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.