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Q.10 section 1.3 in Allan Pollack and Guillemin is the following:

"Generalization of the inverse function theorem.let $f:X \rightarrow Y$ be a smooth map that is 1-1 on a compact submanifold Z of X. Suppose that for all $x \in Z$, $$df_{x}: T_{x}(X) \rightarrow T_{f(x)}(Y) ,$$

is an isomorphism. then $f$ maps $Z$ diffeomorphically onto $f(Z)$. why?

prove that $f$, in fact, maps an open neighborhood of Z in X diffeomorphically onto an open neighborhood of $f(Z)$ in Y. Note that when Z is a single point, this specializes to the inverse function theorem."

The author gives a hint mentioning in it exercise 5 which is:

"Prove that a local diffeomorphism $f: X \rightarrow Y $"is actually a diffeomorphism of $X$ onto an open subset of $Y$, provided that $f$ is 1-1.

The hint says:

"Prove that, by Excercise 5, you need only show $f$ to be 1-1 on some neighborhood of Z. Now if $f$ is not so, construct sequences ${a_{i}}$ and ${b_{i}}$ in $X$ both converging to a point $z \in Z$, with $a_{i} \neq b_{i}$, but $f(a_{i}) = f(b_{i})$. Show that this contradicts the nonsingularity of $df_{z}$."

I have found an answer for this question on the internet in page 8 in the following link:

https://math.berkeley.edu/~ceur/notes_pdf/Eur_GPDiffTopSolns.pdf

But I could not understand the first line in the answer which states that:

"Since $df_{x}$ is an isomorphism for all $ x \in Z$, for each $x \in Z$ there exists $U_{x}$ on which $f$ restricted on $U_{x}$ is a diffeomorphism"

There is Excercise 4 in the book in section 2 which states that:

"Suppose that $f: X \rightarrow Y$ is a diffeomorphism, and prove that at each x its derivative $df_{x}$ is an isomorphism of tangent spaces."

I think the solution uses the other side of this exercise, but how can I proof that it is correct?

Could anyone help me answering this question please?

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    $\begingroup$ HINT: If you have a continuous function $A$ from $U$ to the set of $n\times n$ matrices, and $A(u_0)$ is invertible, what is true of $A(u)$ for all $u$ sufficiently near $u_0$? [P.S. Don't accept answers if you aren't fully satisfied.] $\endgroup$ – Ted Shifrin Sep 19 '18 at 18:17
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Take a chart at $x$ and $f(x)$, then this reduces to the usual inverse function theorem on $\mathbb{R}^n$.

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  • $\begingroup$ What do you mean by taking a chart ..... could you provide more details please? $\endgroup$ – user593485 Sep 19 '18 at 17:59
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    $\begingroup$ Pick a chart $\varphi\colon U\ni x\xrightarrow{\cong}\mathbb{R}^n$ and a chart $\psi\colon V\ni f(x)\xrightarrow{\cong}\mathbb{R}^n$. Then $df_x$ being an isomorphism means $(\psi\circ f\circ\varphi^{-1})'(\varphi(x))$ is an invertible linear map $\mathbb{R}^n\to\mathbb{R}^n$. Apply the usual inverse function theorem for $\mathbb{R}^n$ tells you there is an open neighbourhood of $\varphi(x)$ for which $\psi\circ f\circ\varphi^{-1}$ is a diffeomorphism, i,e. the desired result. $\endgroup$ – user10354138 Sep 19 '18 at 19:02

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