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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.

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Who is the `` U ``? I mean: is it another manifold? – Poli Tolstov Mar 9 '14 at 2:07
@PoliTolstov $U$ is an open set in $\mathbb{R^k}$ – the8thone Mar 9 '14 at 2:08
Usually, if we already have a parametrization $\phi:U\rightarrow X$, we write points of $X$ in the form $(x_1,\ldots,x_k)$ instead of $\phi(x_1,\ldots,x_k)$. I guess that this is what the formula $f(x_1,\ldots,x_k)=(x_1,\ldots,x_k)$ means. – Luiz Cordeiro Mar 9 '14 at 2:48

Take neighborhoods $X_1$ of $x$, $Y_1$ of $f(x)$, such that $f:X_1\to Y_1$ is a diffeomorphism. Take $\phi_1:U_1\to X$, $\psi_1:V_1\to Y$ parametrizations of neighborhoods of $x,y$ (resp.). Let $X_2=\phi_1(U_1)\cap X_1$, $U=\phi_1^{-1}(X_2)$, $Y_2=\psi_1(V_1)\cap Y_1$, $V=\phi_1^{-1}(Y_2)$. Then $F:U\to V$, $u\mapsto \psi_1^{-1}( f(\phi_1(u)))$ is a diffeomorphism. Now take $\phi=\phi_1|_{U},$ $\psi=\phi\circ F^{-1}.$

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