Gluing locally defined smooth maps to get a globally defined smooth map

Question

Please find the discussion underlined with red, which is to be clarified. $\tilde{f_p}$ is a smooth vector-valued function defined on an open subset $W_p$ of a smooth manifold $M$, and $\psi_p$ is a smooth real-valued function defined on the whole $M$ with $\mathrm{supp}(\psi_p)\subseteq W_p$. My question is how to extend the scalar product $\psi_p\tilde{f_p}$ defined on $W_p$ to a globally defined smooth function. I don't understand what John M. Lee, the author, means when he says

if we interpret it to be zero on $M\setminus\mathrm{supp}(\psi_p)$

What exactly is "it"? I have no idea what's happening to $\tilde{f_p}$ around $\partial W_p$. Is there any easy way to understand the trick Lee is using? Thank you.

Answer 1

For a real-valued function $f: U \to \mathbb{R}$ and a vector-valued function $g : V \to \mathbb{R}^k$, their product is the vector-valued function $fg : U\cap V \to \mathbb{R}^k$ given by $(fg)(x) = f(x)g(x)$; note, we had to restrict the domain of $fg$ to be the intersection of the domains of $f$ and $g$ for this to make sense.

In this case we have $\psi_p : M \to \mathbb{R}$ and $\tilde{f_p} : W_p \to \mathbb{R}^k$, so the product is $\psi_p\tilde{f_p} : W_p \to \mathbb{R}^k$; note, the domain is $M\cap W_p = W_p$, not $M$. What Lee is saying is that we can extend this to a smooth function on $M$ by defining $(\psi_p\tilde{f_p})(x) = 0$ if $x \in M\setminus W_p$.