The following statement is in page 14 of Guillemin & Pollack Differential Topology:
Let $f:X \to Y$ be a smooth map, and suppose that $df_x$ is an isomorphism, show that we can find parametrizations $\phi : U \to X$ and $\psi : U \to Y$ such that $f(x_1,x_2,\ldots,x_k)=(x_1,x_2,\ldots,x_k)$.
I know that by inverse function theorem, $f$ is a local diffeomorphism at $x$.
1-$\phi^{-1}x=(x_1,x_2,\ldots,x_k)$, so if I choose $\psi$ such that $\psi^{-1} f(x) = (x_1,x_2,\ldots,x_k)$ I am done, then both $x$ and $f(x)$ have the same coordinates via charts $\phi$ and $\psi$. am I right ?
But then, $f(x_1,x_2,\ldots,x_k)$ does not make sense, since domain of $f$ is the manifold $X$ , not the Euclidean coordinate space!!!
2-The other thing that can be done : since $X \in \mathbb{R^N}$ is a $k-$dimensional manifold , I can think of $(x_1,x_2,\ldots,x_k)$ as a point in the manifold $X$ (not in the coordinate chart), then suppose $\phi$ be a parametrization such that $\phi(0)=(x_1,x_2,...,x_k) :=x$, then $f(x)=(f_1(x),f_2(x),\ldots,f_k(x)) :=y \in Y$, Is this sufficient to choose $\psi$, such that $\psi^{-1}y=0$ ? why ? I am confused....
I already appreciate your help.