Ambiguity in definition of $C^r$ maps between manifolds

Let $M$ and $N$ be smooth manifolds with corresponding maximal atlases $A_M$ and $A_N$. We say that a map $f : M \to N$ is of class $C^r$ (or $r$-times continuously differentiable) at $p \in M$ if there exists a chart $(V, y)$ from $A_N$ with $f(p) \in V$, and a chart $(U, x)$ from $A_M$ with $p \in U$, such that $f (U) \subset V$ and $y\circ f \circ x^{-1}$ is of class $C^r$.

The above appears ambiguous to me. What if there are two other charts $(U',x')$ and $(V',y')$ with $p \in U'$ and $f(U') \subset V'$ but $y'\circ f \circ x'^{-1}$ not of class $C^r$? Will $f$ still be of class $C^r$ at $p$?

It may be the case that $y'\circ f\circ x'^{-1} : U' \to V'$ is not $C^r$, but its restriction to a sufficiently small open neighbourhood of $p$ will be $C^r$. To see this, note that (on an appropriate domain containing $p$)
The maps $y'\circ y^{-1}$ and $x\circ x'^{-1}$ are transition maps for $N$ and $M$ respectively. As the atlases are smooth, $y'\circ y^{-1}$ and $x\circ x'^{-1}$ are smooth functions (and therefore $C^r$). As the composition of $C^r$ functions is a $C^r$ function, and $y\circ f\circ x^{-1}$ is $C^r$, $y'\circ f\circ x'^{-1}$ is $C^r$.
• This proves that map $y' \circ f \circ x,^{-1}$ is $C^r$ on the domain $U \cap U'$. But what about whole $U'$? Is it not possible that $y' \circ f \circ x,^{-1}$ in not $C^r$ on the whole $U'$? – Math Learner Apr 5 '15 at 2:52
• There is a proof that all differential objects of class $C^r$ for $c\geq1$ can be extended to smooth counterparts, which in turn are unique up to diffeomorphisms. Basically, the only thing you need to change is the maximal atlas to a smooth one (which is always possible if the manifolds are of $C^1$). And the rest follows from smooth transition maps. Actually this is the whole point of manifolds: you prove that everything is chart independent, since you can exchange them for other charts and the properties still hold! – mike May 19 '16 at 16:28
Observe $$y \circ f \circ x^{-1} = (y \circ y'^{-1}) \circ (y' \circ f \circ x'^{-1}) \circ (x' \circ x^{-1})$$ but the functions in parenthesis are at least of class $r$ hence the composite is also of class $r$.