Let $ D$ a distribution of dimension $n-1$. I'm trying to prove that $D$ is coorientable (i.e. the bundle $TM/D = \sqcup_{x \in M} T_xM/D_x$ has a global trivialization) if and only if $D$ is the kernel of a non-vanishing $1$-form $\alpha$.
MY ATTEMPT:
$\Leftarrow$ Let us suppose that $D = \text{Ker }\alpha = \bigsqcup_{x \in M}\text{Ker }\alpha(x)$. We will construct a global trivialization $\phi$ from $\alpha$. Since $\alpha(x):T_xM \rightarrow \mathbb R$ is surjective (as $\alpha$ is non-vanishing), $\overline{\alpha(x)}:T_xM/\text{Ker }\alpha(x) \rightarrow \mathbb R$ in a linear iso. Let us define $$\phi: TM/D \rightarrow M \times \mathbb R:(x, [X_x]) \mapsto (x, \overline{\alpha(x)}([X_x])).$$ We clearly see that $\pi_1 \circ \phi = \pi$ (where $\pi_1: M\times \mathbb R \rightarrow M$ is the projection on the first component and $\pi:TM/D \rightarrow M$ is the projection bundle of $TM/D$). Here it only remain to show that $\phi$ is a diffeo, here is my first question, see below. (1)
$\Rightarrow$ Since there exists a global trivialization $$\phi: TM/D \rightarrow M \times \mathbb R:(x, [X_x]) \mapsto (x, \phi_x([X_x])),$$ we can define $\alpha(x)(X_x) = \phi_x([X_x])$. We just have to prove that $\alpha$ is an $1$-form and that $D = \text{Ker }\alpha$. First, since $\phi$ is a trivialization, $\phi_x$ is a linear iso therefore $\alpha(x)$ is a linear functionnal on $T_xM$. Moreover it is clear that $\text{Ker }\alpha(x) = D_x$ because $\phi_x$ is an iso, hence the only vectors that are send to zero are the ones in $D_x$. Now we have to prove that $\alpha$ is a smooth section of $M$. We have that $(\pi' \circ \alpha)(x) = \pi'(\phi_x) = x$. Here is my second question (2).
QUESTIONS:
Here are my questions:
(1): I really do not see how to show that it is a diffeo. I even have some difficulties to show that $\phi$ is a smooth function. If both component of $\phi$ are smooth we are good but how can I show that $\overline{\alpha}$ is smooth in $x$ from the fact that $\alpha$ is ?
(2): Here is a bit the same question as above, how can I show that $\alpha:M \rightarrow T^*M$ is smooth from the fact that $\phi_x$ is ?
Is this a right way to show this equivalence ? It seems to work but I have some difficulties with details..
English is not my mother tongue, so there must be a lot of mistakes, sorry for that.
Let's cheat: fix a Riemannian metric ${\tt g}$ for $M$, so that $TM/\mathcal{D} \cong \mathcal{D}^\perp$.
Assume that $\mathcal{D} = \ker \alpha$ for non-vanishing $\alpha \in \Omega^1(M)$. Use ${\tt g}$ to convert $\alpha$ to a non-vanishing vector field $\alpha^\sharp \in \mathfrak{X}(M)$, normal to $\mathcal{D}$ --- then $\alpha^\sharp \in \Gamma(\mathcal{D}^\perp)$ gives the global trivialization $\Phi\colon M \times \Bbb R \to \mathcal{D}^\perp$ by $\Phi(x,\lambda) = \lambda \alpha^\sharp_x$.
Conversely, assume that $\Phi\colon M\times \Bbb R \to \mathcal{D}^\perp$ is a given global trivialization for $\mathcal{D}^\perp$. Define a vector field $X \in \mathfrak{X}(M)$ by $X_x = \Phi(x,1)$, and note that $X$ never vanishes. In particular, $X \in \Gamma(\mathcal{D}^\perp)$. Then use ${\tt g}$ to convert $X$ to a non-vanishing $1$-form $\alpha = X_\flat \in \Omega^1(M)$. Clearly $\mathcal{D} = \ker \alpha$.