# The orientation induced on the boundary of a manifold.

I just learned about the notion of orientability of a manifold which is difficult and abstract for me. If we consider all basis of a vector space, the matrix that transforms one basis in another basis can have either positive or negative determinant. So we can partition the set of basis in two equivalent classes where in each equivalent class the determinant of the matrix of change of basis in an equivalent class is positive. So the pointwise orientation of a manifold is the choice for each $$p \in M$$ of one of the two equivalence classes. Then an orientation is a pointwise orientation which is continuous (i.e. for each point $$p$$, there is an open $$U$$ that contains $$p$$ and a smooth vector field $$(X_1,...,X_n)$$ such that the equivalent class $$[(X_{1p},...,X_{np})]$$ gives the pointwise orientation at $$p$$. Now if we consider a manifold $$M$$ with boundary, given an orientation $$\mathcal{O}$$ on it (i.e. $$\mathcal{O}_p$$ is one of the two equivalent classes for each $$p$$), it defines an orientation on the boundary : In each point $$p \in \partial M$$, apparently we can choose $$\mathcal{O}_p=[(Y_{1p},...,Y_{np})]$$ where $$Y_{1p}$$ is an outward vector(i.e. the pushforward of a chart $$\phi_*Y_{1p}$$ has a négative component along $$\frac{\partial}{\partial x_n}\rvert_{\phi(p)}$$) and the $$Y_{2p},...,Y_{np}$$ are in $$Tp(\partial M)$$, i.e. the subspace of $$T_pM$$ genereated by $$(\frac{\partial}{\partial \phi_1}\rvert_p,...,\frac{\partial}{\partial \phi_{n-1}}\rvert_p)$$. Then $$[(Y_{2p},...,Y_{np})]$$ is an orientation on $$T_p\partial M$$.
Question : How do we know that such an $$(Y_{1p},...,Y_{np})$$ exists?
For $$[(Y_{2p},...,Y_{np})]$$ to be an orientation, we need that for $$p \in \partial M$$ there is an open $$U \in \partial M$$ that contains $$p$$ and vector fields $$(Z_2,...,Z_{n-1})$$ that are smooth and when evaluated in p give us the same orientation : $$[(Z_{2p},...,Z_{n-1p })]=[(Y_{2p},...,Y_{np})]$$,why is this true?