Does anyone have a proof of problem 14, on page 56 of Guillemin and Pollack? I meant to do it as an exercise (I'm teaching myself the subject) but I'm struggling with the last step. Suggestions?
Suppose that the derivative of $f: X \rightarrow Y$ is an isomorphism whenever $x$ lies in the submanifold $Z \subset X$, and assume that $f$ maps $Z$ diffeomorphically onto $f(Z)$. Prove that $f$ maps a neighborhood of $Z$ diffeomorphically onto a neighborhood of $f(Z)$. Essentially, this is the inverse function theorem but without assumptions on the compactness of $Z$.
So far I have local inverses $g_i : U_i \rightarrow X$ where $\{U_i\}$ is a locally finite collection of open subsets of $Y$ covering $f(Z)$. Define $W = \{y \in U_i : g_i(y) = g_j(y)\mbox{ whenever }y \in U_i \cap U_j\}$. We then have a smooth inverse $g: W \rightarrow X$.
How do I show that $W$ then contains an open neighborhood of $f(Z)$ (using local finiteness of $\{U_i\}$, presumably)?
Best Answer
Eric: Here's the idea. A good example I've found helpful when presenting this is the spiral $f\colon\mathbb R^2\to\mathbb R^2$, $f(t) = (e^x\cos y,e^x\sin y)$, with $Z$ the diagonal. So $f(Z)$ is the spiral $\{(e^t\cos t,e^t\sin t): t\in\mathbb R\}$. You can imagine overlapping but non-adjacent little boxes on which $f$ is a local diffeomorphism, so that the local inverses don't quite glue.
As G&P suggest, cover $f(Z)$ with open sets $U_i$ on which we have a local inverse $g_i$ of $f$. The key point is that we may assume this is a locally finite open cover. (In our example, note that it's crucial that $0\notin f(Z)$.) Let $V_{ij} = \overline{\{v\in U_i\cap U_j: g_i(v)\ne g_j(v)\}}$. Then let $\tilde U_i = U_i - \cup_j V_{ij}$, and let $W=\cup_i \tilde U_i$. Check that $W$ works.
Let me know if this works better than what G&P have.