A disjoint cover for a compact manifold Suppose $M$ is a smooth manifold that is also compact. So, if there is some open cover then there is a finite subcover that covers $M$ by compactness. Now my question is, in general, can I shrink this finite subcover to make it a disjoint finite subcover? Or is does there always exist a disjoint finite subcover for a compact smooth manifold? Thank you.
 A: No, this is almost never possible.  If a topological space $M$ is a disjoint union of open sets $U_i$, then each $U_i$ is also closed, since its complement is the union of the sets $U_j$ for $j\neq i$.  Thus each $U_i$ is clopen, and so $M$ is disconnected as long as there is more than one $U_i$ that is nonempty.
So if $M$ is any connected manifold, it admits no finite cover by disjoint nonempty open sets at all other than the trivial cover (consisting of just $M$ itself).
A: Since $M$ is a manifold, you can choose the cover $\{U_i\}$ so that each $U_i$ is homeomorphic to $\Bbb R^n$. In particular each $U_i$ is connected. If you could get a finite subcover $U_1,\dots,U_m$, where $m>1$, such that the $U_i$ are disjoint, then $M$ would be disconnected as a topological space. Hence you cannot do this if $M$ is connected.
More generally, if $M$ has $k$ connected components, then you can't find a disjoint subcover consisting of more than $k$ subsets, and you can only do this if each connected component was in the original open cover.
A: No. Take $M$ to be the circle $S^1$. Any atlas on $M$ must contain at least $2$ charts (otherwise $M$ would be homeomorphic to a subset of the real line $\Bbb{R}$, which is not the case).
