Tubular Neighborhoods vs. Regular Neighborhoods I am trying to understand the difference between a tubular neighborhood and a regular neighborhood. I have a theorem I want to use that applies to regular neighborhoods, and I am unsure if it will still apply to a tubular neighborhood.
Hudson's definition for a regular neighborhood is: Let $X$ be a polyhedron contained in the p.l. manifold $M$, $N \subset M$. $N$ is a regular neighborhood of $X$ in $M$ if:
1. $N$ is a closed neighborhood of $X$ in $M$.
2. $N$ is an m-manifold
3. $N$ collapses to $X$
(From Hudson, Piecewise Linear Topology)
Rolfsens definition for a tubular neighborhood is: The tubular neighborhood of a submanifold $M^m \subset N^n$ of another manifold $N^n$ is an embedding $t: M \times B^{n-m} \rightarrow N$ such that $t(x,0)=x$ whenever $x \in M$
(From Rolfsen, Knots and Links)
My issue is that tubular neighborhoods are open, whereas regular neighborhoods are closed. My understanding is that the two concepts are the same and tubular neighborhoods are a special case of regular neighborhoods.
For reference: I am working in a compact, orientable 3 manifold.
 A: Given an open tubular neighborhood $t:M\times B^{n-m}\to N$, then given a closed disk $D^{n-m}\subset B^{n-m}$ centered at $0$, you can restrict to a closed tubular neighborhood $t:M\times D^{n-m}\to N$.  (By the way, this definition of tubular neighborhood is more restrictive than usual; the normal bundle of $M$ in $N$ need not be a product $M\times B^{n-m}$.  One example is a knot in a non-orientable $3$-manifold, where a tubular neighborhood might be a solid Klein bottle.)
But, to be a regular neighborhood, it really ought to be a polyhedron.  For $3$-manifolds, every triangulation may be regarded as being a smooth triangulation in a unique way (up to diffeomorphism).  Supposing you're talking about a knot $K$ in an oriented $3$-manifold $M$, then if $N(K)$ is a closed tubular neighborhood of $K$, one can arrange for a triangulation of $M$ where $K$ lies in the $1$-skeleton, $\partial N(K)$ lies in the $2$-skeleton, and $N(K)$ collapses to $K$.  Hence, $N(K)$ is a regular neighborhood with respect to this triangulation.
