Generalization of Inverse Function Theorem to Noncompact Submanifolds

differential-topologysmooth-manifolds

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 $f(Z)$?

Best Answer

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.