I am trying to show by induction that pullbacks and the exterior derivative commute. I know that this question has been discussed on this site, eg. here and here. However, none of these questions employ the approach that I'm interested in:
So if $f: X \rightarrow Y$ is a smooth map of manifolds and $\alpha \in \Omega^p(Y)$ is a differential p-form on $Y$, then I want to show that
$$f^*(d\alpha) = d(f^*(\alpha)) \textit{ in } \Omega^{p+1}(X)$$
So my approach is via induction on the $p$, exploiting that
- Pullbacks commute with wedge products $f^{*}(\alpha \wedge \beta) = f^{*}(\alpha)\wedge f^{*}(\beta)$
- And that the exterior derivative is unique under the conditions that (i) $d^2 = 0$, (ii) $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$, (iii) $dg$ for a smooth map $g \in C^{\infty}(X)$ is just the usual derivative
So the induction step is via the commutativity with wedge products, but I'm struggling to unwrap the definitions for the base case. Here I just need to verify that for a smooth map $a\in C^{\infty}(Y)$ the following holds:
$$f^* (da) = d(f^*(a))$$
I think this should just be a notation chase. Let $f(x) = y$, so the LHS is just
$$f^*(da)\rvert_x = (T^*f)(da\rvert_y) \in T_x^*X$$
Whereas the RHS is
$$d(f^*a)\rvert_x = d(a\circ f)\rvert_x \in T_x^{*} X$$
How can I show that these two expressions are equal? Did I get something wrong?
The chain rule tells you that $d(a\circ f)_x=da_{f(x)}\circ df_x$, which is what you need to conclude.