# Local diffeomorphism on a neighborhood of an embedding

In my reading of the (excellent!) paper of Grabowski and Rotkiewicz on higher vector bundles (https://arxiv.org/abs/math/0702772), I have encountered the following argument which I do not understand. See proof of Theorem 2.1 in the above reference, paragraph starting with "Hence, due to the Implicit function theorem....", they use the following:

Let $$N \subseteq F$$ be a connected closed embedded submanifold and $$\mathcal{V}: F \rightarrow E$$ a smooth map (of connected smooth manifolds without boundary), such that:

1) $$\mathcal{V}$$ restricted to $$N$$ defines an embedding $$\mathcal{V}|_{N} \rightarrow E$$ whose image is a closed embedded submanifold of $$E$$;

2) $$\mathcal{V}$$ is has a one-to-one tangent map $$D_{n}\mathcal{V}$$ at all points $$n \in N$$, that is it is a local diffeomorphism at all points $$n \in N$$.

3) Maybe it is not very important, maybe it is: The manifold $$E$$ is in fact a vector bundle over $$N$$ and the image $$\mathcal{V}(N)$$ is preciesely the (image of) the zero section $$0_{N} \subseteq E$$.

The claim: There is an open neighborhood $$U_{N} \subseteq F$$ of $$N$$ and an open neighborhood $$W_{0} \subseteq E$$ of $$0_{N} \equiv \mathcal{V}(N)$$, such that $$\mathcal{V}: U_{N} \rightarrow W_{0}$$ is a (global) diffeomorphism.

Remark: They specifically say "Since $$\mathcal{V}|_{N}$$ is an embedding, we can say...", so the point 1) is supposedly crucial in this statement.

My ideas: Due to 2), for every $$n \in N$$, there is an open neighborhood $$U_{n}$$ of $$n$$ and $$W_{n}$$ of $$\mathcal{V}(n)$$, such that $$\mathcal{V}: U_{n} \rightarrow W_{n}$$ is a diffeomorphism. Then take $$U_{N} = \cup_{n \in N} U_{n}$$ and $$W_{0} = \cup_{n \in N} W_{n}$$. Then $$\mathcal{V}: U_{N} \rightarrow W_{0}$$ is clearly surjective. However, I struggle to prove that it is injective. Clearly one has to use 1) somehow, or maybe even 3). The only thing I was able to show was that one can choose $$U$$ and $$V$$ to be connected.

Any help, someone? I know that in general, surjective local diffeomorphisms are not diffeomorphisms (not even covering maps).

Thanks, Jan Vysoký

Lemma 1: Let $$N\subset F$$ be a compact embedded submanifold of a smooth manifold $$F$$, and let $$\nu: F \rightarrow E$$ be a smooth map which is injective on $$N$$ and a local diffeomorphism at every $$n\in N$$. Then there is a neighborhood $$U\subset F$$ of $$N$$ such that $$\nu: U \rightarrow E$$ is injective.

Proof: Let $$T(N)\subset F$$ be a tubular neighborhood of $$N$$. It is diffeomorphic to the normal bundle of $$N$$ in $$F$$, and hence we can pick a bundle metric on it. Also, we denote the smooth base point projection by $$\pi: T(N)\rightarrow N$$. For every $$n\in\mathbb{N}$$, set $$T_n:=\{v\in T(N) \mid |v|<\frac{1}{n}\}$$; this is an open neighborhood of $$N$$. For a contradiction, suppose that for every $$n\in\mathbb{N}$$ there are $$u_n$$, $$v_n\in T_n$$ such that $$u_n\neq v_n$$ and $$\nu(u_n)=\nu(v_n)$$ (i.e., $$\nu$$ is not injective on $$T_n$$). Since $$N$$ is compact, we can assume that $$\pi(u_n)\to u \in N$$ and $$\pi(v_n)\to v\in N$$ in $$N$$ as $$n\to \infty$$ (possibly after picking a subsequence). By construction, we have $$|u_n|$$, $$|v_n| \to 0$$, and hence $$u_n\to u$$ and $$v_n\to v$$ in $$F$$ (using the product structure of $$T(N)$$ in a neighbothood of $$u$$ and $$v$$, respectively). Because $$\nu$$ is continuous and $$\nu(u_n)=\nu(v_n)$$ for all $$n$$, we get $$\nu(u) = \nu(v)$$. Because $$\nu$$ is injective on $$N$$, it follows that $$u=v$$. We denote $$w:= u = v\in N$$. Let $$W\subset F$$ be an open neighborhood of $$w$$ such that $$\nu(W)\subset E$$ is open and $$\nu: W \rightarrow \nu(W)$$ is a diffeomorphism. There is an $$n_0\in \mathbb{N}$$ such that $$u_{n_0}$$, $$v_{n_0}\in W$$. By construction, we have $$\nu(u_{n_0})\neq \nu(v_{n_0})$$, which is a contradiction. QED.

Intersecting your $$U_N$$ and my $$U$$, we get a neighborhood of $$N$$ such that $$\nu: U_N\cap U \rightarrow \nu(U_N\cap U)$$ is a diffeomorphism (it is a bijective local diffeomorphism).

EDIT: It holds for non-compact manifolds as well using some topological tricks with compact exhaustions. I wonder if there is a better geometrical construction...

Lemma 2: Let $$\nu: F \rightarrow E$$ be a smooth map which is injective on a compact subset $$K\subset N$$ and a local diffeomorphism at every $$k\in K$$. Then there is a neighborhood $$U\subset F$$ of $$K$$ such that $$\nu: U \rightarrow E$$ is injective.

Proof: This is a variation of Lemma 1. In fact, one does not need the embedded submanifold $$N$$. One picks a system of neighborhoods $$U_n$$ of $$K$$ in $$F$$ such that $$\bar{U}_n$$ is compact, $$\bar{U}_{n+1} \subset U_n$$ and $$\bigcap_n U_n = K$$. This is possible since $$F$$ is a metric space. Suppose, for the contradiction, that there are $$u_n\neq v_n$$ in $$U_n$$ such that $$\nu(u_n)= \nu(v_n)$$. Because $$\bar{U}_1$$ is compact, we get $$\pi(u_n)\to u$$ and $$\pi(v_n)\to v$$ for some subsequence and some $$u$$, $$v\in \bar{U}_1$$. Because of $$(\bar{U}_{n+1})$$ being nested in $$(U_{n})$$, we have $$u$$, $$v\in \bigcap \bar{U}_n = \overline{\bigcap U_n} = K$$. One then proceeds as in the proof of Lemma 1. QED.

Lemma 3: Let $$\nu: F \rightarrow E$$ be a continuous map which restricts to a homeomorphism of embedded submanifolds $$N\subset F$$ and $$\nu(N)\subset E$$. Let $$(N_n)$$ be a compact exhaustion of $$N$$, i.e., $$N_n$$ for $$n\in \mathbb{N}$$ are compact sets such that $$N_n \subset \mathrm{int}(N_{n+1})$$ and $$\bigcup_n N_n = N$$. We set $$S_n:= N_n\backslash \mathrm{int}(N_{n-1})$$, where $$N_0 := \emptyset$$. Then there are neighborhoods $$U_n$$ of $$S_n$$ in $$F$$ such that $$\nu(U_n) \cap \nu(\bigcup_{|m-n|>1} U_m) = \emptyset.$$

Proof: Firstly, because $$\nu$$ is a homeomorphism, it maps the compact exhaustion of $$N$$ to a compact exhaustion of $$\nu(N)$$, and it intertwines the construction of $$S_n$$. Now, let $$T(\nu(N))$$ be a tubular neighborhood of $$\nu(N)$$ in $$E$$ isomorphic to the normal bundle. Because the manifold $$N$$, resp. $$\nu(N)$$ is a normal topological space, we can inductively construct neighborhoods $$V_n$$ of $$\nu(S_n)$$ such that $$V_n \cap \bigcup_{|m-n|>1} V_m = \emptyset$$ (we construct them such that $$\bar{V}_n\subset \mathrm{int}(\nu(N_{n+1}))$$ in every step). Now, let $$W_n:= T(V_n)$$, where $$T(V_n)$$ is the restriction of the normal bundle, resp. tubular neighborhood to $$V_n$$. Clearly, the family $$(W_n)$$ also satisfy the intersection property. It is easy to see that $$U_n:= \nu^{-1}(W_n)$$ have the desired properties. QED.

Lemma 4 (non-compact version of Lemma 1): Let $$N\subset F$$ be an embedded submanifold of a smooth manifold $$F$$, and let $$\nu: F \rightarrow E$$ be a smooth map which restricts to an embedding of $$N$$ and which is a local diffeomorphism at every $$n\in N$$. Then there is a neighborhood $$U\subset F$$ of $$N$$ such that $$\nu: U \rightarrow E$$ is injective.

Proof: Pick an exhaustion of $$N$$ by compact sets as in Lemma 3. Let $$U_n$$ be the neighborhoods of $$S_n$$ such that $$\nu(U_n)\cap \nu(\bigcup_{|n-m|>1}U_m) = \emptyset$$ for every $$n$$. By Lemma 2, we can find beighborhoods $$U_n'$$ of $$S_{n-1}\cup S_n \cup S_{n+1}$$ such that the restriction of $$\nu$$ to $$U_n'$$ is injective. We set $$W_n := U_n \cap U_n'$$. It is easy to check that $$\nu$$ is injective on $$U:= \bigcup W_n$$. QED

Now, as in the compact case, one intersects my $$U$$ with your $$U_N$$ and obtains the following:

PROPOSITION: Let $$\nu: F\rightarrow E$$ be a smooth map, and let $$N$$ be an embedded submanifold of $$F$$ such that $$\nu$$ restricts to an embedding of $$N$$ and such that $$\nu$$ is a local diffeomorphism at every $$n\in N$$. Then $$\nu$$ extends to a diffeomorphism of neighborhoods of $$N$$ and $$\nu(N)$$. (All manifolds are assumed to be Hausdorff and paracompact.)

• That is very nice, thank you! I understand why you need compactness in your proof, and I see not way around this at first glance. Unfortunately, I need it without this restriction - but I will definitely take note of your approach, thanks. – Jan Vysoky May 10 '19 at 20:57
• Ah, I see, I misunderstood "closed". – Pavel May 10 '19 at 21:00
• Btw., I think that 3) is irrelevant because you can always shrink E to a tubular neighborhood $T$of $\nu(N)$ and set $F=\nu^{-1}(T)$. – Pavel May 10 '19 at 21:20
• Yeah, I also think it is not important, that is why added this separately. However, it is entirely possible that the statement holds only for the particular $\mathcal{V}$ they are considering (although I don't think so), so it does not hold in full generality... – Jan Vysoky May 10 '19 at 21:36
• I have found a reference with a very similar ideas here: staff.ustc.edu.cn/~wangzuoq/Courses/18F-Manifolds/Notes/… – Jan Vysoky May 13 '19 at 5:46