For any $n$ dimensional closed manifold $M^n$, can we find an open covering $\{U_i\}_{i\in[2^n]}$ such that $M=\cup U_i$ and each $U_i\cong \mathbb R^n$? How about complex manifolds (replacing $\mathbb R^n$ by $\mathbb C^n$)?
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$\begingroup$ In the smooth and PL cases, $n+1$ (rather than $2^n$) suffices, assuming the manifold is connected. $\endgroup$– Kevin WalkerCommented Jun 24, 2015 at 16:36
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$\begingroup$ Thank you! Could you show me the reason? $\endgroup$– A.T.SaakiCommented Jun 24, 2015 at 16:38
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1 Answer
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In the smooth, PL or topological category, you can cover a $n$-dimensional connected manifold by $n+1$ charts. Furthermore, for $k \leq n-3$ a $k$-connected $n$-dimensional manifold can be covered by at most $\lceil (n+1)/(k+1) \rceil$ charts. The reference for both of these statements is Luft's "Covering of manifolds with open cells". The proof of the first statement is elementary, the latter requires engulfing.
