For polar coordinates calculate the Pullback $P^{*}dx$ and $P^{*}dy$ as well as $P^{*}(dx \land dy)$ $P : (0, \infty) \times \mathbb R \to \mathbb R^{2}\setminus\{0\}$ where $(r, \varphi)\mapsto (r\cos(\varphi),r\sin(\varphi)) $


*

*Calculate the differential $dP$

*Calculate the Pullback $P^{*}dx$ and $P^{*}dy$ as well as $P^{*}(dx \land dy)$

*Determine 1-Forms on $\mathbb R^{2}\setminus\{0\}$ that can be pulled back on $dr$ and $d\varphi$ 
I have just started learning about "abstract" differentials and am struggling to come grips with it. 
My attempts: 
$1.$ $dP: T_{(0,\infty)\times \mathbb R} \to T_{\mathbb R^{2}\setminus\{0\}}$ where the $T$'s represent the tangent bundles of the respective sets. 
$dP= (cos(\varphi)dr -r\sin(\varphi)d\varphi) \wedge(\sin(\varphi)dr + r\cos(\varphi))$ 
Not sure whether this is how the differential is calculated. 
$2.$ If $1.$ is correct, then I would simply consider the inverse of $dP$ but how do I invert with the wedge product involved?
$3.$ No idea how to do this.
I have a bunch of notes with abstract notation but I can get nowhere when it comes to dealing with actual problems. Can anyone help me?
 A: I'll address your question 2.  This is essentially a matter of remembering and concatenating several facts about functions (also called 0-forms), pullbacks, differentials, and wedge products. Here goes.
First, $x$ here denotes the function assigning to each point in the plane its $x$-coordinate.
Second, when applied to functions, $P^*$ just means composition with $P$. In particular, $P^*(x)$ is the function assigning to each $(r,\phi)$ the first coordinate (i.e., $x$) of $P(r,\phi)$. In other words, $P^*(x)$ sends $(r,\phi)$ to $r\cos\phi$.
Third, pullbacks commute with $d$. So
$$
P^*(dx)(r,\phi)=d(P^*(x))(r,\phi)=d(r\cos\phi)(r,\phi)=(\cos\phi)(dr)-(r\sin\phi)(d\phi).
$$
Similarly, $P^*(dy)=(\sin\phi)(dr)+(r\cos\phi)(d\phi)$.
Fourth, pullbacks commute with $\land$. Applying this to the $P^*(dx)$ and $P^*(dy)$ computed above, remembering that $dr\land dr=d\phi\land d\phi=0$, remembering that $d\phi\land dr=-dr\land d\phi$, and remembering that $\sin^2\phi+\cos^2\phi=1$, we get
$$
P^*(dx\land dy)=r\,(dr)\land(d\phi).
$$
(That last expression should be familiar from the change-of-variables formula between polar and cartesian coordinates, often written in the abbreviated form $dx\,dy=r\,dr\,d\phi$.)
