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From Do Carmo's book (Riemannian Geometry, P. 19)

If M can be covered by two coordinate neighborhoods $V_1$ and $V_2$ in such a way that the intersection $V_1\cap V_2$ is connected, then $M$ is orientable. Indeed, since the determinant of the differential of the coordinate change is $\neq 0$, 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 $V_1\cap V_2$.

Why the determinant does not change sign in $V_1\cap V_2$ ? It surely related to the connected assumption, but I miss the argument.

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Let $\varphi_i:V_i\to\mathbb R^n$ be the given coordinate maps. Then the map $f:V_1\cap V_2\to\mathbb R$, given by $$f(x)= \det(d(\varphi_2^{-1}\varphi_1)_x)$$ is continuous, and $f(x)\neq0$ for all $x\in V_1\cap V_2$ (since $d(\varphi_2^{-1}\varphi_1)_x$ is always invertible). Since $V_1\cap V_2$ is assumed to be connected, the image of $f$ is connected, hence must lie in some connected subset of $\mathbb R\setminus\{0\}$. Thus the image of $f$ is a subset of either $(-\infty,0)$ or $(0,\infty)$, i.e., there is no sign change.

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  • $\begingroup$ How do we know that we won't change the sign of the determinant of the derivative of the other transition map to be negative if we change the sign of one of the coordinates? $\endgroup$
    – INQUISITOR
    Mar 19 at 2:15

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