In a Riemannian manifold consider two compact smooth submanifolds $S$, $S^\prime$ that intersect transversely. It seems intuitively obvious that for a sufficiently small number $r$, the union of $r$-neighborhoods of $S$ and $S^\prime$ is a regular neighborhood of $S\cup S^\prime$ in some triangulation of the ambient manifold. Is this written somewhere?

By the way, a standard reference for regular neighborhoods is this paper by Marshall Cohen, published in Trans. Amer. Math. Soc. 136, 1969 189--229.

EDIT: it may be worth discussing why any small $r$-neighborhood $N_r(S)$ of a smooth compact submanifold $S$ is a regular neighborhood. This is written on page 527 of [M. Hirsch, Smooth regular neighborhoods., Ann. of Math. (2) 76 1962 524--530]. First, one uses results of J. Whitehead to show that $N_r(S)$ is a subcomplex in some smooth triangulation. By choosing $r$ small enough we arrange that $N_r(S)$ is inside the second derived neighborhood $R$ of $S$, i.e. the star of $S$ in the second barycentric subdivision. The second derived neighborhood is always a regular neighborhood. Furthermore, Hirsch says $R$ is the mapping cylinder of a retraction $\partial R\to S$, and each radial segment in the cylinder is transverse both to $S$ and to $\partial R$. This transversality implies that for small $r$ removing the interior of $N_r(S)$ from $R$ gives $\partial R\times I$, so that $N_r$ is also a regular neighborhood of $S$.

The same argument may work for $S\cup S^\prime$. In fact it is believable that the second derived neighborhood $R$ of any subcomplex $K$ can be written as a mapping cylinder whose radial segments are PL-transverse to $\partial R$ and $K$. If true, we just need to see that the segments are also transverse to the boundary of the union of $r$-neighborhoods of $S$ and $S^\prime$.

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    $\begingroup$ What definition of regular neighbourhood do you want to use? It sounds like you're in a mixed smooth and PL category? In the smooth category (where I think your question might have a simple answer) a regular neighbouhood of a subset $X$ can be taken to be a smooth neighbourhood $U$ that is a mapping cylinder from the boundary manifold $\partial U$ to the set $X$. If that's all you're after, then I think you can construct the regular neighbourhood from the tubular neighbourhood of your indivitual manifolds. $\endgroup$ – Ryan Budney Jul 19 '10 at 4:12
  • $\begingroup$ The submanifolds are smooth, but the union of their $r$-neighborhoods is not a smooth submanifold, hence we are forced to leave smooth category. I am using the usual PL regular neighborhoods. You are correct that it is enough to realize the neighborhood as a mapping cylinder of some retraction; how do you do that? $\endgroup$ – Igor Belegradek Jul 19 '10 at 4:21

You're not really forced to leave the smooth category at an earlier stage than before. There is a good pseudogroup for the category of "smooth manifolds with transverse submanifolds". If $r$ is small enough, then $S$, $S'$, $\partial N_r(S)$, $\partial N_r(S')$ are all transverse, so what you have exists in that category. Of course there is more work to do than just this remark.

For simplicity, suppose that the normal exponential maps of $S$ and $S'$ commute up to some radius $r$. Then the local structure works out exactly as you expect, and it should be routine to generalize Moe Hirsch's argument. Suppose instead that the normal exponential maps do not commute. Then using the implicit function theorem, you can bend the normal exponential map of $S'$ out to a radius of $r' < r$ to make it commute with normal exponential map of $S$. You can do this in such a way that $\partial N_{r'}(S')$ is not moved. (Well, I think so. I admit that I'm handwaving some.) Then it should work as in the simpler case.

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