"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 (imagine) the second paragraph of the solution given in the image below:
Could anyone explain this for me please?
Also in the second line in the solution I do not know why he take the intersection on $U_{i}'s$ and not the union?Could anyone explain this for me please?
Best Answer
OK, let's break the exercise into steps. First, the "then $f$ maps $Z$ diffeomorphically onto $f(Z)$" part.
Since you have that $f$ in one-one on $Z$, and immersion at every $x\in Z$, you only need to prove that the map $f|_Z: Z \rightarrow f(Z)$ is a proper map, because then it would be an embedding, and an embedding is a diffeomorphism onto its image (pages 17, 18). Just take a compact $K \subset f(Z)$, since $f(Z)$ is given the subspace topology, it is also compact in $Y$, and since $Y$ is Hausdorff, it is closed in $Y$. Then since $f|Z$ is continuous as a map with image in $Y$ (just the restriction of a continuous map), $f^{-1}(K)$ is closed in $Z$, but $Z$ compact implies it is compact in $Z$. So $f$ is also proper, hence a diffeomorphism.
Now, we need to prove that $f$ is a little more, is a diffeomorphism of an open neighborhood of $Z$ in $X$ onto an open neighborhood of $f(Z)$ in $Y$.
Why is it that it's enought to prove $f$ to be one-one is an open neighborhood of $Z$ in $X$ (Exercise 5)? Because Exercise 5 tells you that a local diffeomorphism that is one-one is actually a diffeomorphism. Suppose you find an open nighborhood $U$ of $Z$ in $Y$ with $f|_U$ one-one, we could (if necessary) intersect $U$ with a neighborhood $V$ of $Z$ such that $f|V$ is a local diffeomorphism (this $V$ you get it by just taking a finite union of neighborhoods in $X$ of points in $Z$ restricted to wich $f$ is a diffeomorphism, local condition). Then you can assume there's a $U$ neighborhood of $Z$ in $X$ such that $f|_U: U\rightarrow f(U)$ is a local diffeo and one-one, applying Exercise 5 we get that $f|_U$ is a diffeomorphism.
There are only two things left to prove: Exercise 5, and the injectivity in a neighborhood of $Z$ in $X$.
For the first one, just observe that if $f: A\rightarrow B$ is a one-one local diffeo, then exists $f^{-1}: f(A)\rightarrow A$ and both $f$, $f^{-1}$ are smooth by just using their local expression, the identity. Then $f$ is a diffeomorphism onto its image, that hence is going to be open in $B$.
For the second one, suppose $f$ is not one-one restricted to any neighborhood of $Z$ in $X$.
Define $U_n=$ {$x \in X| d(x,Z)<\frac{1}{n}$} (your manifold X is sitting inside some $\mathbb{R}^k$ by definition, and if you don't take Guillemin's convention, just use Whitney's Embedding Theorem). There are {$a_n$}, {$b_n$} such that $a_n, b_n \in U_n$ and $f(a_n)=f(b_n)$. Clearly $d(a_n, Z), d(b_n, Z)$ tend to zero. Since the distance is continuous and $Z$ is closed in $X$ (again $Z$ compact inside $X$ Hausdorff is closed), then there are $z_1, z_2\in Z$ with $z_1=\lim_{n\to\infty} a_n$, $z_2=\lim_{n\to\infty} b_n$. Since $f$ is continuous, $f(z_1)=f(z_2)$. This contradicts the injectivity of $f$ restricted to $Z$ if $z_1\neq z_2$. So that $z_1=z_2=z$.
Now that contradicts the fact that $df_z$ is an isomorphism, since this implies that there's a neighborhood of $z$ in $X$ restricted to which $f$ is a diffeomorphism, in particular one-one.