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Consider a connected manifold $M$ and two distinct points $p,q$ on the manifold. Is it true that there exists a chart containing both? How can one prove this result?

OBS: I've seen an answer on the internet which relied on taking a path from $p$ to $q$, using a tubular neighbourhood and arguing that two charts on $p$ and $q$ together with the neighbourhood would give something diffeomorphic to $\mathbb{R}^n$, but this isn't clear to me at all.

Aloizio Macedo
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  • We can have S^2 with 6 charts(all 6 hemispheres) but north pole and south pole are not in any single chart. Maybe this is true then if we have a maximal atlas? – Aman Dec 26 '15 at 04:22
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    @Aman By "exists a chart" I'm asking if "there exists an open set around $p,q$ which is diffeomorphic to an open subset of $\mathbb{R}^n$", not fixing an atlas and asking if there exists a chart in that specific atlas, of course. Sorry if this was not clear. – Aloizio Macedo Dec 26 '15 at 04:29
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    Your sketch of a proof isn't convincing to me either, at least without some restrictions on the path. A tubular neighborhood of a simple loop is not simply connected. – Matt Samuel Dec 26 '15 at 04:57
  • @MattSamuel Yes, that was one issue I thought about... but a loop can be "canceled" when going from a point to a different one. So I thought maybe the argument could be made to work, but I don't know how. Ipsis litteris, the argument given is this: "Then take a chart containing the first point, take a chart containing the second point, and take a path connecting the two points. Consider a small contractible neighborhood of the path.

    The union of the two charts with this small neighborhood of the path is then clearly diffeomorphic to R^n"

     – Aloizio Macedo Dec 26 '15 at 05:02
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    @AloizioMacedo That's the hand-waviest proof I've ever seen. For one, I'm pretty sure there are contractible open subsets of $\mathbb{R}^n$ that are not homeomorphic to $\mathbb{R}^n$. – Matt Samuel Dec 26 '15 at 05:05
  • @MattSamuel I aggree with you completely. – Aloizio Macedo Dec 26 '15 at 05:09
  • If we allow self intersections, a path can even pass through every point of the manifold if it is compact. Despite my wealth of counterexamples though I don't think I'll be of much help with the proof. Good luck. – Matt Samuel Dec 26 '15 at 05:16
  • Yes, you pick a path from one to the other and take a tubular neighborhood. The submanifold here is the unit interval, so any bundle over it is trivial, including the normal bundle. If you're jncomfortable with this, pick a simple loop that includes the two points and take a tubular neighborhood; this is diffeomorphic to the normal bundle of the loop; now pick some other point $t$ in the loop and delete $p^{-1}(t)$ to get the desired chart. –  Dec 27 '15 at 02:40
  • @MattS: What's worse, there are contractible manifolds that aren't $\Bbb R^n$ that can nonetheless be written as the Union of two open sets, each diffeomorphic to $\Bbb R^n$. –  Dec 27 '15 at 02:41
  • @MikeMiller Thanks! That makes sense to me : ).

    But this is not what the answer given is arguing, correct? For instance, you take no two charts and "glue" them: you take directly a chart through the tubular neighbourhood.

    Also, could you please write that as an answer?

     – Aloizio Macedo Dec 27 '15 at 03:24
  • Unfortunately I don't have computer access for about a week, and don't really want to write an answer version of this on my phone. I find the given answer a little light on the details - not obvious to me how they would fill them in. –  Dec 27 '15 at 03:26

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