1
$\begingroup$

If $M$ is a Hausdorff $n$-manifold (without further assumption like paracompactness), given $x,y$ in $M$ is there a smooth function $f$ such that $f(x) \ne f(y)$?

  • 0
    What's your target space?2012-03-05
  • 0
    First, assume $\dim M > 0$ (otherwise this is either false or easy). Use bump functions, and divide into cases depending on whether $x$ and $y$ are in the same coordinate chart.2012-03-05
  • 0
    @Zhen Lin: so, assuming the target is $R^m$ (m is the dimension of M) , if x,y not in the same chart, you would use bump functions to extend the chart map?2012-03-05
  • 1
    The target should be $\mathbb{R}$, not $\mathbb{R}^n$. If $x,y$ not in the same chart, separate them by a function with support on chart $\ni x$ but not supported on chart $\ni y$. Otherwise, use a bump function on $\mathbb{R}^n$ to separate them.2012-03-05
  • 0
    Not assuming paracompactness seems odd; I've always seen it built into the definition of a smooth manifold.2012-03-05

0 Answers 0