Smooth maps considered as locally ringed space morphisms?

For a morphism $f: (X,{mathcal O}_X)to (Y,{mathcal O}_Y)$ of locally ringed spaces the locality of the maps $f^{sharp}_x: {mathcal O} _{Y,f(x)}to {mathcal O} _{X,x}$ implies that for each function $lambdain{mathcal O}_Y(U)$ we have $v(f^{sharp}_U(lambda)) = f^{-1}(v(lambda))$ (here $v(lambda)$ denotes the zero set of $lambda$, i.e. the set of all points $yin U$ such that $lambda_yin{mathfrak m}_y$.

(*) Therefore the zero set of $f^{sharp}_U (lambda)$ is determined by the topological component of $f$ alone, and it equals $f^{-1}(v(lambda))$

From now on, let $X,Y$ be topological spaces and ${mathcal O} _X$, ${mathcal O} _Y$ denote the sheaves of continuous real-valued functions on $X$ and $Y$, respectively. If you know that $f^{sharp}$ preserves constant functions, then by translating a function $lambdain{mathcal O}_Y (U)$ by a real number $alphain{mathbb R}$, you see by (*) that the niveau set $v_alpha(f^{sharp}_U(lambda)) := {xin f^{-1}(U) | f^{sharp}_U(lambda)(x) = alpha}$ equals $f^{-1}(v_alpha(lambda))$, so $f^{sharp}_U(lambda) = lambdacirc f$ as suspected.

Now $f^{sharp}$ has to preserve constant functions, because each map $overline{f^{sharp}_x}: {mathbb R} = {mathcal O}_{Y,f(x)}/{mathfrak m}_{Y,f(x)}to{mathcal O}_{X,x}/{mathfrak m}_{X,x} = { mathbb R}$ is an automorphism of ${mathbb R}$, thus equals the identity. (Ok, one could use this to argue directly that $f^{sharp}_U(lambda)(x)=lambda(f(x))$ - didn't see this when starting to write)

For a field different from ${mathbb R}$, this argument doesn't work any more, because there may be nontrivial automorphims. For example, if you take ${mathbb C}$-valued continuous functions, you could set $f^{sharp}_U(lambda) := text{conj}circ lambdacirc f$ for the algebraic component of $f$.