I am trying to understand the definition of irreducible $3$-manifold. Let $M$ be a connected $3$-manifold.
Definition 1: We say $M$ is irreducible if given any smoothly embedded submanifold $S$ of $M$ with $S\cong \Bbb S^2$, there exists a smoothly embedded submanifold $B$ of $M$ such that $B\cong \Bbb B^3$ and $\partial B=S$.
Here, $\cong$ stands for being homeomorphic, $\Bbb S^2:=\{x\in \Bbb R^3:||x||=1\}$, and $\Bbb B^3:=\{x\in \Bbb R^3:||x||\leq1\}$. Note that for $\dim\leq 3$, the notion of being homeomorphic and being diffeomorphic are the same.
Definition 2: We say $M$ is irreducible if given any smoothly embedded submanifold $S$ of $M$ with $S\cong \Bbb S^2$, there exists a subset $B$ of $M$ such that $B\cong \Bbb B^3$ and $\partial B=S$.
See this https://en.wikipedia.org/wiki/Prime_manifold#Irreducible_manifold
Now, note that if $M$ is irreducible in the sense of Definition 1 (for example $\Bbb R^3$), then it's irreducible in the sense of Definition 2. What about other direction?
Question: Let $M$ be irreducible in the sense of Definition 2. Is it then irreducible in the sense of Definition 1? In other words, Suppose we are given a three-manifold $M$ and a smoothly embedded submanifold $S$ of $M$ such that the following hold: $(1)$ $S$ is homeomorphic to $\Bbb S^2$, $(2)$ There is a subspace $B$ of $M$ so that $B$ is homeomorphic to $\Bbb B^3$ with $\partial B=S$. Is $B$ a smoothly embedded submanifold $M$?
I guess that these two definitions are equivalent, because if I have a subset $B$ with $B\cong \Bbb B^3$ and $\partial B=S$, then taking smooth tubular neighborhood of $S$ in $M$, one can give a smooth structure so that $B$ is a smoothly embedded submanifold of $M$.