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I received the following question. It's also in Chapter 2 of Spivak's book on Differential Geometry, Volume 1.

Given a smooth connected manifold $M$, prove that between any two points $p,q \in M$, there exists a smooth curve $c : [0,1] \to M$ such that $c(0) = p, c(1) = q$. Furthermore, $c$ can be chosen to be one-one.

How I propose to do this problem is as follows : I would like to find a common chart(open neighbourhood homeomorphic to $\mathbb R^n$) containing $p$ and $q$. This is done as follows : take a chart $C_p$ containing $p$, and take a path connecting $p$ and $q$ (This must exist by the definition of a manifold as a second countable Hausdorff space, but I'm unable to say why). Around this path, consider a contractible neighbourhood $N$ (Can I do this?). Now, $N \cup C_p$ is a chart containing $p$ and $q$.

Therefore, since this neighbourhood is homeomorphic to $\mathbb R^n$, I can find a 1-1 smooth path connecting the images of $p$ and $q$ in $\mathbb R^n$ (this is a standard property of $\mathbb R^n$, easily provable), and pulling this back into $M$ gives me the $c$ I desire, which is also $1-1$.

Is this correct?

EDIT : This question is different from a proposed duplicate, given in the comment below, since here I am attempting the question in a way different from how it been done in the duplicate, and furthermore I am asking for a critique of my attempts, not the answer itself. When I am confident, I will answer this question myself and close it.

Sarvesh Ravichandran Iyer
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    Charts are local but the points $p$ and $q$ may not be close together. Maybe you should fix one point, $p$, and consider the collection of points of $M$ that can be connected to $p$ by means of a path. – user99163 Aug 19 '17 at 05:46
  • @user99163 Thank you. Can you explain the meaning of $p$ and $q$ not being close together? Does it mean there is no path joining the two? – Sarvesh Ravichandran Iyer Aug 19 '17 at 05:54
  • I have another question : is this kind of smooth-path connectivity a transitive property? For then, I have an answer, although I cannot guarantee that $c$ is 1-1 there. – Sarvesh Ravichandran Iyer Aug 19 '17 at 05:58
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    The reason I mention that $p$ and $q$ may not be close together is because you suggested finding a chart containing both points. This may not be possible as the definition of a manifold only gives a chart existing locally. – user99163 Aug 19 '17 at 06:03
  • @user99163 You are correct, in that the definition only asserts the existence of a chart locally. However, to see the flaw in my argument, I suggest you come up with a counterexample i.e. a smooth path connected manifold in which there exist two points which cannot be together in a coordinate chart. In that case, I will explore the angle you have suggested i.e. the collection of points that can be connected to $p$ by a smooth path. – Sarvesh Ravichandran Iyer Aug 19 '17 at 06:07
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    Any 2 points can be contained in a single chart. But the construction of that chart is probably more difficult than finding a smooth path. @user99163 –  Aug 19 '17 at 06:16
  • @JohnMa Thank you. What is possibly wrong in my construction? Is one of the italic statements untrue for a manifold of the kind defined? – Sarvesh Ravichandran Iyer Aug 19 '17 at 06:17
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    I was wondering what your contractible nbd means here, if the path is e.g. space filling –  Aug 19 '17 at 06:19
  • @JohnMa Ouch. I did not think about that! How is this approach : If we can prove that 1-1 smooth-path connectivity is a equivalence relation, then since all the equivalence classes are open, and their union is a connected set, it follows that the whole manifold is an equivalence class, which is the desired result. Can we do this? – Sarvesh Ravichandran Iyer Aug 19 '17 at 06:22
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    sounds good. Indeed it is easy to show that's an equivalence relation. –  Aug 19 '17 at 06:27
  • @JohnMa Can you give a few hints to that in an answer? Because I am unable to get transitivity. How would you assert that the usual "concatenation of paths" operator preserves smoothness? – Sarvesh Ravichandran Iyer Aug 19 '17 at 06:29
  • How about a partition of unity,to go from local to global? .. –  Aug 19 '17 at 06:31
  • @ChrisCuster Thank you for the reply. Can you explain what you mean? More specifically, I know that there exists a partition of unity on $M$, but how would I use this to provide a smooth path? Can I do this : provide an arbitrary path between $p$ and $q$, and then smooth it out with bump functions here and there? I need you to help me make this rigorous. Also, what about injectivity of the path? – Sarvesh Ravichandran Iyer Aug 19 '17 at 06:33
  • @MoisheCohen I am happy to see answers. However, I want to know if the above approach that I have suggested in my comment, and in the body of my question, is also viable. I believe that my attempt is different from the answers in the duplicate question, therefore the questions are different, since I am asking for a critique of my attempts. That's why I would like to see in which direction this conversation goes. But thank you for the link. – Sarvesh Ravichandran Iyer Aug 19 '17 at 09:38
  • @астонвіллаолофмэллбэрг: Your proposed approach can be made to work, but it is considerably harder than the problem you are trying to solve. In particular, you first prove the existence of a smooth path and only then the existence of a common open subset containing the two points and diffeomorphic to an open ball. – Moishe Kohan Aug 19 '17 at 09:52
  • @MoisheCohen Ok, I think I have understood. I will answer this question myself and close it as soon as possible. Thank you – Sarvesh Ravichandran Iyer Aug 19 '17 at 09:55

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