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 inverse function theorem that is stated in Allan Pollack differential Topology before this question is:
Suppose that $f:X \rightarrow Y$ is a smooth map whose derivative $df_{x}$ at the point $x$ is an isomorphism. Then $f$ is a local diffeomorphism at $x$.
Does the answer is "by the inverse function theorm"? am I correct?
Best Answer
Yes, but you must be careful since you only know that the derivative is an isomorphism over the points of $Z$. However, the compactness will give you a nice neighborhood of $Z$ where everything works.
For each point of $Z$ you'll find, by the inverse function theorem, a open neighborhood (in $X$) where the restriction of $f$ is a diffeomorphism. By the compactness of $Z$ we have an open subset $U \subset X$ that contains $Z$ and $\left. f \right|_U$ is 1-1 and a local diffeo, hence a diffeomorphism onto its image. In particular, $\left. f \right|_U$ maps $Z$ diffeomorphically onto $f(Z)$.