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I'm wondering if it's possible to represent a class of in one point $x \in M$ smooth maps $f:M \to N$ by smooth maps on an open nbhd. $U(x) \subset M$, s.t. the equivalence relation is given by equality on an open nbhd. (not only on the point). I'm asking that, because I found it in the definition of smooth Alexander-Spanier cohomology here: http://arxiv.org/pdf/math.GR/0402303.pdf (Appendix A, Def. A1)

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1 Answer 1

So, your equivalence is: $f\sim g$ if $f=g$ on a neighborhood of $x$. If $f$ is smooth at $x$ only (but not on any neighborhood of $x$) and $f\sim g$, then $g$ can't be smooth on any neighborhood of $x$, either. Two things can't be equal if one is smooth and the other isn't.

The following is true: if $f$ is smooth at $x$, then there exists a map $g$, smooth in a neighborhood of $x$, such that $f(x)=g(x)$ and the derivatives of $f$ and $g$ at $x$ are equal (of all orders). Proof:

  1. Since the issue is local, move to Euclidean space
  2. Smoothness is coordinate-wise with respect to the target, so we can focus on real-valued $f$.
  3. By Borel's theorem, there is a smooth function whose partial derivatives are the same as the partial derivatives of $f$, to all orders.
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thx, so in general it is possible for a map to be smooth in just one point and not in any neighbourhood, right? –  user83496 Nov 25 '13 at 10:27
@user83496 Depending on your interpretation of "smooth at a point". If it requires the first derivative to exist, then take $f(x)=x^2$ if $x$ is rational, and $f(x)=0$ if $f$ is irrational. This $f$ is differentiable at $0$ and only there. On the other hand, if "smooth" requires derivatives of all orders to exist, then they automatically exist in some neighborhood, since the existence of $k$th derivative at a point requires $(k-1)$th derivative to exist in a neighborhood of that point. –  Post No Bulls Dec 7 '13 at 21:30

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