Exercise about differential forms and pull-back: $\omega_p=0$ if and only if $(\omega|_U)_p=0$

Question

Let $M$ be a manifold, $\omega\in\Omega^q(M)$. Let $U\subset M$ be an open subset embedded as manifold in $M$. Let $p\in U$. Show that $\omega_p=0$ if and only if $(\omega|_U)_p=0$.

Notation: Let $F:U\to M$ be the inclusion map which is an embedding. $F^*:\Omega^*(M)\to\Omega^*(U)$ be its pull-back. We denote $F^*\omega$ by $\omega|_U$.

Suppose $\omega_p =0$. Then $(\omega|_U)_p=(F^*\omega)_p=dF^*_p(\omega_{F(p)})=dF^*_p(\omega_{p})=0$, where the last equality follows by linearity of $dF^*_p$.

I can't prove the reverse implication. Any suggestion?

Answer 1

Weird notation, usually people write $F^* \omega = \omega \circ F $... which is $\omega | _U$ in this case since $F|_U = id$... Your proof of the one direction doesn't make sense, how did the exterior differential come into play? Nevertheless, note that $F$ is the inclusion map, so we obviously have $F(p) = p$ since $p \in U$. Thus $$ (F^* \omega)_p = \omega \circ F(p) = (\omega)_p$$ The claim now follows.

Answer 2

Possible answer.

Does it make sense? $$ 0=(\omega|_U)_p(v_1,\dots,v_q)=(F^*\omega)_p(v_1,\dots,v_q)=dF^*_p(\omega_{F(p)})(v_1,\dots,v_q)=dF^*_p(\omega_{p})(v_1,\dots,v_q)=\omega_p(dF_p(v_1,\dots,v_q))=\omega_p(v_1,\dots,v_q) $$