I have a question related to a proof in Spivaks "Calculus on Manifolds". On page 117 he defines what $d\omega$ ought to mean on a manifold:
$$ d\omega(x)(v_1, \ldots, v_{p+1}) = d(f^*\omega)(a)(w_1,\ldots,w_{p+1}) $$
where $f: W \to \mathbb{R}^n$ is a coordinate system with $f(x)=a$, $v_1,\ldots,v_{p+1} \in M_x$, $w_1,\ldots,w_{p+1} \in {\mathbb{R}^k}_a$, and $f_*(w_i)=v_i$.
In the proof of Theorem 5-3 he states that "it is clear" that the form $d\omega$ defined in the way explained in this proof is unique. I must admit, that I can't see why it is unique. Do you have an idea?
Thank you for your help!
The uniqueness claim here says that no other form $d\omega$ satisfies the desired property $f^*(d\omega) = d(f^*\omega)$ for every coordinate system $f$. This is one of the proofs where nothing really clever has been done; it's all just working through the appropriate definitions.
We already have definitions of $f^*$ in this context, and the definition of $d$ in the right-side context $d(f^* \omega)$, since $f^* \omega \in \Lambda^p(\mathbb{R}^n)$. So the only thing remaining to define in the equation $f^*(d\omega) = d(f^*\omega)$ is $d\omega$, and the definition essentially says that $d\omega$ is the form which makes that equation true. Spivak skims over the proof that this is well-defined: At a particular $x \in M$ and considering different coordinate systems $f_1$ and $f_2$ around $x$, the values $(d\omega)(x)(v_1,\ldots,v_{p+1})$ are the same when determined using $d(f_1^* \omega)$ or using $d(f_2^* \omega)$. With that proved, we know that we have a definition for $d\omega$ on all of $M$, and it satisfies $f^*(d\omega) = d(f^*\omega)$ for every $f$, and any other form $d\omega$ would not satisfy the equation for at least one combination of $x \in M$ and $v_1,\ldots,v_{p+1} \in M_x$.