Recall that, given $f:\mathbb{R}^n \to \mathbb{R}^m$ a differentiable function, then we define the pullback function $f^* : \Lambda ^k \left(\mathbb{R_{f(p)}} ^m\right) \to \Lambda^k \left(\mathbb{R_{p}} ^n\right)$ by $\left(f^* \omega \right) (p) \left(v_1, \ldots , v_k \right) = \omega(f(p)) = (f_* (v_1) , \ldots , f_* (v_k)$ and where $f_* : \mathbb{R_{p}} ^n \to \mathbb{R_{f(p)}} ^m$ is the push forward given by $f_* (v_p) = \left(Df(p)(v) \right)_{f(p)}$.
Now, for a problem I am solving, I am evaluating $f^* (d \omega)$ where $\omega$ is a closed form. By definition, this means that $f^* (d\omega) = f^* (0)$. My book is saying that it must therefore be the case that $f^* (d\omega) = f^* (0) = 0$. However, I am not sure I fully understand the part where we have that $f^* (0) = 0$.
I think my misunderstanding is primarily notational, but I would appreciate it if someone could help me understand why this is true. My intuition tells me that we know that since $\omega$ is some $k$-form, then we must have that $d \omega = 0$ is a $k+1$ form (the definition of $d$).
However, how do we define this form? Do we just say that since $0$ is a $k+1$ form, then $0(f(p)) = 0$ for all $p$, and thus it follows trivially that $$f^* 0 = 0(f(p))(f_* (v_1) , \ldots , f_* (v_k)) = 0 \cdot (f_* (v_1) , \ldots , f_* (v_k)) = 0$$
(since $0$ must map to a real number as a $k+1$ form, and by definition it always maps to zero)?
Is this, in essence, the proof that $f^* (0) = 0$?
Apologies if this seems trivial, I am learning about this tensor stuff for the first time, and I am not acquainted with the notation yet.
Yes, of course when you pull back the zero form along a smooth map you find out the zero form of the starting vector space.
To answer your question about the definition of the exterior derivative, you can see it axiomatically as the unique extension of the differential of functions to an odd derivation on smooth $k$-forms which is also nilpotent. Explicitly you require the properties $$d^{(k+1)} \circ d^{(k)} = 0$$ and $$ d^{(k+l)}(\eta \wedge \theta) = d^{(k)} \eta \wedge \theta + (-1)^k \eta \wedge d^{(l)}\theta $$ where $\eta$ is a $k$-form and $\theta$ an $l$-form.
For 1-forms you may also write a nice explicit expression as $$ d \omega(X,Y) = d(\omega(Y))X - d(\omega(X))Y - \omega([X,Y]) $$ where $X,Y$ are vector fields over some open subset of $\mathbb{R}^n$. Of course this can be generalized for forms of any order, but it gets more complicated. Usually, you don't need such a formula in real-world computations.