I'm looking at Exercise 2.6b (p.58), which is to prove that for smooth manifolds $M$ and $N$ and a continuous map $F:M\rightarrow N$, we have that $F$ is smooth iff $F^*(C^\infty(N))\subseteq C^\infty(M)$, where $F^*(f)=f\circ F$ is the pullback by $F$. The forwards direction is obvious, but I'm stuck on the backwards direction (i.e. that $F^*(C^\infty(N))\subseteq C^\infty(M)$ implies $F$ is smooth). It seems clear that the right approach should something like "the component functions of each chart on $N$ are smooth functions to $\mathbb{R}$, hence so are their pullbacks, and a function to $\mathbb{R}^n$ is smooth iff its component functions are smooth." But I'm having trouble making that work, because the component functions of a chart $\psi:V\rightarrow\mathbb{R}^n$ of $N$ are, of course, functions from $V$ to $\mathbb{R}$, and I don't see a way of showing that $F^*(C^\infty(N))\subseteq C^\infty(M)$ implies $(F|_{F^{-1}(V)})^*(C^\infty(V))\subseteq C^\infty(F^{-1}(V))$ for an arbitrary open $V\subset N$. If I could get that to work, then because $F$ being smooth is a property that can be checked locally (i.e., $F$ is smooth iff for any $p\in M$, there is an open $U\subset M$, $p\in U$ such that $F|_U$ is smooth) I think I would be done.
Any hints in the right direction?