I'm wondering if the nice pullback formula for top-degree forms (Prop. 14.20 from Lee's Intro. to Smooth manifolds) 
holds more generally for frame fields and not just coordinate fields. What I mean is suppose $F : M \rightarrow N$, and let $Z_1, \cdots, Z_n$ be a frame on $U\subset M$ and $Z_1^*, \cdots, Z_n^*$ the corresponding dual frame. Similarly, let $W_1, \cdots, W_n$ be a frame on $V\subset N$ and $W_1^*, \cdots, W_n^*$ the corresponding dual frame.
Since $DF$ is just the map that takes the tangent space at each point $p$ in $U$ to the corresponding tangent space at $F(p)$ in $V$, we can write this map in terms of the the $Z_1, \cdots, Z_n$ and $W_1, \cdots, W_n$ bases. Call this map $DF_{ZW}$. Then it seems like the following should hold (ignoring the function $u$ for the moment): $$ F^* (W_1^* \wedge \cdots \wedge W_n^*) = (\det DF_{ZW} ) Z_1^* \wedge \cdots \wedge Z_n^*. $$
Is this correct?