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Suppose that $M$ is a path connected smooth manifold, so any two points $p,q\in M$ can be joined with a continuous curve on $M$. Is it true that any two points can be joined with a smooth (I mean $C^{\infty}$) curve on $M$?

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Use finitely many local charts and approximate the given curve with a smooth one locally. – Hans Engler Nov 23 '12 at 17:36
Do I need partition of unity? – Dubious Nov 23 '12 at 17:44
up vote 3 down vote accepted

No need for a partition of unity. Let $\gamma$ be a path from $p$ to $q$, not necessarily smooth. Take finitely many local charts that cover the compact image of $\gamma([0,1])$. Call these local charts $U_1, \cdots, U_n$. Let us have ordered these charts so that the overlap between $U_{i}$ and $U_{i+1}$ is nonempty, but rather some point $r_i$. Pragmatically, you can do this by pulling back the $U_i$ to a cover of $[0,1]$ by the $\gamma^{-1}(U_i)$, and then ordering them by left-endpoints. Now, in each of these local charts the manifold looks like $\mathbb{R}^n$, so we can find a smooth path from $p$ to $r_1$ in $U_1$, then a smooth path from $r_1$ to $r_2$ in $U_2$, etc. Connecting all these smooth paths gives us a smooth path from $p$ to $q$.

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if the domain of the curve is $(0,1)$ then the image is not compact, but the proof works. Is it right? – Dubious Nov 23 '12 at 20:59
You generally need endpoints so that one goes to $p$ and the other goes to $q$. That's the definition of a continuous path, no? – Isaac Solomon Nov 23 '12 at 21:01
yes, you're right! I'm sorry. – Dubious Nov 23 '12 at 21:04
You also might specify that within a neighborhood of the intersection of two paths the restriction of the path to these two segments can also be made smooth by this neighborhood also being diffeomorphic to $R^n$ – Squirtle Apr 24 '13 at 14:39
How do we know that when we patch these paths together the result will remain smooth at the $r_i$? – Eric Auld Aug 29 '15 at 6:54

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