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Let $M$ be a surface and $(g_1,U_1), (g_2,U_2)$ two charts that cover M. Additionally, $U_1\cap U_2$ has two connected components $W_1$ and $W_2$. Such that the jacobian of the transition map has positive determinant in $W_1$ and negative in $W_2$.


Proof that $M$ is not orientable.

What I got so far: The transition map is a continious function and therefore $W_1$ and $W_2$ are open sets. How can I use that to prove that $M$ is not orientable ?

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    Doesn't it follow immediately from the fact that the Jacobian has positive determinant in $W_1$ and negative in $W_2$? Thus there is no way to extend either $g_1$ or $g_2$ to all of $U_1 \cup U_2 = M$ such that the sign of the determinant doesn't change? In terms of vector bundles (see here: https://en.wikipedia.org/wiki/Orientability#Orientation_of_vector_bundles) the fact that $U_1 \cap U_2$ has two components means that the structure group of the tangent bundle can't be made to be $GL^+(n,R)$, thus $M$ has to be non-orientable. (Since GL(n,R) has 2 components, but $GL^+(n,R)$ only has one) – 2017-01-09
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    $M$ is supposed to be connected? Otherwise it isn't true. – 2017-01-09
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    @juanarroyo: I forgot to mention that $M$ is a surface and not any manifold. It didnt explicitly say that $M$ is connected but if it doesnt work otherwise I should assume that it is in fact connected. – 2017-01-09
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    Your charts also need to be connected. What definition of orientation are you using? – 2017-01-09
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    @juanarroyo: I am using the definiton that for two charts $g_i:U_i\rightarrow M,i=1,2$ the jacobian has positive determinant on $U_1 \cap U_2$ – 2017-01-09

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First of all consider the surface $S=(-3,3)\times\mathbb{R}$. Let $U_1=(-3,-1)\times\mathbb{R}\ \cup\ (1,3)\times\mathbb{R}$ and $U_2=(-2,2)\times\mathbb{R}$. Let the coordinate function on $U_1$ be the identity on the left component, and on the right component send $(x,y)\to(x,-y)$. For $U_2$ use the identity. You can check that this cover satisfies all the conditions in your problem but $S$ is orientable.

If you require $U_1$ and $U_2$ to be connected then it is true (this automatically requires $M$ to be connected). By continuity the transition map for any two charts must have either strictly positive or negative Jacobian in a small neighborhood of any point. So if $M$ is orientable and $U$ is connected coordinate neighborhood, by connectedness $(U,\phi)$ has a well defined orientation.

Suppose $U_1$ in your problem is positively oriented. Then the transition function Jacobian at $W_1$ says that $U_2$ is positively oriented, but the transition function at $W_2$ says the opposite.