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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

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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.)

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  • $\begingroup$ 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. $\endgroup$ – Jan Vysoky May 10 at 20:57
  • $\begingroup$ Ah, I see, I misunderstood "closed". $\endgroup$ – Pavel May 10 at 21:00
  • $\begingroup$ Btw., I think that 3) is irrelevant because you can always shrink E to a tubular neighborhood $T$of $\nu(N)$ and set $F=\nu^{-1}(T)$. $\endgroup$ – Pavel May 10 at 21:20
  • $\begingroup$ 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... $\endgroup$ – Jan Vysoky May 10 at 21:36
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    $\begingroup$ I have found a reference with a very similar ideas here: staff.ustc.edu.cn/~wangzuoq/Courses/18F-Manifolds/Notes/… $\endgroup$ – Jan Vysoky May 13 at 5:46

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