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I'm reading Warner. "Foundations of Differentiable Manifolds and Lie Groups." p. 138. I don't get the statement in the definition of orientable manifolds.

4.1 Definitions $\;$ (the preface omitted) Let $M$ be a connected differentiable manifold of dimension $n$. We chall call $M$ orientable if it is possible to choose in a consistent way an orientation on $M_m^*$ for each $m\in M$. More precisely, let $O$ be the "$0$-section" of the exterior $n$-bundle $\Lambda_n^*(M)$; that is, $$O = \cup_{m\in M} \{0\in\Lambda_n(M_m^*)\}.$$ Then since each $\Lambda_n(M_m^*)-\{0\}$ has exactly two components, it follows easily that $\Lambda_n^*(M)-O$ has at most two components. We say that $M$ is orientable if $\Lambda_n^*(M)-O$ has two components; and if $M$ is orientable, an orientation on $M$ is a choice of one of the two components of $\Lambda_n^*(M)-O$. (the rest omitted)

Why does $\Lambda_n^*(M)-O$ have at most two components?

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  • $\begingroup$ Try 3 points p,q and r in the fibers above m,n and o which are points of M. M is connected so you can give paths between all 3. Try to lift those paths to paths between p,q and r. See if it is possible for none of them to be connected to any other. $\endgroup$ – AHusain Feb 22 at 3:34
  • $\begingroup$ @AHusain Okay, I see that if $\Lambda_n^*(M)-O$ has three or more components, then there should be three points $p, q, r\in\Lambda_n^*(M)-O$ such that there is no paths connecting any two points in $\{p, q, r\}$. But how do I lift the paths between $m, n, o$? $\endgroup$ – zxcv Feb 22 at 6:39
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Let there be 3 points p,q and r in the fibers above s,t and u which are points of M. M is connected so you can give paths between all 3.

There is a path from s to t. This lifts to a path from p to something projecting to t. Call this w. You can build this path by using the local triviality over small neighborhoods $U \subset M$. If we get something in the same component of $\Lambda_n (M_t^*)$. If this is in the same component as q then we could get a path from p to q.

So if we want p,q and r to all be in different components, we can assume that we landed in the wrong component.

Repeat this for t to u. But this time start from w instead of q. If we can connect from w to r, then we get a path from p to r. If we are in the wrong component for r, then do a fiberwise reflection. Now it is a path from $-w$ to the correct component for r. $-w$ is in the same component as q so this gives a path from q to r.

So there is no way for 3 separate path components.

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