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In the definition of Manifold, we are given countable atlases $(U_i, \psi_i)$, such that every point $p$ is contained in one such $U_i$ and $\psi_i(U_i)$ is homeomorphic to some subset in $\mathbb{R}^n$.

Now I want to reverse this process. Suppose I choose a countable subset in $\mathbb{R}^n$ and also assume I specify the individual intersection between any two subsets. Now I move around those subsets to satisfy the intersection rule I specified above, then will the resultant figure be a manifold. Also, can every manifold be deconstructed and then reconstructed in the way I mentioned.

Note:

(1) I am allowed to work in higher $\mathbb{R}^n$, as I figured the intersection that I specified might be unattainable in lower dimension but possible in a higher dimension (I will further clarify on this remark if needed. Let me know in the comment).

(2) The subsets are of the same dimension. For example, a circle has dimension 1 because it is essentially a length, the disk has dimension 2 because it is an area and so on.

henceproved
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  • Your question lacks context and rigour. But I think Exercise 5, page 14 of Hirsch, Differential Topology may be related. – Laz Jun 01 '19 at 16:42
  • @Laz Yeah that question does sort of look similar to my line of thinking. Out of curiosity, how did you recall this particular problem which was buried in a 200 page book. Seeing you, I am really feeling inadequate. – henceproved Jun 01 '19 at 17:05
  • I just remember virtually every exercise I solve. I guess that in this particular case is because I really like Hirsch's book. – Laz Jun 01 '19 at 17:21

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