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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)?

Mario Carneiro
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Eric
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    Could you post the problem statement? – Mario Carneiro May 16 '13 at 21:04
  • 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 $Z$ – Eric May 16 '13 at 21:06
  • @Eric Please edit your question and add it there. – Ayman Hourieh May 16 '13 at 21:08

1 Answers1

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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.

Ted Shifrin
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  • That seems reasonable, but my problem is with how to check that $W$ in fact works. How exactly do we employ the local finiteness of the open cover? Perhaps I haven't thought about it long enough? – Eric May 17 '13 at 01:21
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    When is a (perhaps) infinite union of closed sets closed? – Ted Shifrin May 17 '13 at 02:36
  • Is the point just that $W$ is a nonempty open set, and so therefore contains some open neighborhood of $f(Z)$? – Eric May 17 '13 at 03:22
  • But to argue that $W$ is open requires the local finiteness. The rest is easy, yes, as the $g_i$'s glue to make a well-defined smooth map. – Ted Shifrin May 17 '13 at 03:34
  • Hi, Ted, I was working this problem and I came across your old answer here. How can we be sure that $V_{ij}$ does not meet $f(Z)$? Suppose that $y_i$ are a sequence are points in $U_i \cap U_j$ such that $g_i(y_i) \neq g_j(y_i)$ and $y_i \to y \in f(Z)$. This is impossible if $y$ is in $U_i \cap U_j$, by continuity of the inverses. But what if it is not? – Eric Auld Aug 28 '15 at 01:48
  • Eric: I'm two years rusty on this. But isn't it the point that $f|_Z$ has an inverse, so on points of $f(Z)$ the various $g_i$ (where they're defined) must of course agree?... Or am I missing something? – Ted Shifrin Aug 28 '15 at 02:07
  • Ted: definitely the $g_i$ must agree on $f(Z)$. But when we remove the $V_{ij}$, which are the closures of the sets where the $g_i$ disagree, we want to end up with an open set containing $f(Z)$. Definitely ${g_i \neq g_j}$ does not meet $f(Z)$, but how can we be sure its closure does not meet $f(Z)$? In particular, what if a sequence of points in ${g_i \neq g_j}$ converges to a point in $f(Z) \setminus (U_i \cap U_j)$? – Eric Auld Aug 28 '15 at 09:25
  • Eric, the point is that because of local finiteness, $\cup V_{ij}$ is closed, so you're left with an open set! – Ted Shifrin Aug 28 '15 at 15:21
  • Ted, I agree it's open, but why does it not meet $f(Z)$? Certainly ${g_i \neq g_j}$ does not meet $f(Z)$, but why does its closure not meet $f(Z)$? – Eric Auld Aug 28 '15 at 18:30
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    We should discuss this in person in 2 weeks! :) $W$ is open and contains $f(Z)$. And $f$ has a well-defined inverse on $W$. (I still like my spiral example to understand what's going on.) What am I missing? – Ted Shifrin Aug 28 '15 at 18:50
  • Great, would be lovely to meet you. There is something I'm not getting, because I can't see why $W$ contains $f(Z)$ (although certainly $g_i = g_j$ on $f(Z)$). I think I'll ask it as a separate question. – Eric Auld Aug 28 '15 at 19:17
  • Great, would be lovely to meet you. There is something I'm not getting, because I can't see why $W$ contains $f(Z)$ (although certainly $g_i = g_j$ on $f(Z)$). I asked it as a separate question here: http://math.stackexchange.com/questions/1413022/generalization-of-inverse-function-theorem-to-noncompact-submanifolds – Eric Auld Aug 28 '15 at 20:03