I'm studying the article "An alternative proof of Lickorish–Wallace theorem" (doi link) and I got stuck in a problem.

Let $H_g$ be a 3 dimensional handlebody of genus $g$, a primate curve in $H_g$ is a knot in $\partial H_g$ that intersects an essential disk of $H_g$ in a single point. Let $c$ be a primitive curve, pushing $c$ in the interior of $H_g$ we obtain the knot $c'$. Now consider a spanning annulus $A$ in $H_g \setminus \eta(c')$ with $c \subset \partial A$, and the other boundary component of $A$ is called $c''$ and lies in $\partial \eta(c')$. How can I prove that if I perform a surgery on $c'$ along $c''$ I obtain a genus $g$ handlebody?

According to my notations, a surgery on $c'$ along $c''$ means glueing the meridian $\{x\} \times \partial D^2 \subset S^1 \times D^2$ on $c''$.

I found a similar question (Dehn surgery on handlebody), the answers (in particular the one by Ian Agol) seems to confirm that my statement is true, but there are no details.


1 Answer 1


Since $c\subset H_g$ intersects an essential disc $D$ in a single point, the boundary of a regular neighbourhood of $D\cup c$ is another disc $D'$, which splits $H_g$ into a solid torus containing $D\cup c$ and the rest. You can forget about the rest (this is a $\partial$-connected sum) and consider the solid torus alone. Here, if you push $c$ inside the solid torus, the complement will be diffeomorphic to $T \times [0,1]$ for a torus $T$, hence any Dehn surgery on one component will give you a solid torus back.


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