# Why the boundary of $M_1\# M_2$ is incompressible if the boundaries of $M_i$ are incompressible?

This may be a stupid question, because I am completely new to $3$-manifolds and get stuck (and honestly, a bit confused) in the following problem.

Let $M_i$ ($i=1,2$) be $3$-manifolds with boundary and $\partial M_i$ is incompressible in $M_i$. Then $\partial M$ is incompressible in $M$, where $M=M_1\# M_2$.

Since $\partial M_i$ is incompressible in $M_i$, from the definition we know $\pi_1(\partial M_i)\hookrightarrow \pi_1(M_i)$, which is induced by the inclusion $\partial M_i\hookrightarrow M_i$. For the connected sum of two $3$-manifolds, the fundamental group is their free product $\pi_1(M)=\pi_1(M_1)\ast\pi_1(M_2)$. To show that $\partial M$ is incompressible in $M$, I need to show the inclusion $\partial M\hookrightarrow M$ induces a monomorphism $\pi_1(\partial M)\hookrightarrow\pi_1(M)$. However, I don't know what $\pi_1(\partial M)$ looks like? Is $\pi_1(\partial M)$ simply the free product of $\pi_1(\partial M_1)$ and $\pi_1(\partial M_2)$? As I remembered, the fundamental group of the connected sum of two $2$-manifolds need not be the free product of their fundamental groups. Or how can I link the monomorphisms $\pi_1(\partial M_i)\hookrightarrow \pi_1(M_i)$ to the verification of $\pi_1(\partial M)\hookrightarrow\pi_1(M)$?

Any help is appreciated.

• use loop theorem – Anubhav Mukherjee Jul 13 '18 at 14:34

## 1 Answer

This can be done with a cut and paste argument. If $\Sigma$ is the copy of $S^2$ separating punctured copies of $M_1$ and $M_2$, and $D$ is a compression disk for the boundary of $M$, then consider the intersection $D \cap \Sigma$. If this intersection is generic, it is a union of simple closed curves, and we can find such a curve $C$ that is innermost on $\Sigma$. This curve also bounds a disk $D'$ in the interior of $D$, so we can remove $D'$ and replace it with the inner disk of $\Sigma$ bounded by $C$ to get a new disk $D$. Pushing the new $D$ off of $\Sigma$ a little bit then reduces the number of intersection curves, and this process can be repeated until there are no curves of intersection. But if there are no intersection curves, $D$ lies entirely in the punctured $M_1$ or $M_2$, so it gives a compressing disk for $M_1$ or $M_2$, which is a contradiction.

As for doing it with $\pi_1$ injectivity, the reason you are having trouble is that that boundary is not connected in general (in fact, it may not have been connected for $M_1$ and $M_2$). $M$ has a boundary component for each boundary component of $M_1$ and $M_2$, and they don't get connect summed together because the connect summing happens away from the boundary. For a given component, say assuming $M_1$ and $M_2$ have one boundary component each, it injects in the "obvious" way like $\pi_1(\partial M_1)\hookrightarrow \pi_1(M_1) \hookrightarrow \pi_1(M_1) \ast\pi_1(M_2)$

• I think I have understood what to do:) Thanks! – josephz Jul 13 '18 at 10:39