# After Whitney embedding

Let $M$ be a $d$-dimensional compact manifold and $f:M\to M$ be a diffeomorphism on $M$. Whitney embedding theorem says that we can embed $M$ into $\mathbb{R}^{2d+1}$. Let $T$ be a tubular neighborhood of $M$ with respect to the embedding.

In some papers they say that there exists a diffeomorphism $F:T\to F(T)\Subset T$ with $F|_M=f$. Is there a theorem about this?

Any reference will be great!

You're asking for an extension of the diffeomorphism to a tubular neighbourhood of the manifold. The simplest proof I can think of would be to realize $f$ as an isotopy, and then apply the isotopy extension theorem. That $f$ is isotopic to the identity (as an embedding $M \to \mathbb R^{2d+1}$) isn't always true but if you consider $M$ to be in $\mathbb R^{2d+2}$, $f$ is isotopic to the identity -- this is another observation of Whitney's and has a similar proof.
In more generality this is an argument that tells you when two homotopic submanifolds of a given manifold are isotopic. Do you really need the $2d+1$ dimension range? It might be true there but a proof would likely be more subtle.
(in fact Whitney theorem says that we can embed $M$ to ${R}^{2d}$)
Let $N\to M$ be the normal bundle of the embedding $M\subset {R}^k$. We know that $N\oplus TM={R}^k$ is a trivial bundle on $M$. You want to know why the bundles $N$ and $f^*N$ are isomorphic. Unfortunately I can't prove this. However they are stably isomorphic: there in $n$ s.t. $N\oplus\mathbb{R}^n$ and $f^*N\oplus {R}^n$ are isomorphic. For $n$ we can take $k$. Namely, $f^*TM\cong TM$, hence $R^k\oplus N\cong f^*R^k\oplus N\cong f^*N\oplus f^*TM\oplus N\cong f^*N\oplus TM\oplus N\cong f^* N \oplus R^k$.
Altogether it means that if you embed $M$ into $R^k$ and $R^k$ standardly into $R^{2k}$ then you have your tubular neighbourhood isomorphism for the resulting embedding $M\to R^{2k}$. Perhaps it's sufficient for your purposes.