I am learning how to compute a pullback of a 1-form. My reference does not provide many examples, and those few there are utterly trivial that reduce to 0... I am wondering how one might compute something like: \begin{equation} \phi^*(x^2dy + y^2dx), \end{equation} where $\phi:\mathbb{R}\to\mathbb{R}^2$, $t\mapsto (t,t^2)$.
Much thanks in advance!
Pullbacks of forms work exactly like variable substitution.
For example, if we have standard variables $r, \theta$ on one copy of $\mathbb{R}^2$ and $x,y$ on a second copy of $\mathbb{R}^2$, then the transformation $\phi : \mathbb{R}^2 \to \mathbb{R}^2 : (a,b) \mapsto (a \cos(b), a \sin(b))$ has pullback given by
and pullbacks commute with exterior derivatives: $\mathrm{d} \phi^*(s) = \phi^*(\mathrm{d} s)$. e.g.
$$ \phi^*(\mathrm{d}x) = \mathrm{d}(r \cos(\theta))$$
For a proof that this works, suppose you're given a differentiable map $\phi : M \to N$.
Consider the manifold $M \times N$, along with the pais $p : M \times N \to M$ and $q : M \times N \to N$ defined by simply taking the first or second coordinate.
The point of doing this is to construct a space in which we can simultaneously work with both the coordinate functions of $M$ and the coordinate functions of $N$.
This manifold has a submanifold $S \subseteq M \times N$ consisting only of the points $(m, \phi(m))$; this is nothing more than the graph of the function $\phi$!
But the point is that $S$ is diffeomorphic to $M$ in the obvious way, so we can work with $S$ as if it were the same thing as $M$.
Going back to the example above, the point is that on the manifold $M$, we have the four scalar fields $x,y,r,\theta$. When restricted to the submanifold $S$, the relationship between them is literally just equations: $x = r \cos(\theta)$ and $y = r \sin(\theta)$.