I sincerely apologize if MathOverflow is not the appropriate place to ask this question. I also tried consulting M.SE but it seems that this question gained little to no interest .




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Consider a Heegaard splitting $M = V\cup\_F W$ of a $3$-manifold $M$ with splitting surface $F = \partial V = \partial W$. Suppose there is a meridian pair $\{D\_1,D\_2\}$, that is, $D\_1$ is a meridian disk in $V$ and $D\_2$ a meridian disk in $W$. Both disks intersect transversaly so that their boundary curves $\partial D\_1$ and $\partial D\_2$ intersect in exactly one point.


Now I would like to verify that $V\setminus N(D\_1)$ is ambient isotopic to $V\cup N(D\_2)$ where $N(\cdot)$ denotes a regular neighbourhood. But I'm somewhat new to geometric topology and I am not sure on how to proceed.




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My attempt:
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What I know is that both disks $D\_i$ are compression disks in $V$ respectively $W$, thus we can compress along $D\_1$ in $V$, that is, we cut at the regular neighbourhood $N(D\_1)$ of $D\_1$ in order to obtain $V\setminus N(D\_1)$.


Now the union of $N(D\_1)$ and $N(D\_2)$ along the square in which they intersect is a $3$-ball and so we can isotope "through" that ball.


But now I'm unsure if this is the correct attempt and if this already shows that $V\setminus N(D\_1)$ is ambient isotopic to $V\cup N(D\_2)$.


Another Idea was to somehow construct the deired isotopy in a "pedestrian way" by considering a collar of the boundary $\partial V$ and somehow extending an isotopy from the boundary to the entire handlebody $V$ but so far I haven't had much success with this approach either.


Any feedback would be highly appreciated! Thanks in advance.



Given a Heegaard splitting $M = V\cup_F W$, then $V\setminus N(D_1)$ is ambient isotopic to $V\cup N(D_2)$ for a meridian pair $\{D_1,D_2\}$
Yes, you are correct that the two handlebodies ($V$ cut along $D$ and $V$ plus a neighbourhood of $E$) are isotopic. This is discussed when defining *stabilisation* and the inverse operation *destabilisation* of Heegaard splittings. I don’t have a copy of Hempel’s book with me, but I am fairly sure it is discussed there.


