TITLE: How to compute the pullback of $(2xy+x^{2}+1)dx+(x^{2}-y)dy$ along $f(u,v,w)=(u-v,v^{2}-w)$?
QUESTION [5 upvotes]: I'm trying to do my first pull-back of a differential form.
I know that $\omega=(2xy+x^{2}+1)dx+(x^{2}-y)dy$ is a differential form on $\mathbb{R}^{2}$.
I have $f : \mathbb{R}^{3} \to \mathbb{R}^{2}$ which is
$$f(u,v,w)=(u-v,v^{2}-w)$$
and I have to calculate the pullback. I was told that by definition
$$(f^{*}\omega)(X) = \omega(f_{*}(X)),$$
and so I calculated 
$$f_{*}=\begin{pmatrix}
1 & -1 & 0\\
0 & 2v & 1
\end{pmatrix}$$
But then I don't really know how to proceed. Should I take a general vector and calculate the form, should I substitute $x,y$ with $u,v,w$? Do you have a general recipe to proceed?

REPLY [4 votes]: The nice thing about forms is that you are intuitively doing the right thing, which is: just plug in $x=u-v$ etc. as Andrew already wrote.
To see the connection with the formal definitions, note that the exterior derivative and pullbacks commute and hence $f^*dx=d(f^*x)=d(x\circ f)=d(u-v)$. Furthermore, pullbacks respect wedge products, so e.g.
$$f^*(x^2dx)=f^*(x^2 \wedge dx)=(f^*x^2) \wedge (f^*dx)=(u-v)^2d(u-v)$$