If exist orientable atlas then exist nondegenerate n-form

Let $X$ is $n$-manifold with smooth atlas $\phi_\alpha : U_\alpha \to \mathbb{R}^n$ such that, for every $a \in U_\alpha \cap U_\beta$, $\phi_{\alpha \beta}' (\phi(a)) > 0$. I am need to proof, that there exist $n$-form $\omega \in \bigwedge^n T^* X$ such that $\forall a \in X$ $\omega(a)$ nondegerate on $T_a X$.

I am tried to define $\omega$ on every chart $U_\alpha$ in such way: $$\omega(a)(\gamma_1'(0),...,\gamma_n'(0)) = \det((\phi_\alpha \circ \gamma_1)'(0),...,(\phi_\alpha \circ \gamma_n)'(0))$$ but this definition isn't agreed on intersections, because: $$\det((\phi_\alpha \circ \gamma_1)'(0),...,(\phi_\alpha \circ \gamma_n)'(0)) = \det(\phi_{\beta \alpha}' (\phi_\beta(a)) \det ((\phi_\beta \circ \gamma_1)' (0),...,(\phi_\beta \circ \gamma_n)') \neq \det ((\phi_\beta \circ \gamma_1)' (0),...,(\phi_\beta \circ \gamma_n)'(0))$$ for some $a \in U_\alpha \cap U_\beta$ so, I am a little confused, how I can define that form? Thank you!

• Are you familiar with partitions of unity? – Eric Wofsey Nov 25 '16 at 3:40
• Yes, but I don't understand how can I use it for that problem. – kp9r4d Nov 25 '16 at 3:46
• Hint: Observe that non-degenerate simply means non-zero in this case, so you can glue using partitions of unity. – Andreas Cap Nov 25 '16 at 7:53
• Sorry, but I dont understand how I need to use partitions of unity for gluing it. Can you describe your answer more widely? – kp9r4d Dec 5 '16 at 1:31