How to prove this problem about orientation on manifold?

Question

Let $M$ be a smooth manifold. If $M$ can be covered by two charts $(U,\varphi)$ and $(V,\phi)$ in such a way that the intersection $U\cap V$ is a connected subset of $M$, then $M$ is orientable.

This is an example in Do Carmo's Riemannian Geometry, but how to prove it? I have no ideas. Can anyone help me?

Answer 1

do Carmo proves it. Quotation from Example 4.5:

Indeed, since the determinant of the differential of the coordinate change is $\ne 0$, it does not change sign in $V_1 \cap V_2$; if it is negative at a single point, it suffices to change the sign of one of the coordinates to make it positive at that point, hence on $V_1 \cap V_2$.

Here are some clarifying remarks. Let us adopt all notation from do Carmo. The concept of orientation is defined in definition 4.4. There are systems of coordinates $x_i : U_i \to V_i$, $U_ i \subset \mathbb R^n$ open and $V_i \subset M$ open, which belong to the differentiable structure of $M$. Then $\{ x_1, x_2 \}$ is a smooth atlas for $M$.

Note that most authors work with charts instead of systems of coordinates. A chart is nothing else than the inverse $x^{-1} : V \to U$ of a system of coordinates $x : U \to V$. In that sense do Carmo has an unusual approach, but clearly his approach is equivalent to the standard one.

It is a philosophical question whether the empty set is regarded as connected. There are authors who do and authors who don't. Anyway, if the intersection is empty, then the assertion is trivial.

If $V_1 \cap V_2 \ne \emptyset$, then the coordinate change $\gamma = x_2^{-1} \circ x_1 : O \to O'$ is a smooth map between connected open subsets of $\mathbb R^n$. Let $J(\gamma,\xi)$ be the Jacobian of $\gamma$ in $\xi \in O$. The determinant $\det J(\gamma,-) : O \to \mathbb R$ is continuous (since all partial derivatives of $\gamma$ are continuous). Hence it cannot change sign on $O$ because $O$ is connected. Pick $\xi \in O$. If $\det J(\gamma,\xi) > 0$, we are done. If $\det J(\gamma,\xi) < 0$, replace $x_1$ by the system of coordinates $x_1': R(U_1) \to V_1$, where $R :\mathbb R^n \to \mathbb R^n$ is a reflection (e.g. $R(x_1,\ldots,x_n) = (-x_1,x_2,\ldots,x_n)$). Note that $x'_1$ also belongs to the the differentiable structure of $M$ because $R$ is a diffeomorphism. Then $\{ x'_1, x_2 \}$ is a new smooth atlas for $M$, but now $\gamma' = x_2^{-1} \circ x_1'$ has positive determinant at $\xi$ and we are done again.

Answer 2

First, by Van Kampen's theorem, $M$ is simply connected.

Second, any simply connected smooth manifold is orientable as explained here.

Answer 3

You need to show that $\det \mathcal J(\phi^{-1}\circ \psi)$ does not change sign on $U \cap V$. But this is clear, because $\phi^{-1}\circ \psi$ is a diffeomorphism and $U \cap V$ is connected.