A difficulty in understanding a part of the solution of Q.10 section 1.3 in Allan Pollack and Guillemin(3).

122 Views Asked by At

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?

1

There are 1 best solutions below

2
On BEST ANSWER

Take a chart at $x$ and $f(x)$, then this reduces to the usual inverse function theorem on $\mathbb{R}^n$.