# Glueing manifolds with boundaries and Seifert-Van Kampen theorem

I've seen many times the following application of the SVK theorem:

Let $M$ and $N$ two smooth $n$-manifolds ($n\ge 3$) with boundary and suppose that they have the same boundary $B$. Now, after glueing $M$ and $N$ along $B$ we obtain: $$X=M\cup_{B} N$$ At this point we can apply the SVK to the triple $M$, $N$, $M\cap N=B$ in order to calculate the fundamental group of $X$.

The problem is that $M$, $N$ and $B$ are in general not open in $X$ (once that they are glued together). On the other hand we know that the openness condition is necessary in the proof of SVK theorem.

• While the openness condition in SvK's theorem is essential (and there are counterexamples otherwise), but unless you're in a pathological case, it's often possible to find a slightly bigger open set that deformation retracts onto the subspace you want, just like in Michael Albanese's answer. Jan 8 '16 at 9:35

Let $M$ be a smooth manifold with boundary. A neighborhood of $\partial M$ is called a collar neighbourhood if it is the image of a smooth embedding $[0, 1)\times\partial M \to M$ that restricts to the identification $\{0\}\times\partial M \to \partial M$.

Every smooth manifold with nonempty boundary has a collar neighbourhood - see Lee's Introduction to Smooth Manifolds (second edition), Theorem $9.25$.

As $B$ is the boundary of $M$, it has a collar neighbourhood $C \subset M$ and as $B$ is the boundary of $N$, it has a collar neighbourhood $D \subset N$. Then $C\cup D$ becomes an open neighbourhood of $B$ in $X = M\cup_B N$. Furthermore, $C\cup D$ deformation retracts onto $B$, so $\pi_1(C\cup D, b_0) \cong \pi_1(B, b_0)$ for all $b_0 \in B$.

So we really want to apply the Seifert-van Kampen Theorem to the following open sets of $X$: $M\cup D$ (which deformation retracts to $M$), $C\cup N$ (which deformation retracts to $N$), and $C\cup D$ (which deformation retracts to $B$).

Note, it doesn't matter which collar neighbourhoods we take as the fundamental groups of the resulting open sets are independent of $C$ and $D$.

• Wait a moment, now $C\cup D$ is open but $M$ and $N$ are not open. Maybe you want to apply SVK to $M\cup D$, $N\cup C$, $C\cup D$ Jan 8 '16 at 1:37
• Yes, I just realised I was being a bit sloppy. What you suggest should be fine. Jan 8 '16 at 1:38