I am reading Dehn filling which is defined as follows in this lecture note(page 20): Let $M$ be a $3$-manifold and $T\subseteq \partial M$ be an embedded torus. For a homeomorphism $\varphi\colon \partial(\Bbb S^1\times \Bbb D^2)\to T$ define the Dehn filling $M(\varphi):=\frac{\displaystyle M\sqcup (\Bbb S^1\times \Bbb D^2)}{\displaystyle \varphi(p)\sim p}$.
I have a question given below related to showing that the homeomorphism type of $M(\varphi)$ depends only on the isotopy class of $\varphi(\mu)$ where $\mu:=\{\text{pt}\}\times \partial \Bbb D^2$.
The proof goes as follows: Take two homeomorphisms $\varphi_1,\varphi_2\colon \partial(\Bbb S^1\times \Bbb D^2)\to T$, let $\alpha_1:=\varphi_1(\mu),\ \alpha_2:=\varphi_2(\mu)$. Suppose, $\alpha_1$ is isotopic to $\alpha_2$. So, isotop $\varphi_1$ so that $\varphi_1=\varphi_2$ on a tubular neighborhood of $\mu$. .......
Question: How can one construct this isotopy? (So, isotop $\varphi_1$ so that $\varphi_1=\varphi_2$ on a tubular neighborhood of $\mu$.)
My Idea: $\alpha_1$ is isotopic to $\alpha_2$ implies there are homeomorphisms $H_s\colon T\to T,\ s\in [0,1]$ with $H_0=\text{Id}$ and $H_1(\alpha_1)=\alpha_2$. Now, $H_s\circ \varphi_1, s\in [0,1]$ gives an isotopy from $\varphi_1$ to $H_1\circ \varphi_1$ so that $H_1\circ \varphi_1(\mu)=\alpha_2$. So, without loss of generality we may assume $\varphi_1=\varphi_2$ on $\mu$, say $\alpha:=\varphi_1(\mu)=\varphi_2(\mu)$. Choose a tubular neighborhood $\mathcal N:=\mu\times [-2,2]\subseteq \partial(\Bbb S^1\times \Bbb D^2)$ of $\mu$, and $\varphi_1(\mathcal N)\cong \alpha\times [-2,2]$ gives a neighborhood $\alpha$. Now, by an isotopy push $\varphi_1\big|\mu\times [-2,0]$ to $\varphi_1\big|\mu\times [-2,-1]$ relative to $\mu\times \{-2\}$ and map $\mu\times [-1,0]$ onto $\alpha\times [-1,0]$ label-preserving fashion. Similarly, for $\varphi_1\big|\mu\times [0,2]$. Now, one has to do the same thing for $\varphi_2$.
Is my (vague)idea correct?