A vector field on a manifold $M$ is a linear map $X:C^\infty(M)\longrightarrow C^\infty(M)$ with an additional property.
The set $\mathfrak{X}(M)$ of all vector fields on $M$ is a $C^\infty(M)$-module with the product $$(f\cdot X)(g):=fX(g),\ f, g\in C^\infty(M), X\in\mathfrak{X}(M),$$ where $fX(g)$ is the point product of functions.
Given an open subset $U\subset M$ we might define a restriction map $\mathfrak{X}(M)\longrightarrow \mathfrak{X}(U)$, $X\longmapsto X_U$, where $X_U$ satisfies $$X_U(f|_U)=X(f)|_U.$$ Now I'm working through a similar construction for differential $1$-forms: A differential $1$-form on $M$ is an element $\omega\in \displaystyle\Omega^1(M):=\textrm{Hom}_{C^\infty(M)}(\mathfrak{X}(M), C^\infty(M))$.
To get such a restriction map I must show the following:
Let $X\in\mathfrak{X}(M)$ and $V\subset M$ be open such that $X_V=0$. Then $\omega(X)|_V=0$.
Sketch of Proof: My book does the following: Let $p\in V$ and choose $g\in C^\infty(M, \mathbb R)$ such that $g(p)=1$ and $g|_{M\setminus V}=0$. Then it says $X=(1-g)\cdot X$ so that
$$\begin{align} \omega(X)(p)&=\omega((1-g)\cdot X)(p)\\ &=(1-g)(p)\omega(X)(p)\\ &=0.\end{align}$$
How to justify the equality $$X=(1-g)\cdot X?$$
Thanks
Since $X_V = 0$, the equation is true at any point in $V$. Since $g = 0$ outside $V$, the equation reduces to $X = (1-0)X$ away from $V$, so it is true at any point not in $V$.