Let $\Sigma$ be a topological surface (compact and orientable, if necessary). Let $C$ be a subspace of $\Sigma$ homeomorphic to the annulus $A := [0,1] \times S^1$, together with a homeomorphism $h: A \to C$. We define the Dehn twist supported on $C$ to be the homeomorphism $T_C: \Sigma \to \Sigma$ defined by $T_C(p) = h \circ \tau \circ h^{-1}(p)$ if $p \in C$, and $T_C(p) = p$ otherwise. Here, the map $\tau: A \to A$ is defined as $\tau(r,\theta) = (r,\theta e^{2 \pi i r})$.
In turn, we define the characteristic loop of $C$ to be the map $\gamma_C: S^1 \to \Sigma$ given by $\gamma_C(\theta)=h(0,\theta)$.
I would like to show the following basic result:
Given two annuli $C_0$ and $C_1$ of $\Sigma$ as above, $T_{C_0}$ and $T_{C_1}$ are isotopic if and only if $\gamma_{C_0}$ and $\gamma_{C_1}$ are isotopic.
However, I was unable to give a proof, and in fact, I think the result relies on nontrivial topological fact. For instance, to prove the sufficient condition I would argue as follows. Let $g_t$ be an isotopy from $T_{C_0}$ to $T_{C_1}$. Let $C_t$ be the support of $g_t$. If each $C_t$ was homeomorphic to $A$, and moreover, such homeomorphisms could be chosen to be continuous in $t$, then we would be done, since we would have a map
$$\tilde{g}: I \times A \to \Sigma,$$
and restricting to $r=0$ would give us the desired isotopy between $\gamma_{C_0}$ and $\gamma_{C_1}$. However, I think that claim is highly nontrivial.
I also checked the book A Prior on Mapping Class Groups, by Farb and Margalit, but their argument is not elaborated.
In summary, I would like a reference or a proof of the result mentioned above.
Thank you in advance for reading!
Here is a proof of the sufficient condition.
For each homeomorphism $f : \Sigma \to \Sigma$ and each nontrivial simple closed curve $\gamma \subset \Sigma$ one may ask whether $f(\gamma)$ is isotopic to $\gamma$. In other words, one may ask whether the isotopy class of $\gamma$, which I'll denote $[\gamma]$, is fixed by $f$. Let me use $Fix(f)$ to denote the set of all such $[\gamma]$ that are fixed by $f$.
Next, one shows that if $f,g : \Sigma \to \Sigma$ are isotopic homeomorphisms then $Fix(f) = Fix(g)$.
Next, one shows that for any annulus $C \subset \Sigma$, the set $Fix(T_C)$ consists of those $[\gamma]$ such that the minimal geometric intersection number of the core curve of $C$ with curves in the class $[\gamma]$, a number that I will denote $i(C,\gamma)$, is equal to zero.
Next, one shows that $C$ is an inessential annulus (one that is isotopic to an annulus contained in a disc) if and only if $T_C$ is isotopic to the identity if and only if $T_C$ fixes every $[\gamma]$.
Finally, and here is the heart of the matter: one shows that if $C_0,C_1$ are two inessential annuli and if $C_0$ is not isotopic to $C_1$ then there exists a nontrivial simple closed curve $\gamma$ such that $i(C_0,\gamma)=0$ and $i(C_1,\gamma)>0$. This is best done using a case analysis. First one isotopes $C_0,C_1$ so that they are in efficient position, meaning that their core curves have minimal geometric intersection number; denote that number by $i(C_0,C_1)$. Now one breaks into cases.
Let me start with the case that $i(C_0,C_1)=0$; up to isotopy this is equivalent to saying that $C_0,C_1$ are disjoint. Since $C_0$, $C_1$ are essential, no component of $S-C_0$ nor of $S-C_1$ is a disc. Since we have assumed the cores of $C_0$ and $C_1$ are not isotopic, no component of $\Sigma - (C_0 \cup C_1)$ is an annulus. Let $\Sigma' \subset \Sigma$ be the component of $\Sigma - C_0$ that contains $C_1$. One can construct $\gamma$ so that $\gamma \subset \Sigma'$ and hence $i(C_0,\gamma)=0$, and so that one of the following subcases holds: $C_1$ does not separate $\Sigma'$ and $i(C_1,\gamma')=1$; or $C_1$ does separate $\Sigma'$ and $i(C_1,\gamma')=2$.
In the other case that $i(C_0,C_1) > 0$, one starts by picking $\Sigma'$ to be any component of $S-C_0$. In this case the core curve of $C_1$ intersects $\Sigma'$ in an arc system, meaning a pairwise disoint system of properly embedded arcs, each of which is essential (i.e. not homotopic rel endoints into the boundary of $\Sigma'$). The argument now comes down to showing that for any essential arc system in $\Sigma'$ there exists an essential curve $\gamma$ in $\Sigma'$ which intersects that arc system essentially in at least one point. If any single one of these arcs is nonseparating then we can simply choose $\gamma$ to cross that arc in a single point. If all of those arcs are separating, then one can choose $\gamma$ to intersect that arc essentially in two points.