I'm now trying to prove Gauss' divergence theorem with a lecture note by Professor B.K. Driver. But I'm facing a problem with outward pointing condition on page 530.
Let $\Omega \subset \mathbb R^n$ be a manifold with boundary, which means, for each point $x_0 \in \Omega \setminus \Omega^{\circ}$, there exist $\varepsilon > 0$, an open set $D \subset \mathbb R^n$ containing $0$ and a diffeomorphism $\psi: D \to B_{\varepsilon}(p)$ such that
$$\psi( D \cap \{ y_n \geq 0 \} ) = \Omega \cap B_{\varepsilon}(x_0).$$
(For this definition, see page 528.)
Then, a vector $n(x_0) \in \mathbb R^n$ is said to be outward pointing if
$$\psi'(0) e_n \cdot n(x_0) < 0$$
equivalently, letting $\phi = \psi^{-1}$,
$$\phi'(x_0) n(x_0) \cdot e_n < 0.$$
The latter condition is natural since the condition means $\phi'(x_0) n(x_0)$ is a vector pointing $\{ y_n \geq 0 \}$ which corresponds to the outside of the $\Omega$.
But, the former condition is mysterious for me. Can anyone prove equivalency?