Pullback of differential form of degree 1 Good evening,
In differential forms (in the proof of the naturality of the exterior derivative), I don't get why if $h\in \Lambda^0(U)$ and $f^*$ is the pullback then,
$f^*dh=d(f^*h)$.
I wrote $dh=\frac{\partial h}{\partial x^i}dx^i$
so (?) $f^*dh(v)=\frac{\partial h}{\partial f^i}df^i(df(v))$ but I don't know if this is correct and I can't continue...
Thanks a lot for your help
 A: Good evening!
If we define the pullback map as $$f^*(\sum g_I dx_{i_1} \ldots dx_{i_p}):= \sum (g_I \circ f) df_{i_1} \ldots df_{i_p}$$
Then on one hand we have:
$$d(f^*h)=d(h \circ f) $$    
And on the other hand (using the $\textit{Chain Rule}$):
$$f^*(dh)= f^*(\sum \frac{\partial h}{\partial x_i} dx_i)= \sum ((\frac{\partial h}{ \partial x_i} \circ f) df_i) = d(h \circ f)$$
As we wanted.
This result can be generalised to n-forms and interpreted as ''the pullback commutes with the exterior derivative''.
If you need more help, consult Bott and Lu's Differential Forms book.
A: Let $f: M \longrightarrow N$ be a smooth map between the manifolds $M$ and $N$, and let $h: N \longrightarrow \Bbb R$ be a smooth map. Then by definition, for any point $x \in M$ and any vector $V \in T_x M$, we have
$$(f^\ast dh)_x(V) = dh_{f(x)}(df_x(V)).$$
On the other hand, since $f^\ast h = h \circ f$, by the chain rule we have that
$$d(f^\ast h)_x(V) = d(h \circ f)_x(V) = (dh_{f(x)} \circ df_x)(V) = dh_{f(x)}(df_x(V)).$$
Hence pullbacks commute with the exterior derivative in the case asked about in your question (of course this holds for forms of arbitrary degree as well).
