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'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: (Appendix A, Def. A1)

share|cite|improve this question

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.
share|cite|improve this answer
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

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.