Let $M$ be a finite-dimensional smooth manifold and $E=\bigwedge^rT^\star M$ be the vector bundle of $r$-forms in $M$. The smooth $r$-forms in $M$ constitutes the vector space smooth sections of $E$: $$\Omega^r(M)=\Gamma(E).$$ I was wondering if the exterior differential $d:\Omega^r(M)\to\Omega^{r+1}(M)$ defines a fibrewise $\mathbb R$-linear mapping $d_x:E_x\to E_x$ for every $x\in M$. However, I am not able to convince myself that this is the case.
Any help would be appreciated.
The exterior derivative is not a fiber-wise map but a differential operator. If it were a fiber-wise linear map, then for any form $\phi\in\Omega^r(M)$ and any point $x\in M$, $\phi(x)=0$ would imply $d\phi(x)=0$ and this is not true. Similarly, coming from a fiber-wise map would imply that for any smooth function $f$, one would have $d(f\phi)=fd\phi$ which is not true either.