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Part of 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?"

The rest of the question is: "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."

I feel like he want me to prove that $f$ is locally diffeomorphic onto an open neighborhood of $f(Z)$ ... am I correct?

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 $df_{z}$."

Does this hint mean that I will not be able to use exercise 5?

I also could not understand the note in the question which say:

"Note that when Z is a single point, this specializes to the inverse function theorem."

Could anyone give a concrete example for me on this?

It is not clear for me how can I show that this constructed sequence contradicts the nonsingularity of $df_{z}$.?

Related Question:First generalization of the inverse function theorem Q.10 section 1.3 in Allan Pollack and Guillemin(1).

Could anyone help me answering those questions?

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Let's see. When $Z = \{x\}\subset X$ is a point, then $f$ is automatically one-to-one on $Z$, and the hypothesis that $df_x\colon T_xX\to T_{f(x)}Y$ is an isomorphism is precisely the hypothesis of the inverse function theorem. You then conclude that $f$ maps a neighborhood $U$ of $x$ diffeomorphically to a neighborhood $f(U)$ of $y=f(x)$.

You are going to apply Exercise 5 to an appropriate neighborhood $U$ of $Z$. Provided you show $f$ is one-to-one on $U$, you'll deduce that $f$ maps $U$ diffeomorphically to $f(U)$. (You will know $f$ is a local diffeomorphism if you start by taking $U$ to be a union of open sets around points $x\in Z$ on which $f$ is a local diffeomorphism. You will then likely have to shrink $U$ a bit to make sure $f$ is one-to-one on $U$.)

Finally, here's a hint for the hint. Consider $U_i = \{x\in X: d(x,Z)<1/i\}$. If $f$ fails to be one-to-one for every $i\in\Bbb N$, there are points $a_i,b_i\in U_i$ with $f(a_i)=f(b_i)$. All these sets are contained in the compact set $\overline U_1$, and so, passing to convergent subsequences carefully, we may assume that $a_i,b_i\to z\in Z$. (Here you use the fact that $f$ is one-to-one on $Z$.) But what do you conclude from the fact that $df_z$ is an isomorphism?

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  • $\begingroup$ Could you please explain this statement in details "(You will know $f$ is a local diffeomorphism if you start by taking $U$ to be a union of open sets around points $x\in Z$ on which $f$ is a local diffeomorphism. You will then likely have to shrink $U$ a bit to make sure $f$ is one-to-one on $U$.) " ....... specifically how will I shrink $U$? $\endgroup$ – hopefully Sep 23 '18 at 17:41
  • $\begingroup$ for me, I conclude that $df_{z}$ can not be onto ..... correct? $\endgroup$ – hopefully Sep 23 '18 at 17:54
  • $\begingroup$ No, the shrinking comes in the third paragraph. The contradiction will tell us that some $f$ must in fact be one-to-one on some $U_i$ (indeed, then, on all $U_j$ with $j\ge i$). As I said in the first paragraph, the fact that $df_z$ is an isomorphism tells you that $f$ must be a diffeomorphism (in particular, one-to-one) on some neighborhood $U$ of $z$; but for large enough $i$, both $a_i$ and $b_i$ must be in $U$ (why?). $\endgroup$ – Ted Shifrin Sep 23 '18 at 21:24

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