I am reading about derivations of the algebra of differential forms and there is an argument used to prove a proposition that I find hard to understand. The author wants to prove that a derivation $D$ satisfies $D \omega = 0$ for all $1$-fom $\omega$. In order to do that, he states that since $D$ can be localized (i.e., if a tensor field $K$ vanishes on an open set $U$, then $DK$ vanishes on $U$), it is sufficient to show that $D\omega = 0$ when $\omega$ is of the form $f dg$, $f,g \in \mathcal{C}^\infty(M)$.
I understand the rest of the proof, but I think there is a subtlety I am missing. Why is so important the part about 'being localized'?
The point is that you know that locally, any one-form can be written as a finite sum of one-forms that can be written as $fdg$. (Just go to a chart and write $\omega$ as $\sum \omega_idx^i$ there.) Edit: Having proved that a derivation can be localized, you consider two tensor fields $K_1$, $K_2$ which satisfy $K_1|_U=K_2|_U$ for some open subset $U$. Then $K_1-K_2$ vanishes on $U$, so $D(K_1-K_2)=D(K_1)-D(K_2)$ vanishes on $U$, too, and thus $D(K_1)|_U=D(K_2)|_U$. Puting this together, you see that $D(\omega)|_U=\sum_i D(\omega_idx^i)$ and hence it suffices to deal with one-forms that can be written as $fdg$. It is also possible to get similar statements globally, but this needs more heavy tools like finite atlas theorems.