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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 $f(x) \ne f(y)$?

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What's your target space? – AQP Mar 5 '12 at 7:45
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. – Zhen Lin Mar 5 '12 at 7:48
@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? – AQP Mar 5 '12 at 7:58
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. – Neal Mar 5 '12 at 8:33
Not assuming paracompactness seems odd; I've always seen it built into the definition of a smooth manifold. – Neal Mar 5 '12 at 8:34

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