# Connected intersection of a manifold and orientation

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.

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.