Guillemin & Pollack Exercise 1.8.14 (Generalized inverse function theorem)

Question

I have a question regarding Exercise 1.8.14 in Guillemin & Pollack. (I think this question also applies to the answers here.) Here's the exercise:

Inverse Function Theorem Revisited. Use a partition-of-unity technique to prove a noncompact version of Exercise 10, Section 3. Suppose that the derivative of $f\colon X\to Y$ 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 $f(Z)$. [Outline: Find local inverses $g_i\colon U_i\to X$, where $\{U_i\}$ is a locally finite collection of open subsets of $Y$ covering $f(Z)$. Define $$W=\{y\in U_i: g_i(y)=g_j(y)\text{ whenever }y\in U_i\cap U_j\}.$$ The maps $g_i$ "patch together" to define a smooth inverse $g\colon W\to X$. Finish by proving that $W$ contains an open neighborhood of $f(Z)$; this is where local finiteness is needed.]

I think the approach suggested in the problem does not work if one chooses the local inverses $g_i\colon U_i\to X$ without additional constraints. Here is the picture I have in mind:

Above $Z$ is the open interval drawn in black. Using the inverse function theorem, we can find open covers $\{V_i\}_{i\in I}$ of $Z$ in $X$, and $\{U_i\}_{i\in I}$ of $f(Z)$ in $Y$, such that $f$ restricts to a diffeomorphism $f\colon V_i\xrightarrow{\,\cong\,}U_i$ for each $i\in I$. The idea is to then replace $\{U_i\}_{i\in I}$ by a locally-finite open refinement which still covers $f(Z)$. In the example above, we can take take the entire cover $\{U_1,U_2,U_3\}$ (thanks to @EricWofsey for the simplifying my argument). If one constructs $W$ as in the problem statement, then $W$ misses all of $U_3$, so does not cover $f(Z)$.

Is there a way to remedy this? It feels like one approach might be to use a tubular neighborhood of $f(Z)$ in $Y$, however in G&P one is asked to prove the existence of such a neighborhood using this very question (c.f. Exercise 2.3.16).

Answer 1

As suggested by your counterexample, you need to choose the sets $U_i$ more carefully so that they (and their closures) only contain the parts of $f(Z)$ that they are "supposed to" and don't "wrap around" like your $U_1$ does.

Specifically, you can choose the $U_i$ as follows. For each $z\in Z$, there is an open neighborhood $V_z\subseteq X$ of $z$ on which $f$ is invertible. Since $f|_Z:Z\to f(Z)$ is a homeomorphism, there exists an open subset $T\subseteq Y$ such that $T\cap f(Z)=f(V_z\cap Z)$. Let $U'_z=f(V_z)\cap T$, so $U'_z\cap f(Z)=f(V_z\cap Z)$ and $f$ has an inverse defined on $U'_z$ that maps into $V_z$. Now take the $U_i$ to be a locally finite refinement of these sets $U'_z$ by open sets whose closures are contained in the corresponding $U'_z$'s and still cover $f(Z)$.

[The motivation here is that by intersecting with $T$, we make sure that $U'_z$ contains only the parts of $f(Z)$ it is "supposed to". Then, by taking the $U_i$ have closures contained in the $U'_z$, we make sure that also their closures only contain the parts of $f(Z)$ they are "supposed to". This is needed to make sure that $W$ contains not just $f(Z)$ but a neighborhood of $f(Z)$.]

With these carefully chosen $U_i$, the set $W$ will indeed contain all of $f(Z)$, since if $U_i\subseteq U_z$ then $U_i$ can only intersect $f(Z)$ on $f(V_z\cap Z)$ so the chosen inverse $g_i$ will map that intersection back to $V_z \cap Z$. Moreover, $W$ will actually contain an open neighborhood of $f(Z)$. To prove this, fix $x\in Z$ and consider a $U_i$ whose closure contains $f(x)$. Then $\overline{U_i}\subseteq U'_z$ for some $z$ with $g_i$ being the inverse that maps $U'_z$ into $V_z$. Then $f(x)\in U'_z$ so $g_i(y)$ must approach $x$ as $y$ approaches $f(x)$ (since $U'_z\cap f(Z)=f(V_z\cap Z)$ so the inverse of $f$ on $U'_z$ must map all of $U'_z\cap f(Z)$ back to $V_z\cap Z$). This implies that any two of the $g_i$ for $U_i$ whose closures contain $f(x)$ must agree locally near $f(x)$. By local finiteness of the $U_i$, this means there is a neighborhood of $f(x)$ on which the $g_i$ always agree with each other, so that neighborhood is in $W$.

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More generally, this argument should prove the following statement. Suppose $X$ and $Y$ are topological spaces with $Y$ paracompact Hausdorff, $f:X\to Y$ is continuous, and $Z\subseteq X$ is such that $f|_Z$ is an embedding and each point of $Z$ has a neighborhood on which $f$ restricts to an open embedding. Then there is a neighborhood of the entire set $Z$ on which $f$ restricts to an embedding.