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.

  • $\begingroup$ Who is the `` U ``? I mean: is it another manifold? $\endgroup$ – Poli Tolstov Mar 9 '14 at 2:07
  • $\begingroup$ @PoliTolstov $U$ is an open set in $\mathbb{R^k}$ $\endgroup$ – the8thone Mar 9 '14 at 2:08
  • $\begingroup$ 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. $\endgroup$ – 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 $\phi=\phi_1|_{U},$ $\psi=\phi\circ F^{-1}.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.