Apparent contradiction in properties of the pullback of forms

Question

I think this question is better illustrated with an example.

Let $f:\mathbb{R}^3\rightarrow \mathbb{R}^2$ and $\theta=xydx+y^2dy\ \in\Omega^1(\mathbb{R}^2)$.

We know that $f^*:\Omega(\mathbb{R}^2)\rightarrow\Omega(\mathbb{R}^3)$ is linear and $f^*(\omega\wedge\alpha)=f^*(\omega)\wedge f^*(\alpha)$.
Using linearity: $f^*\theta=xyf^*dx+y^2f^*dy\ $. Using the second property ($xy$ and $y^2$ are 0-forms): $f^*\theta=f^*xy\wedge f^*dx+f^*y^2\wedge f^*dy\ $

So I get two different things. What is going on?

Answer 1

Pay attention! The pull-back is linear over $\Bbb R$ (or $\Bbb C$ if you work with complex objects), but not over $\mathcal C ^\infty (M, \Bbb R)$ (or $\mathcal C ^\infty (M, \Bbb R)$). In other words, you may take numbers out of $f^*$, but not functions!

Your first approach is mistaken because $f^* (xy \ \Bbb dx) = f^*(xy) \ f^* (\Bbb dx) = [(xy) \circ f] \ \Bbb d (x \circ f)$.

Answer 2

Remember that you have a 1-form field: at each point of $\mathbb R^3$ you may evaluate to obtain a 1-form. And to evaluate the coefficients (which are functions on the domain) you must first precompose $f$