What is a 2-surgery on a disk? I am confused by a certain point in Scharlemann's paper "Sutured Manifolds and Generalized Thurston Norms", which seems important enough to not just skip it. I mean the "2-surgery on disks" in the first paragraph on page 7 (or 563). I will try to explain what is going on for those who don't have this paper, but first the questions I have:


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*What is this 2-surgery geometrically? Is there a picture somewhere?

*Why does it preserve homology classes?

*Why does it not increase the generalized Thurston norm?


So the situation is the following. We have two properly embedded surfaces in a 3-manifold that minimize the generalised Thurston norm. If there is a curve on one of the surfaces that bounds a disk in the surface without points of intersection of the $\beta$-graph in the disk (or the curve can be boundary parallel, but still without points of intersection with graph in the disk), we want to do this 2-surgery on this disk to eventually arrive at two surfaces that have no such curves of intersection. But we also want to not increase the generalised Thurston norm and preserve the homology classes. Can anybody help me to understand what is going on here? Thanks!
 A: Intuition / Motivation
Letting $S$ and $T$ be our surfaces, the point of the 2-surgery operation is to eliminate intersection curves $\gamma \subset S \cap T$ that bound disks in $S$ or $T$ in such a way that the resulting surface $\Sigma$ represents the same homology class as $S\cup T$. This is a step in showing that the generalized Thurston (semi-) norm satisfies the triangle inequality
$$\lVert a+b \rVert_{\beta}\leq \lVert a \rVert_{\beta} + \lVert b \rVert_{\beta}$$
So how do we show that $[S\cup T]$ has an embedded representative? The first step is to use the so-called "innermost disk" argument sketched in Thurston, William P., A norm for the homology of 3-manifolds, Mem. Am. Math. Soc. 339, 99-130 (1986). ZBL0585.57006.
Per question 3 in your list (on why the generalized Thurston norm doesn't change under 2-surgery on innermost disks): remember that the Thurston norm of an immersed surface $\Sigma$ is defined in terms of the homology class $[\Sigma]$ carried by $\Sigma$, so any operation $\tau$ on our surface that doesn't change the underlying homology class will not change the Thurston norm, i.e. $$[\Sigma]=[\tau(\Sigma)] \Rightarrow \lVert \Sigma \rVert_{\beta} = \lVert \tau(\Sigma) \rVert_{\beta} $$
Our goal is to find a compact, embedded surface that represents the "sum" (union) of two compact embedded surfaces $S$ and $T$ at the level of homology. 2-surgery on innermost disks and the double-curve sum are rather violent operations (they need to be since the goal is to turn a gnarly immersed surface into an embedded one) but they don't do anything at the level of homology representatives and the surface we get at the end of these two steps will be embedded.
I'll try to give a self-contained account for why the innermost disk procedure preserves the underlying homology class of a surface.
Innermost Disk Argument
Let $\Sigma := S \cup T$ where $S$ and $T$ are compact, embedded surfaces. If $\Sigma$ is embedded, we're done. Otherwise, we may generically assume that they intersect transversely in a compact 1-dimensional submanifold, i.e. a union of circles. The goal of 2-surgery is to replace $\Sigma$ with a new surface $\Sigma'$ which carries the same homology class but differs from $\Sigma$ in the sense that all circles in $S\cap T$ which bound disks in either $S$ or $T$ have been eliminated.
First, suppose I have a closed curve $\gamma\subset S\cap T$ which bounds a disk in either $S$ or $T$, say, $D\subset S$. This disk could contain another closed curve in its interior, so what we do is we go to the "innermost disk", by looking for a $\gamma' \subset (S\cap T)\cap D^{\circ}$ which itslef bounds a disk $D'\subset D \subset S$ such that no other component of $S\cap T$ is contained in the interior of $D'$ (that such a disk exists follows from compactness of $S$ and $T$).
2-surgery on the curve $\gamma'$ means (in Thurston's words) that we modify $T$ by attaching a copy of $D'$ on "either side" of $S$ so that the resulting surface, $T'$, no longer has the component $\gamma'$ in its intesection with $S$.
More explicitly: let $D'\times [-\epsilon, \epsilon]$ be a bicollar of $D'$ in the normal direction to $S$. Let $A(\gamma'):=\gamma' \times (-\epsilon,\epsilon)$ be a bicollar of $\gamma$ in $T$.
For $\alpha\in \{ \pm \epsilon\}$, let $\phi_{\alpha}$ be an attaching map taking $D'\times \{\alpha\}$ to $\gamma' \times \{\alpha\}$ we can construct
$$T':= \Big(T\setminus A(\gamma) \Big) \sqcup_{\phi_{\epsilon} \sqcup \phi_{-\epsilon}} \Big(D'\times \{-\epsilon\} \sqcup D'\times \{\epsilon\}\Big)$$
Here's the picture I have in mind:

The idea is that once we've done 2-surgery to eradicate the $\gamma'$ bounding the innermost disk, we can do 2-surgery again on the intersection curve bounding the next-to-innermost disk and so on until there are no more curves in $S\cap T$ bounding a disk in either $S$ or $T$.
In the end, all homologically trivial components of $S\cap T$ will be gone. If there are any components of $S\cap T$ left, then we take the "double-curve sum" to resolve the remaining components of $S\cap T$, so the result will be an embedded surface.
Invariance of homology class
Note that outside of a neighborhood $N$ of $\gamma'$, $S\cup T$ and $S\cup T'$ are identical, so if we can show that no cycles or boundaries were introduced in the interior of $N$, then we're done. Neither of the disks we attached contribute a cycle: they are bounded by $\gamma' \times \{-\epsilon\}$ and $\gamma' \times \{\epsilon\}$ respectively. Likewise, no 3-cell boundaries have been introduced.
