Let $M$ be a compact smooth manifold (closed, for simplicity), $n\in\mathbb{N}$ and equip the space of embeddings $Emb(M,\mathbb{R^n})$ wih the Whitney-$C^{\infty}$-topology. (The weak and the strong one are the same as $M$ is assumed to be compact.) Fix an embedding $j\in Emb(M,\mathbb{R}^n)$ and a tubular neighbourhood of the embedded $j(M)\subseteq\mathbb{R}^n$.

Are small normal deformations of $j$ open in $Emb(M,\mathbb{R}^n)$?

To be more precise: Equip the normal sections $\Gamma(N(j(M))$ with the Whitney-$C^{\infty}$ topology and let $U\subset\Gamma(N(j(M))$ be the open (!) set of all sections $s$ such that the induced normal deformation $\{m+s(m)|m\in j(M)\}$ of $M$ lies in the tubular neighbourhood. Every such section $s\in U$ gives me an embedding $[m\mapsto j(m)+s(j(m)]\in Emb(M,\mathbb{R}^n)$. Do these embeddings form an open subset of $Emb(M,\mathbb{R}^n)$?

Some naive thoughts:

Consider a unit circle in the plane. Every open subset of the standard embedding of the circle should contain a small rotation of the circle, which does not come from a normal deformation of the circle. This suggests that the claim is false, but I believe to remember that I saw a flavor of this kind ("small normal deformations of embeddings are open") somewhere, although I cannot figure out where it was.

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    $\begingroup$ Perhaps you saw a statement claiming that the image of a set of small normal deformations is open in $Emb(M,\mathbb{R}^n)/Diff(M)$, where $Diff(M)$ acts by precomposition. That sounds more plausible. Otherwise, your counterexample shows that the other version of that statement is false. $\endgroup$ – Igor Khavkine Mar 25 '15 at 15:37
  • $\begingroup$ @IgorKhavkine That's it, thank you. Do you or somebody else have a reference or even a proof of this result? $\endgroup$ – Tom Mar 25 '15 at 15:43
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    $\begingroup$ @Tom A good reference is Theorem 42.1 and Theorem 44.1 in 'Kriegl, Michor: The convenient setting of global analysis'. You can download the book freely from Peter Michor's website. I know it's a bit scary book, but the geometric reasoning is, I believe, clear-ish from the proofs of the two theorems I mentioned. $\endgroup$ – Oldřich Spáčil Mar 25 '15 at 17:25
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    $\begingroup$ @OldřichSpáčil Thank you, I managed to prove the result with inspiration from the proof of Theorem 44.1 of 'Kriegl, Michor'. (Which has some flaws by the way...) $\endgroup$ – Tom Mar 26 '15 at 16:39

Not to leave this question open:

The question as posed is wrong by the example I gave. However the example I gave is in some sense the maximum which can go wrong. More precisely small normal deformations as defined in the question are open if one factors out diffeomorphisms of $M$, e.g. they are open in $Emb(M,\mathbb{R}^n)/Diff(M)$ as Igor Khavkine kindly pointed out to me.

A proof of the new claim can be constructed along the lines of the proof of Theorem 44.1 in 'Kriegl, Michor: The convenient setting of global analysis', which shows that $$Diff(M)\rightarrow Emb(M,\mathbb{R}^n)\rightarrow Emb(M,\mathbb{R}^n)/Diff(M)$$ is a fiber bundle. The local sections are exactly the normal deformations I considered. I am grateful to Oldřich Spáčil who gave me the hint digging into the proof referenced.

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    $\begingroup$ Maybe with less formalism, one can use the nearest neighbor projection $\pi : T\to j(M)$, well-defined from a tubular neighborhood $T$ of $j(M)$. Let $\ell$ be any embedding of $M$ which is $C^1$ close to $j$: then $\ell(M)\subset T$, and by the inverse function theorem $\pi\circ \ell$ is an embedding; since it has the same image as $j$, it differs from it by a diffeomorphism and you are done. $\endgroup$ – Benoît Kloeckner Mar 27 '15 at 9:59

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