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

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?

## 1 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 $f(Z)$.

• why $f|_{U}$ is 1-1, and what is the importance of $Z$ being a diffeomorphism ? did you apply exercise 1.3.5 in the sentence before last in your proof above? thanks in advance. – hopefully Sep 23 '18 at 12:44