Attaching a thickened annulus between two 3-manifold

Question

Let $X$ and $Y$ be compact, orientable 3-manifolds both with incompressible boundary. Pick a non-contractible simple closed curve on a boundary component of both $X$ and $Y$ and attach a thickened annulus $A \times I$ between them. Is it true that the boundary of the resulting manifold $X \cup Y \cup (A \times I)$ is incompressible? Any help is appreciated.

Answer 1

The answer is 'yes'. Furthermore, this is more or less equivalent to a group-theoretic fact, which applies in much greater geneality, called Shenitzer's Lemma.

First, note that we may assume that $X$ and $Y$ are irreducible, since a standard innermost disc argument shows that any compressing disc can be made disjoint from any essential sphere.

We can now translate into group theory using the following lemma. Recall that, by Stallings' Ends Theorem, a finitely generated group $G$ is one-ended if and only if $G$ is infinite and does not split non-trivially (as an amalgamated product or HNN extension) over a finite subgroup.

Lemma: Let $M$ be a compact, orientable 3-manifold with non-empty boundary and infinite $\pi_1M$. Then $\pi_1M$ is one-ended if and only if $\partial M$ is incompressible.

Proof: If $\partial M$ is compressible, then cutting along the compressing disc realizes a splitting of $\pi_1M$ over the trivial subgroup. Conversely, if $\pi_1M$ splits over a finite subgroup, then by a standard argument of Stallings--Epstein--Waldhausen, the splitting can be realized by cutting $M$ along some properly embedded, essential surface $\Sigma$. Since $\pi_1\Sigma$ is finite, $\Sigma$ is either a disc or a 2-sphere; but since $M$ is irreducible, $\Sigma$ must be a disc, and the boundary is incompressible. QED

Since we may assume that $X$ and $Y$ are irreducible, and since $\pi_1X$ and $\pi_1Y$ are infinite, the answer now follows immediately from Shenitzer's Lemma, a very useful fact in group theory which explains exactly when an amalgam over $\mathbb{Z}$ is one-ended. See, for instance, Theorem 18 here, or this paper of Touikan for a comprehensive modern treatment.

In fact, you only need the following weaker fact, which can be proved directly using Bass--Serre theory.

Weak Shenitzer's Lemma: If $G=A*_{\mathbb{Z}} B$ and $A,B$ are one-ended then so is $G$.

Proof: It's enough to show that $G$ does not split non-trivially over a finite subgroup. Suppose therefore there is such a non-trivial splitting, and let $T$ be the corresponding Bass--Serre tree. Since $A$ and $B$ are one-ended and infinite, they each fix unique vertices $a$,$b$. But then $\mathbb{Z}$ fixes every edge on the geodesic between $a$ and $b$, so since edge stabilizers are finite, $a=b$. Therefore this is a global fixed point for $G$, and the splitting was trivial. QED

The full force of Shenizer's Lemma would give you the slightly stronger statement that it's sufficient to have every compressing disc crossed by your closed curve on the boundary.

The proof of Shenitzer's lemma is basically a group-theoretic version of the kind of geometric argument that @ThiKu outlines in the other answer.

Answer 2

Let me try to give a more elementary argument.

Assume there is a compression disk. One can assume it is transversal to $A\times\left\{\frac{1}{2}\right\}$, thus it intersects $A\times\left\{\frac{1}{2}\right\}$ in a collection of circles and arcs.

The complement of these circles and arcs in the disk is a union of topological subdisks whose boundaries all lie in $\partial X^\prime$ and $\partial Y^\prime$ for $X^\prime=X \cup A\times\left[0,\frac{1}{2}\right]$ and $Y^\prime=Y\cup A\times\left[\frac{1}{2},1\right]$.

The obvious deformation retractions $X^\prime\simeq X, Y^\prime \simeq Y$ gives incompressibility of $\partial X^\prime,\partial Y^\prime$ as a consequence of incompressibility of $\partial X$ and $\partial Y$.

This means that all the subdisks can be homotoped to lie either in $\partial X^\prime$ or in $\partial Y^\prime$. Some inductive argument (induction over the number of subdisks) then gives that the whole disk can be homotoped into $\partial (X\cup Y\cup A\times I)$, which proves incompressiblity.