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Let $M$ be a smooth manifold.

Then, is any open covering $\{U_i\}$ of $M$ is contractible?, i.e. each $U_i$ is contractible?, where $\{(U_i, \phi_a)\}$ is an atlas of $M$, i.e. $M = \bigcup_i U_i$ and $\phi_a:U_a \mapsto \phi_a(U_a)$ is a homeomorphic.

math student
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  • Not unless it's part of the definition you're using. For example, let $M=\Bbb R^2\setminus{0}$ and just take the open covering consisting of a single open set $U_1=M$ with $\phi_1$ being the identity function. – Greg Martin Feb 15 '20 at 07:06
  • If M is simply-connected, then does the argument hold? – math student Feb 15 '20 at 07:13

1 Answers1

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Nothing (other than perhaps the particular definition you are using) is stopping you from covering your manifold with copies of $\Bbb R^n-\{0\}$. This can be done to any manifold.

For instance, if you have a covering using only copies of $\Bbb R^n$, then for each chart map $\phi:\Bbb R^n\to M$ you can split it into two charts by restricting $\phi$. One map $\Bbb R^n-\{0\}\to M$ and one map $\Bbb R^n-\{p\}\to M$ for some non-origin $p$. Doing this for each chart yields an atlas with no contractible $U_i$.

Of course, if your definition requires all charts to use $\Bbb R^n$ as domain, then that's a different story entirely.

Arthur
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  • However, there exists an atlas such that each set is contractible at least in the case of any manifolds? – math student Feb 15 '20 at 08:14
  • @mathstudent Yes, any manifold has an atlas with all charts contractible. A similar construction to mine can be made where you take any non-contractible chart, and split it into contractible ones. – Arthur Feb 15 '20 at 08:17
  • We constract {Ui} by the following: ϕi:Ui→Bi, where Bi is the open unit ball in Rn. Since Bi is contractible, Ui=ϕi(Bi) is also contractible. Hence, {Ui} is constactible.

    This argument is true?

     – math student Feb 15 '20 at 08:34
  • @mathstudent No, that's not right. If that were right, consider how many non-contractible things you know about that can be put inside some ball of some dimension and suddenly become contractible. Like the circle. No, do not put an intermediate $B_i$ between $U_i$ and the manifold. Instead, see how I made a non-contractible cover out of a contractible cover (split each $U_i$ you don't want into smaller $U_i$'s you do want), and apply basically the same technique. – Arthur Feb 16 '20 at 07:26