I was reading "Braid Groups" by Christian Kassel and Vladimir Turaev where I found the following question: Prove that an arbitrary geometric knot $L$ in an orientable 3-dimensional manifold has an open neighbourhood $U \supset L$ such that the pair $(U, L)$ is homeomorphic to ($\mathbb R^{2} \times S^{1}, \{ x\} \times S^{1}$), where $x\in \mathbb R^{2}$.
What I know:
A manifold of dimension $n$ is a Hausdorff, second countable space such that neighborhood of every point of the space is homeomorphic to an open subset of $\mathbb R^{n}$ (This property is called being locally Euclidean of dimension $n$).
A geometric link in a 3-dimensional topological manifold $M$(possibly with boundary $\delta$M) is a locally flat closed 1-dimensional submanifold of $M$.
A closed (compact and having empty boundary)1-dimensional manifold $L \subset M$ is said to be locally flat if every point of $L$ has a neighborhood $U \subset M$ such that the pair $(U , U \cap L)$ is homeomorphic to the pair ($\mathbb R^{3}, \mathbb R \times \{ 0\}\times \{ 0\}$)
Being a closed 1-dimensional submanifold, a geometric link $M$ must consist of a finite number of components homeomorphic to the standard unit circle. Now a geometric link consisting of only one component is said to be a geometric knot.
How do I proceed with the proof? Any hints...