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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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: