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Let $M$ be a connected closed smooth manifold and $B(1)$ the unit ball in $\mathbb R^n$. Suppose $f:M \times B(1) \to M \times \mathbb R^n$ is a smooth embedding.

Can we find an ambient isotopy $F_t$ of $M \times \mathbb R^n$ such that $F_0=\text{id} \times \text{id}$ and $F_1 \circ f=h\times \text{id}$, where $h$ is a self-diffeomorphism of $M$.

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    $\begingroup$ You need some more assumptions there to rule out things like $f(p,x)=(g(p),x)$ for $g\colon M \rightarrow M$ a diffeomorphism that is not homotopic to the identity even after crossing with a large enough euclidean space (e.g. for $g$ that acts nontrivially on the homology of $M$). $\endgroup$
    – Andy Putman
    Commented Jun 11, 2020 at 3:40
  • $\begingroup$ No, you can not. The simplest example I can imagine is $M=S^0$ and $f$ be any embedding which switches the components. $\endgroup$
    – Ryan Budney
    Commented Jun 11, 2020 at 3:43
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    $\begingroup$ @RyanBudney: This is definitely false (see my comment), but the OP did insist that $M$ be connected. $\endgroup$
    – Andy Putman
    Commented Jun 11, 2020 at 3:45
  • $\begingroup$ In view of Andy's comment, you should also allow for $F_1\circ f$ to be of the form $h\times id$, where $h$ is a self-diffeomorphism of $M$. $\endgroup$
    – Moishe Kohan
    Commented Jun 11, 2020 at 4:05
  • $\begingroup$ @AndyPutman: That just pushes the problem one dimension up, let $M=S^1$ and $f$ be conjugation on $S^1$. Your initial response appeared at essentially the same time as my own so I missed it. $\endgroup$
    – Ryan Budney
    Commented Jun 11, 2020 at 7:11

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That's not true as stated. $h$ being a diffeomorphism implies that $F_1 \circ f$ is a homotopy equivalence, thus $F_0 \circ f$ also needs to be a homotopy equivalence. One can easily construct counterexamples, such as $$S^1 \times [-1,1] \to [-1,1] \times [-1,1] \to S^1 \times \mathbb{R}.$$ Thinking of $S^1 \subset \mathbb{C}$, the first map can be $ (z,t) \mapsto (z \cdot e^{t-1})$ and the second one $ (t,x) \mapsto (e^{it},x)$. Since the first map is contractible, the composition is contractible. Both maps are smooth embeddings so the composition is as well.

Here's a picture. Contractible embedding of S^1 \times [-1,1] into S^1 \times \mathbb{R}.

In such case you can't even hope for $h$ to be a topological embedding since that would be a homeomorphism. But in this case you can get some smooth $h$.

And of course asking $f$ to be a homotopy equivalence, or homotopic to identity (both spaces are homotopic to $M$), would yield an interesting question.

EDIT: Even requiring that $f|_{M \times \{0\}} = \mathrm{id} \times \{0\}$ (and thus $f$ is homotopic to identity) would not be sufficient. Say this time that $M=S^1$ and $n=2$, and take the map $ S^1 \times D^2 \owns (z,w) \mapsto_f (z,zw) \in S^1 \times \mathbb{C} = S^1 \times \mathbb{R}^2 $. If you view $S^1 \times \mathbb{R}^2$ as a solid torus embedded in $\mathbb{R}^3$, then $f( S^1 \times \{0\})$ and $f(S^1 \times \{1/2\})$ have linking number $1$ and no isotopy can unlink them.

So, you also need to assume at least that $Df$ as a bundle map is, in a suitable sense, homotopic through embeddings to identity. Precisely, you need to assume that there is a homotopy $f_t:M \times B(1) \to M \times \mathbb{R}^2$ between $f$ and identity and a homotopy $F_t : TM \times \mathbb{R}^n \to TM \times \mathbb{R}^n$ between $Df$ and identity of bundle maps lifting $f_t$.

At this point it sounds plausible to me, probably some $h$-principle machinery may be used to prove this.

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