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.

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.

I'm studying the article "https://www.researchgate.net/publication/267464201_An_alternative_proof_of_Lickorish-Wallace_theorem" 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.

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.