# Jacobian Determinant of Mobius Band

Where is my logic wrong in the paragraph below?

Consider the mobius band in the context of differential geometry. It isn't orientable so there must be two chart transition maps from a differentiable atlas for the mobius band such that they have differing sign for the Jacobian determinant. But the determinant is a continuous map and since its range in this case includes $0$ there must be a point $p=det^{-1}(0)$ on the mobius band such that in a nbhd of $p$ any two charts with intersecting domain are not admissible to the atlas for the mobius band. This would mean any atlas for a mobius band can not completely cover it, right?

EDIT: Sorry, to the people that read this prior to my edits. But, I believe I've formulated my question correctly at this point and hopefully there will be no more edits.

• $f:(-3,-1)\cup(1,3)\to \mathbb{R},x\to x$ is continuous. $f(-2)<0,f(2)>0$ and however $f$ doesn't take the value $0.$ Do you see the reason? – mfl Nov 28 '16 at 21:00
• The domain of $f$ isn't connected. And as Andrew pointed out, in the case of overlapping charts they need not be. – Bob Nov 28 '16 at 21:09
• Exactly that's the reason. But even assuming each overlap is connected, the problem is that if the manifold is non-orientable then you can't make any Jacobian positive. If you change a chart to make it positive then some other overlap becomes negative. As @Bob says, non-orientability is not a local property. – mfl Nov 28 '16 at 21:14
• Do you mean positive everywhere? – Bob Nov 28 '16 at 21:16
• I mean all overlap Jacobians positive. – mfl Nov 28 '16 at 21:18