$\partial$-incompressibility of a surface obtained when attaching a 2-handle to an irreducible 3-manifold produces a reducible 3-manifold

Question

This question arises in my previous question.

Let $M$ be a compact, orientable, irreducible 3-manifold with incompressible boundary. Let $\alpha\subseteq \partial M$ be a simple closed curve, which is essential in $\partial M$. Let $N$ be obtained by attaching a 2-handle to $M$ along $\alpha$.

I notice in many articles the following fact, especially when the component of $\partial M$ containing $\alpha$ is a torus. Suppose $N$ is reducible, then there exists a properly embedded surface $(F,\partial F)\subset (M,\partial M)$ so that $\partial F$ is some parallel copies of $\alpha$ and $F$ is incompressible and $\partial$-incompressible. My question is: how do we guarantee that $F$ is $\partial$-incompressible?

Here is my understanding for the conclusion except for the $\partial$-incompressibility.

In fact, one can choose $\hat F$ among the reducible spheres in $N$ which intersects the 2-handle as some parallel copies of "meridian disks", so that the number of these meridian disks reaches the minimal, and let $F=\hat F\cap M$. The incompressibility of $F$ is easy to understand. Indeed, if $\gamma$ is an essential simple closed curve in $F$ which bounds a disk $D$ in $M\setminus F$, then $\gamma$ cuts $\hat F$ into two disks $D_1$ and $D_2$, each containing at least one component of $\partial F$. In addition, at least one of $D_1\cup D$ and $D_2\cup D$ is a reducing sphere of $N$, which intersects $\partial M$ with less components, and contradicts the "minimality" assumption. Yet, I have no idea why $F$ is $\partial$-incompressible in $M$.

Answer 1

Suppose that $S$, the relevant boundary component of $M$, is a torus. Suppose that $G$ is the given essential two-sphere in the filled manifold $N$. We isotope $G$ to have minimal intersection with $S$. All intersections $S \cap G$ are essential simple closed curves in $S$. So all of them are parallel. Set $F = M \cap G$. As you argue, $F$ is incompressible.

Suppose, for a contradiction, that $F$ is boundary compressible. Let $B \subset M$ be the given bigon boundary compressing $F$. Let $\beta = B \cap F$ and let $\beta' = B \cap S = B \cap \partial M$.

Let $\alpha$ and $\alpha'$ be the curves of $S \cap G$ which meet the corners of $B$ - that is, the points $\beta \cap \beta'$. If $\alpha = \alpha'$ then $\beta'$ is an inessential arc in $(S, \alpha)$. So $\beta'$ cuts a bigon out of $S - G$. Thus $B \cup B'$ is a compression of $F$, a contradiction.

Suppose instead that $\alpha$ and $\alpha'$ are distinct. Thus they co-bound an annulus $A$ in $S$. We choose $A$ so that $A \cap G = \alpha \cap \alpha'$. That is, there are no more curves of $G$ in $A$. Let $D$ be the disk obtained by boundary compressing $A$, along $B$, into $M$. This disk $D$ is a compressing disk for $F$ (or we can further reduce $G \cap S$). This is the desired contradiction.

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If $S$ is not a torus, the argument is more involved. I have given the details for that in my answer to your previous question.