Is $n$-dim manifold which is immersed in $\mathbb{R}^n$ diffeomorphic to a ball?

Question

Let $M$ be an $n$-dimensional smooth compact oriented manifold with boundary which is immersed in $\mathbb{R}^n$ (codimension zero). Must $M$ be diffeomorphic to a ball with boundary (the closed unit ball in $\mathbb{R}^n$)?

Does anything change if we assume $M$ is simply-connected? or that it can be immersed in $\mathbb{R}^n$ without self-intersections?

Edit: There are (probably many) non-simply connected examples. It seems the interesting case is the simply-connected one.

Answer 1

Consider any emebdded compact submanifold of dimanions $k$ in $\bf R^n$, say $V$. Then a tubular neighbourhood $W$ of $V$ is simply connected, but not a ball (it has the same homotopy type than $V$. For instance if $V=S^2$ is the unit two-sphere in $\bf R^3$ the set $1/2\leq x^2+y^2+z^2 \leq 1+1/2$ is a compact simply connected manifold with the same homotopy type than a sphere, hence simply connected and not contractible. If the codimension of $V$ is at least two, the boundary of $W$ is connected.

Answer 2

Consider a closed annulus in $\mathbb{R}^2 $. It is not even homeomorphic to a closed disk.