# Generalization of Inverse Function Theorem to noncompact submanifolds

In Guillemin and Pollack's Differential Topology, they give as an exercise (#1.8.14) to prove the following generalization of the Inverse Function Theorem:

Use a partition-of-unity technique to prove a noncompact version of [the Inverse Function Theorem]. Suppose that the derivative of $f: X \to Y$ 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 $f(Z)$.

There is an answer here, but there is one thing I don't understand. I'll summarize the answer. Take a local diffeomorphism neighborhood $U_z$ around each $z \in Z$, giving an open cover $f(U_z)$ of $f(Z)$. Take a locally finite refinement $V_i$. Let $g_i$ be the local inverses defined on each $V_i$. For each $V_i \cap V_j$, let $W_{ij}:= \{y \in V_i \cap V_j: g_i(y) \neq g_j(y)\}$. The sets $\overline{W_{ij}}$ are locally finite, so their union is closed, so $(\cup_z f(U_z)) \setminus \ (\cup_{i,j}\overline{W_{i,j}})$ is an open set on which an inverse is defined, and it contains $f(Z)$.

My problem is: why does it contain $f(Z)$? Certainly $W_{ij} \cap f(Z) = \emptyset$ for all $i,j$, but why should $\overline{W_{ij}} \cap f(Z) = \emptyset$ for all $i,j$? In particular, why cannot a sequence of points in $W_{ij}$ converge to a point in $f(Z)$?

We can show that any two elements of the refined collection $\{g_i\}$ that are defined at $y$ have a nonempty open set around $y$ on which they agree:

Suppose we have two neighborhoods $V_i=f(U_i)$ and $V_j=f(U_j)$ of $y =f(z) \in f(Z)$ together with local inverses $g_i$ and $g_j$. Then we have $$g_i(V_i) \cap g_j(V_j)=U_i \cap U_j \subset X,$$ which is an open neighborhood of $z$ in $X$. Thus the set $$V=g_i^{-1}(U_i \cap U_j) \cap g_j^{-1}(U_i \cap U_j)$$ is an open neighborhood of $y$ in $Y$ on which $g_i$ and $g_j$ are both defined. Moreover, for $y' \in V$, we have $$f(g_i(y'))=y'=f(g_j(y'))$$ and thus $g_i(y')=g_j(y')$ because $f$ is one-to-one.

• (Update: I just changed the symbols and added in the emphasis to show that the resolution is a nice fact about topological inverses.)
– Kyle
Aug 31, 2015 at 21:09
• Why should $g(U)$ and $h(U)$ be open necessarily? For topological manifolds we can invoke invariance of domain, but why in general? Sep 1, 2015 at 6:26
• Ah, you're right. I've rolled back to my original answer and just tried to make the presentation clearer. That previous claim that $g(U)$ and $h(U)$ are open isn't true (even for manifolds!) as stated. My intuition was telling me that $f(g(U) \cap h(U))$ was open, but I'm not even sure if that's true or how you'd go about proving it. Any issues with the original answer (or it's rephrasing, as it appears now)?
– Kyle
Sep 1, 2015 at 18:56
• But this could be semantics; I don't know for sure which version/aspect of the claim you're responding to.
– Kyle
Sep 6, 2015 at 5:42
• I see. I thought we were assuming an inverse on an open set in the domain. My mistake. Sep 6, 2015 at 7:54

Suppose $f: X \to Y$ maps $Z \subset Y$ diffeomorphically onto $f(Z) \subset Y$.

Since $df_x: T_x(X) \to T_y(y)$ is an isomorphism for each $x \in Z$, there exist open sets $x \in U_x \subset X$ and $V_x \subset Y$ such that $f$ maps $U_x$ diffeomorphically onto $V_x$ by the Inverse Function Theorem presented as Exercise 1.3.10 of Guillemin-Pollack (the compact version).

The $\{V_x\}_{x \in X}$ forms an open cover of $f(Z)$, since $f(x) \in V_x$. Now, apply Exercise 1.8.13 of Guillemin-Pollack (every open cover of $\{V_\alpha\}$ of a manifold $X$ has a locally finite refinement) to take a locally finite refinement, which we will call $V_i$, with local inverse $g_i: V_i \to X$, such that $f \circ g_i = \text{Id}_{V_i}$.

Define $W = \{y \in Y: g_i(y) = g_j(y) \text{ whenever }y \in V_i \cap V_j\}$. On $W$, it is clear we can define a global inverse $g: W \to X$ by taking $g(y) = g_i(y)$ for any $i$ such that $y \in V_i$. This is well-defined on $W$ as $g_i(y) = g_j(y)$ whenever $y \in V_i \cap V_j$.

$W$ contains $f(Z)$, as $f$ maps diffeomorphically on $f(Z)$, so $g_i(y) = g_j(y) = f^{-1}(y)$ for any $y \in Z$. Now fix a $f(x) \in f(Z)$. We wish to show that $W$ contains an open neighborhood of $f(x)$. By the property that $\{V_i\}$ is a locally finite cover of $f(Z)$, there exists a neighborhood $V$ of $f(x)$ that intersects only finitely many of the $V_i$ – by reindexing them, if necessary, call them $V_1, \dots, V_k$.

Then $\widetilde{V} = V \cap V_1 \cap \dots \cap V_k$ is a finite intersection of open sets that contain $f(x)$, so $\widetilde{V}$ is an open neighborhood of $f(x)$. Moreover, on $\widetilde{V}$ each $g_i$ is a local diffeomorphism with $g(\widetilde{V}) \subset U_i \ni x$. Hence, on $\widetilde{V}$, the $g_i$ all agree, so $\widetilde{V} \subset W$. So we are done.

• I am convinced that the $g_i$ must agree on $\cap_i g_i^{-1}(\cap_i g_i(V_i))$, as @squirrel says, but I am not convinced that the $g_i$ agree on $\tilde{V}$. If I'm mistaken, could you clarify this point, perhaps? I.e. I am not convinced by your last sentence "Hence, on $\tilde{V}$...". $f$ is of course not assumed injective, so why could it not be that local diffeomorphisms defined on the same open set, and which are sections of $f$, and which share a point in their image, nevertheless disagree somewhere? Aug 31, 2015 at 20:50
• As a counterexample, take the neighborhood $(-.1, .1) \cup (\pi - .1, \pi + .1)$ on the circle, taken to either $(-.1, .1) \cup (\pi - .1, \pi + .1) \subset \mathbb{R}$ or $(-.1, .1) \cup (3\pi - .1, \pi + .1) \subset \mathbb{R}$. They are both sections of the covering map, both defined on the same open set, both local diffeomorphisms, share a point in their image, and yet do not agree everywhere. Aug 31, 2015 at 21:03

Eric is correct--the answer by user149792 is insufficient. Instead, we can modify his/her arguments easily by taking the $$V_i$$'s that contain the point $$y$$. Then, the local finiteness ensures there are only a finite number of $$V_i$$'s. Then, since the intersection of the images $$U_i$$'s ($$g_i(V_i)=U_i$$) is nonempty, by the injectivity of $$f$$, we conclude that all $$g_i$$ must agree on the intersection $$\cap_i V_i$$.