Let $M$ be a manifold. Let $B \cong S^1$ be its boundary. Assume there exists a diffeomorphism $\phi$ from some neighbourhood $U \subset M$ of $B$ to $S^1 \times [0,1)$.
$M$ can be embedded in four-dimensional space. Does there exist a diffeomorphism from $M$ to $f(M) \subset \mathbb{R}^4$ so that $f(U) = \{ (\cos(\theta), \sin(\theta), z,0) |\theta \in [0, 2 \pi], z \in [0,1) \}$?
I am intuitively completely convinced that such a diffeomorphism must exist, but practically rather stumped how to actually construct it using $\phi$. I.e. I don't know if $\phi$ can be 'extended' to include all of $M$ while retaining its original image on $U$.
It seems to be false, if I am not wrong. Regard the Möbius strip as a real projective plane with an open disc removed. This surface has $S^1$ as border and there is a diffeomorphism $\phi$ as the one you request.
However, the Möbius strip cannot be embedded into $\mathbb R^3$ as a cylinder because the former is non-orientable while the later is orientable. So there cannot be a diffeomorphism like the $f$ in the question.
However, I think we would have quite a different answer if we had more flexibility in the $z$ component of $f$. Were this the case, I am at this moment inclined to believe that all manifolds with $S^1$ as border can be embedded into $\mathbb R^3$, but at this moment I can think of no proof of this statement.
Apologies in advance if I turn out to be wrong.