Assume $X, \mathscr O_X$ and $Y, \mathscr O_Y$ are smooth manifolds with the corresponding structure sheaves. We know that any smooth $f:X\to Y$ defines a morphism of locally ringed spaces $(f, f^\#): (X, \mathscr O_X)\to (Y, \mathscr O_Y)$. I would like to prove the converse.
So, I only need to do it for open subsets $U\subset \mathbb R^n$ and $V\subset \mathbb R^m$. Assume $x_1, \ldots, x_n$ coordinates on $U$ and $y_1, \ldots, y_m$ coordinates on $V$. Also, let $(f, f^\#): (U, \mathscr O_U)\to (V, \mathscr O_V)$ morphism of locally ringed spaces, where $\mathscr O_U, \mathscr O_V$ are the corresponding structure sheaves.
It is easy to show that there is a smooth function $g: U\to V$ such that $g=f$, so $f$ is smooth. Also, we can prove that $f^\#(y_j)= g_j := g^\#(y_j)$. Thus, the same thing holds on polynomials. My question is, how can we generalize this on any smooth function, i.e. for any $W\subseteq V$ and $h: W\to \mathbb R$ smooth, show that $f^\#(W)(h)= g^\#(W)(h):= h\circ g$.
Thanks.
