# Calculation of This Pullback of a Differential Form

Consider the differential form $$f dx$$ in $$\Omega^1 (\mathbb{R}^2)$$, where $$f \in C^{\infty }(M)$$. $$\Omega^1 (\mathbb{R}^2)$$ has basis $$\{ dx, dy \}$$.

I am looking for a rigorous calculation of the pullback of $$dx$$ to $$\mathbb{R}$$ by some smooth map $$\phi : \mathbb{R} \rightarrow \mathbb{R}^2$$. Its basic but I want to do it rigorously.

I think $$f dx$$ should pull back to $$p \mapsto f \circ \phi (p) \frac{\partial \phi_1 } {\partial t}(p) dt$$ in $$\Omega^1 (\mathbb{R})$$, but I'm not sure.

If someone gives me a brief outline of how to do this calculation, I can fill in the details for myself.

Your guess is correct. If you just want a hint, try to use the following:

$$1.$$ Think about how the pullback interacts with an expression like $$fdx$$.

$$2.$$ Think about how the pullback acts on smooth functions.

$$3.$$ Think about how the pullback interacts with the exterior derivative $$d$$.

$$4.$$ Think about what the exterior derivative of a function locally looks like.

Spoiler below:

If we call $$\omega=f dx,$$ then \begin{align*}\phi^*\omega&=\phi^*(f dx)=(\phi^* f)(\phi^* dx)\\ &=(f\circ \varphi)d( \varphi^* x)=(f\circ \varphi)d (x\circ\varphi)\\ &=(f\circ \varphi)\frac{\partial \varphi^1}{\partial t}dt, \end{align*} where $$\varphi^1=x\circ\varphi.$$ Here, we used that $$(1)$$ $$\varphi^*(g\omega)=(\varphi^*g)(\varphi^* \omega)$$ for functions $$g$$ and forms $$\omega$$, $$(2)$$ $$\varphi^*f=f\circ\varphi$$ for functions $$f$$, $$(3)$$ $$\varphi^*$$ commutes with $$d$$, and $$(4)$$ a local expression for $$d$$: $$dg=\sum_i\frac{\partial g}{\partial x^i}dx^i$$. It's especially easy to verify in this case.  In general, you can show that, if $$\omega=\sum\limits_{j} a_jdx^j$$ in a chart, then $$F^*\omega=\sum\limits_{i,j} (a_i\circ F)\frac{\partial F^i}{\partial x^j}dx^j.$$