Manifold is not orientable

Question

Let $M$ be a manifold of dimension $n$ such that there exist two charts $(U_a,\phi_a)$ and $(U_b,\phi_b)$ such that $U_a,U_b$ are connected and $U_a\cap U_b\ne\emptyset$. Moreover the transition function $\phi_{ab}$ neither preserves the orientation nor reverses the orientation. I have to show that $M$ is not orientable.

A manifold is orientable if it admits an oriented atlas, i.e. an atlas in which all the transition functions have Jacobian with positive determinant.

Answer 1

Here's a fairly detailed sketch:

Let $M$ be an $n$-manifold. Suppose, contrapositively, that $M$ is orientable, and fix a maximal oriented atlas, i.e., a maximal atlas for which all the transition maps have Jacobian with positive determinant.

Lemma 1: If $(U, \phi)$ is a chart and $U$ is connected, then $(U, \phi)$ is either compatible with the orientation, or anti-compatible with the orientation.

Proof: Let $(V, \psi)$ be an arbitrary oriented chart with $U \cap V$ non-empty. The sign of the Jacobian of $\psi \circ \phi^{-1}$ in $\phi(U \cap V)$ is independent of $\psi$ because $(V, \psi)$ is selected from an oriented atlas.

Let $U^{+}$ denote the set of points of $U$ for which the Jacobian $\psi \circ \phi^{-1}$ is positive, and let $U^{-}$ denote the set of points of $U$ for which the Jacobian $\psi \circ \phi^{-1}$ is negative. The sets $U^{\pm}$ are obviously disjoint, each is open, and their union is $U$. Since $U$ is connected, one set is empty: Either $U = U^{+}$ or $U = U^{-}$.

Lemma 2: If $(U, \phi)$ is an anti-compatible chart with component functions $(\phi^{1}, \dots, \phi^{n})$, then the chart $(U, \bar{\phi})$ defined by $\bar{\phi} = (\phi^{1}, \dots, \phi^{n-1}, -\phi^{n})$ is compatible, and conversely.

Proof: The linear transformation $T(x^{1}, \dots, x^{n}) = (x^{1}, \dots, x^{n-1}, -x^{n})$ has determinant $-1$.

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Let $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ be arbitrary charts with $U_{a}$ and $U_{b}$ connected, and with $U_{a} \cap U_{b}$ non-empty. If necessary, replace $\phi_{a}$ by $\bar{\phi}_{a}$ to get a compatible chart, and similarly for $\phi_{b}$. Since each chart is compatible with the oriented atlas, the transition map has positive Jacobian. It follows that the transition map $\phi_{ab}$ between the original charts has Jacobian of constant sign, i.e., either preserves orientation or reverses orientation.

Contrapositively, if there exist connected, overlapping charts $(U_{a}, \phi_{a})$ and $(U_{b}, \phi_{b})$ whose transition map $\phi_{ab}$ neither preserves nor reverses orientation, then $M$ is not orientable.