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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 .

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.

My attempt:

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.

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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.

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  • $\begingroup$ Thank you so much! Does my argument happen to be sound? I.e. is my proof-attempt reasonable enough? (I apologize for this question, I'm fairly new to geometric topology). $\endgroup$
    – Zest
    Commented Apr 18, 2022 at 19:00
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    $\begingroup$ Yes. The union of the neighbourhoods of the two disks is a three-ball, and this three-ball contains the support of the desired isotopy. To prove all of this carefully (in particular, to describe the isotopy) requires some work. It is much easier to draw lots of pictures. $\endgroup$
    – Sam Nead
    Commented Apr 18, 2022 at 19:46
  • $\begingroup$ That's all I could have asked for. I really appreciate your feedback! $\endgroup$
    – Zest
    Commented Apr 18, 2022 at 20:00

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