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I came across this Question: Atlas on a smooth manifold that contains 2 charts in which Professor Lee commented that this Proposition is true only if the Intersection of the two Maps is connected, so I've been trying to prove the following:

Let $M$ be a smooth Manifold, covered by an Atlas $\mathcal{A}$ containing two charts $\phi,\psi$ and $M= U_\phi \cup V_\psi$. Show that $M$ is orientable if $U_\phi \cap V_\psi$ is connected.

So in other words I will have to prove that,for $\tau = \phi \circ \psi^{-1}$, the following holds: $det(Jac(\tau))>0$.

First of all I tried showing that $det(Jac(\tau)) > 0 \vee det(Jac(\tau)) < 0$ holds. I reasoned : Since $M$ is smooth it follows that $\tau$ is a Diffeomorphism, which implies $\tau$ is continuous. Suppose $\tau$ would take values both bigger and smaller than $0$, this would imply that there is a point at which $\tau$ becomes $0$ and therefore would no longer be a Diffeomorphism. The next step would be to show that $det(Jac(\tau))> 0$ however I do not see why this is indeed the case.

Is my reasoning so far correct and if so how do I continue?

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scathefire
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If $\det\bigl(\operatorname{Jac}(\tau)\bigr) < 0$, you can re-orient one of your charts (for example, by precomposing with a linear diffeomorphism that multplies the first coordinate by $-1$).

Andrew D. Hwang
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  • Thanks for the answer first of all. But I do not see how this is only possible because the Intersection is connected? – scathefire Jun 20 '14 at 02:18
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    @scathefire: Your argument, which shows that the determinant of the Jacobian has constant sign on the overlap of your charts, uses connectivity of the overlap. (If the overlap were disconnected, the determinant could take opposite signs on distinct components.) – Andrew D. Hwang Jun 20 '14 at 02:36
  • Of course! Thank you very much. This solution seemed so obvious. :) – scathefire Jun 20 '14 at 13:58