basically the existence of a cutoff function. –  Akhil Mathew Dec 6 '09 at 2:12
    
Upvoted since reasonableness should be encouraged. :) –  Harrison Brown Dec 6 '09 at 2:21

1 Answer 1

I think your question is mis-stated, because this map is not onto:

There is no smooth function on the reals whose germ at $0$ is the germ of $x|x|$, a once-differentiable function germ.

Assuming this is not what you wanted, please restate the question, and include an explanation of why you are interested in it ;)

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Thanks, I'll look into your counterexample! –  user2333 Dec 6 '09 at 1:17

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