The set of embeddings is open in the strong Whitney topology

Question

In Hirsch's book "Differential Topology," he claims in Chapter 2, Theorem 1.4 that the set of $C^1$-embeddings is open in the strong Whitney topology $C^1(M, N)$ where $M$ and $N$ are $C^1$ manifolds. My question is: How do you see that the set $\mathcal{N}_1$ is open? See below for the setup and description of $\mathcal{N}_1$.

$M$ and $N$ are $C^1$ manifolds and we have an embedding $f: M \to N$. We equip $M$ with a locally finite atlas $\{\phi_i, U_i\}$ of $M$ which admits compact sets $K_i \subseteq U_i$ whose interiors cover $M$ too. We find a cover $\{\psi_i, V_i\}$ of $N$ so that $f(U_i) \subseteq V_i$. A lemma says we can find numbers $\epsilon_i > 0$ so that any $C^1$ function $g: M \to N$ that is in the neighborhood $$\mathcal{N}_0 = \{ g : M \to N \,\vert\, (\forall i) (\forall x \in K_i)(g(K_i) \subseteq V_i)\text{ and }\, \lvert D^k(\psi_i f\phi_i^{-1} (x) - D^k(\psi_i f \phi_i^{-1}(x) \rvert < \epsilon_i \text{ for } k = 0,1 \} $$
satisfies $g\vert_{\mathrm{Int}(K_i)}$ is an embedding.

Since $f$ is an embedding, we can find open sets $A_i$, and $B_i$ in $N$ so that $f(K_i)\subseteq A_i$ and $f(M\setminus U_i) \subseteq B_i$ and $A_i\cap B_i = \emptyset$. The claim is that there is an open set $\mathcal{N}_1$ in $C^0(M,N)$ about $f$ such that if $g \in \mathcal{N}_1$ then $g(K_i) \subseteq A_i$ and $g(M\setminus U_i) \subseteq B_i$.

The result is certainly true if $M$ is compact, so I believe it is possible to extend it to the case where $M$ is not compact using the strong Whitney topology. I just can't get it to work.

Answer 1

A proof of this is also contained in Michor: Manifolds of mappings. There Proposition 5.3 establishes that the $C^r$-embeddings are $WO^1$ open in the space $C^r (M,N)$ (no restriction on the manifolds $M,N$). Thus if you set $r=1$ this implies that they are open with respect to the strong Whitney topology, as the $WO^1$-topology in Michor is coarser then the strong $C^1$-Whitney topology. For definitions and a comparison of the topologies see the book, it can be downloaded for free here: https://www.mat.univie.ac.at/~michor/manifolds_of_differentiable_mappings.pdf.

Going a bit in the details concerning the proof (answering to the comment by the OP, you have to decide whether this is an easy way to see that it is open): Michor defines the set $$B = \{g \in C^1(X,Y) : d(g(x),f(x)) <\varepsilon (x), d_1 (j_1 f (x),j_1 g (x))< \delta (x), g(K_\alpha) \subseteq A_\alpha , g(X\setminus U_\alpha ) \subseteq B_\alpha \text{for all }x \in X, \text{ for all } \alpha \}.$$

• For the first two conditions $d(g(x),f(x)) <\varepsilon (x), d_1 (j_1 f (x),j_1 g (x))< \delta (x)$ recall that $d,d_1$ are metrics adapted to the jet-bundles and $\varepsilon, \delta$ are continuous functions. Thus these conditions are open in $WO^1$ by the characterisation of the $WO^1$-topology in Michor 4.4.2.
• I claim that the conditions $g(K_\alpha) \subseteq A_\alpha , g(X\setminus U_\alpha ) \subseteq B_\alpha$ are open conditions in the $WO^1$-topology. To see this recall that $K_\alpha$ is compact, the $A_\alpha, B_\alpha$ and $U_\alpha$ are open with $K_\alpha \subseteq U_\alpha$ and the $U_\alpha$ form a locally finite-family. Now $WO^1$ is induced by an embedding from the so called $LO$-topology (recorded as 4.6 in Michor) and for the $LO$-topology, 3.7 Lemma in Michor states that the above conditions are open.

As pointed out by the OP the second point is still missing details. Since the family of closed sets $X \setminus U_\alpha$ will in general not be locally finite. We can not use for this condition as it is written the cited result. The additional information making this possible is that we have an embedding f to construct a family of neighborhoods which is locally finite and implements a condition similar to the one asked in Michors book. A writeup on how to do this can be found her https://www.math.uni-hamburg.de/home/latschev/lehre/ws17/embeddings.pdf

Thus in conclusion the set B is open in $WO^1$. I admit it is not pretty and easy, but that is the shortest explanation I could come up with in limited time.

Answer 2

I think the following works, but it is ad hoc. I would have suspected Hirsch had something more general in mind.

Because there is a tubular neighborhood of the embedding $f$ in $N$, we can work in an open subset of the zero section of the normal bundle $N_f(M)$. Because of this, I'll identify the points $f(x)$ with $x$ in the normal bundle. The open sets $A_i$ and $B_i$ can be taken to be of the form $\pi^{-1}(A'_i)$ and $\pi^{-1}(B'_i)$ where $A'_i$ and $B'_i$ are open subsets in $M$, with $M\setminus U_i \subseteq B'_i$ and $K_i \subseteq A'_i$.

We want to find $\epsilon$'s about each $x \in X$ so that all points in $B(x; \epsilon)$ in the normal bundle stay in $A_i$, whenever $x \in K_i$, and stay in $B_j$ whenever $x \in M \setminus B'_j$.

First step: If $x \in K_i$, there are only finitely many $K_j$ that intersect $K_i$ (if there were infinitely many, we can take a sequence of elements of $K_i$ that lie in some other $K_j$, this has a convergent subsequence with limit in $K_i$, but then this limit point lies in infinitely many other $K_j$, which is prohibited by local finiteness of the cover). So we can now measure the distance to any $x \in K_i$ to $N \setminus A_i$ to get $\epsilon_i$. We require that $x$ can't move more than $\epsilon_i$ then, but we must impose this condition for every $j$ that $x \in K_j$. We just showed there are only finitely many $j$ though, so there is a smallest positive of $\epsilon$ for each $x$.

Second step: Suppose now that $ x \in K_j$ and $x \in X \setminus U_i$. We can then measure the distance from $x$ to $N\setminus B_i$, call it $\epsilon_{i,j}(x)$ which is positive. We require that for a fixed $x$, for every $j$ and $i$ for which $x \in (X \setminus U_i) \cap K_j$ that we do not move from $x$ more than $\epsilon_{i,j}(x)$. The worry is that there is no positive $\epsilon(x)$ that is smaller than all of the $\epsilon_{i,j}(x)$. Let us check if this can happen.

We know that each $x$ lies in only finitely many $K_j$, but $x$ can lie in infinitely many of the sets $X \setminus U_i$. So for each $j$, consider if it is possible that there is a sequence of $i_k$ so that $x \in \setminus U_{i_k}$ with the distance $\epsilon_{j,i_k} \to 0$ as $k \to \infty$. If this were the case, then there would be a sequence of points $z_{i_k}\in N \setminus B_{i_k}$ in the normal bundle, lying over points $y_{i_k} \in M \setminus B'_{i_k}$ so that the distance $d(x, y_{i_k}) \to 0$ as $k \to \infty$. But then, as $x \in K_j \subseteq \mathrm{Int}(U_j)$, there are infinitely many values of $k$ for which $y_{i_k} \in U_j$. That is, local finiteness at $x \in K_j$ is contradicted. We therefore conclude that each $x \in X$ admits an $\epsilon(x)$ for which all points $z \in B(x; \epsilon(x))$ lie in all possible $N \setminus B_i$.

To finish up the argument, we just need to pick the smallest value of the $\epsilon$'s determined above on each compact set $K_j$. This then gives an open set $\mathcal{N}_1$ in the strong Whitney topology $C^0_S(M, N)$ for which all $g$ in this set satisfy $g(K_j) \subseteq A_j$ and $g(M \setminus U_j) \subseteq B_j$ for all $j$.