Claim 1 Any $T \in \mathcal {T}$ becomes compressible in $M(s_i)$ for only finitely many $s_i$ -fillings.

Proof Suppose that some $T \in \mathcal {T}$ becomes compressible after $s_i$ -fillings for infinitely many i. Then, by [Reference Culler, Gordon, Luecke and Shalen2, Theorem 2.4.4], $\partial M$ and T cobound a cable space, which must be $M_0$ . (A cable space is the exterior of a $(p,q)$ -cable of the core of V, where p and q are coprime integers with $q \ge 2$ .) It follows that, after infinitely many $s_i$ -fillings, the number of blocks in the geometric decomposition decreases, and, by comparing the number of blocks in the geometric decomposition, T also becomes compressible after infinitely many $s^{\prime }_j$ -fillings. It follows that $M_0(s_i)$ and $M_0(s^{\prime }_i)$ are all homeomorphic to a solid torus for such i. Then, since the gluing maps of $M_0(s_i)$ and $M_0(s^{\prime }_i)$ to the manifold $M' := M_1 \cup \cdots \cup M_t$ vary with i, this implies that the manifold $M'$ admits infinitely many purely cosmetic fillings. By the induction hypothesis, this gives a contradiction.

Claim 2 For only finitely many i, $M_0(s_i)$ becomes an interval bundle over a torus (actually, the product $T^2 \times I$ ) or over a Klein bottle K.

Proof Suppose that $M_0(s_i)$ yields interval bundles over a torus (actually, products $T^2 \times I$ ) or over a Klein bottle K for infinitely many i. We consider the former case, where some $T, T' \in \mathcal {T}$ become parallel after $s_i$ -fillings for infinitely many i. The latter case can be treated similarly.

As claimed in [Reference Martelli13, Proof of Theorem 3.5, Section 3.12(c)], for infinitely many i, the product $T^2 \times I$ arises as $M_0(s_i)$ only when $M_0 = P \times S^1$ , with P the pair of pants. Set a homology basis $(m,l)$ on $\partial M$ by taking m as a single boundary component of a fixed section of $M_0$ and l as the fiber of $M_0$ . Then, also as in [Reference Martelli13, Proof of Theorem 3.5], the slopes $s_i$ are represented by the homology class $\pm ( m + q_i l)$ for some $q_i \in \mathbb {Z}$ . It follows that $M(s_i)$ is realized by removing the product $M_0(s_i)$ and gluing the adjacent block(s) directly via a map that twists $q_i$ times along the fiber of $M_0 = P \times S^1$ . Thus, for sufficiently large i, the product bundle $M_0(s_i)$ gives a decomposing torus of the geometric decomposition of $M(s_i)$ . Under the assumption that $M(s_i) \cong M(s^{\prime }_i)$ , by considering the geometric decomposition, the same situation occurs for $M_0(s^{\prime }_i)$ and $M(s^{\prime }_i)$ for sufficiently large i.

Let T be the decomposing torus of $M(s_i)$ arising from the product $M_0(s_i)$ . Then, for sufficiently large i, $M(s^{\prime }_i)$ is obtained by cutting along T in $M(s_i)$ and gluing the adjacent block(s) of the geometric decomposition of $M(s_i)$ directly via Dehn twists along the fiber of $M_0 = P \times S^1$ .

If the two boundary tori of the product $M_0(s_i)$ are glued together to obtain $M(s^{\prime }_i)$ , then $M(s_i)$ and $M(s^{\prime }_i)$ are both torus bundles over the circle. Their monodromies differ only by Dehn twists along a simple closed curve. Such torus bundles become homeomorphic for only finitely many i. See [Reference Jaco8] for example. This implies a contradiction.

Suppose otherwise, that is, suppose that each of the two boundary tori of $M_0(s_i)$ is glued to the other block of $M(s_i)$ . Again, by considering the geometric decomposition, also holds for $M(s^{\prime }_i)$ . Consider an orientation-preserving homeomorphism $h: M(s_i) \to M(s^{\prime }_i)$ . From the above arguments, h induces an orientation-preserving self-homeomorphism $h_B := h|_B$ of the block B, which is one of the blocks adjacent to $M_0$ at T, such that $h_T := (h|_B)|_T$ is a Dehn twist along the fiber of $M_0 = P \times S^1$ .

However, this leads to a contradiction as follows. When B is a hyperbolic block, by the Mostow–Prasad rigidity, $h_B$ gives an isometry of the hyperbolic manifold $B - \partial B$ . Then, its restriction to a horotorus corresponding to T is also a Euclidean isometry, which cannot be a Dehn twist along any simple closed curve on T. When B is Seifert fibered, the restriction $h_T$ of an orientation-preserving homeomorphism $h_B$ of B must preserve the Seifert fibration of B, which cannot be a Dehn twist along the fiber of $M_0 = P \times S^1$ , since the fibers of $M_0$ and B do not match.

Claim 3 $M_0$ is not hyperbolic.

Proof Suppose that $M_0$ is hyperbolic. Suppose that $M_0(s_i) \cong M_k$ for some $k \in \{1, \dots , t\}$ for infinitely many i. By the famous Hyperbolic Dehn Surgery Theorem, there are only finitely many exceptional fillings, i.e., fillings yielding non-hyperbolic manifolds, on $\partial M \subset \partial M_0$ . It follows that $M_0(s_i) \cong M_k$ for infinitely many i implies that $M_k$ is hyperbolic. However, since $vol(M_0(s_i)) \to vol(M_0)$ and $vol(M_k) < vol(M_0)$ (see [Reference Martelli13], for example), $M_0(s_i) \cong M_k$ can hold only for finitely many i, contradicting the assumption. Thus, for infinitely many i, $M_0(s_i) \cong M_0(s^{\prime }_i)$ must hold. However, this is impossible by the argument given in [Reference Futer, Purcell and Schleimer3, Theorem 1.13]. Actually, the theorem considers only the case where the manifold is one-cusped, but by using [Reference Futer, Purcell and Schleimer3, Theorem 7.29] instead of [Reference Futer, Purcell and Schleimer3, Theorem 7.30], we see that the same statement holds for the multiple cusp case.

Claim 4 For sufficiently large i, $h_i(M_0(s_i)) = M_0(s^{\prime }_i)$ , or $h_i(M_0(s_i)) = M_l$ and $h_i(M_l) = M_0(s^{\prime }_i)$ for some $l \in \{1, \dots , t\}$ with $M_0 \cap M_l \ne \emptyset $ .

Proof Since the geometric decomposition is unique up to isotopy, for each $i,$ we have either $h_i(M_0(s_i)) = M_0(s^{\prime }_i)$ , or $h_i(M_0(s_i)) = M_k$ and $h_i(M_{k'}) = M_0(s^{\prime }_i)$ for some $k, k' \in \{1, \dots , t\}$ . For each $k \in \{ 1, \dots , t \}$ , if $h_i(M_0(s_i)) = M_k$ for only finitely many i, then $h_i(M_0(s_i)) = M_0(s^{\prime }_i)$ must hold for infinitely many i, and hence the claim follows. Suppose instead that, for some $k \in \{ 1, \dots , t \}$ , $h_i(M_0(s_i)) = M_k$ for infinitely many i. After passing to a subsequence, we may assume that, for some $k'$ , $h_i(M_{k'}) = M_0(s^{\prime }_i)$ also holds for these i.

Note that, in $M_0(s_i)$ , a new singular fiber of index $p_i$ arises for $s_i = q_i/p_i$ . Thus, since $M_k$ has only finitely many singular fibers, the set $\{p_i\}$ must be bounded; that is, $\{p_i\}$ is a finite set of nonnegative integers. It then follows that the set $\{q_i\}$ is unbounded, with $|q_i| \to \infty $ . Similarly, we have $|q^{\prime }_i| \to \infty $ .

Now, let us assign a nonnegative integer $\Delta _T$ to each member T of the collection $\mathcal {T}$ of tori and Klein bottles that define the geometric decomposition of $M^i := M(s_i) \cong M(s^{\prime }_i)$ . (The idea is inspired by the arguments in [Reference Martelli13].)

First, for each block N of the geometric decomposition of $M^i$ and each boundary torus $T \subset \partial N$ , we choose a finite set of preferred slopes, containing at least two elements, as follows. If N is hyperbolic, a preferred slope is defined to be a slope of shortest or second-shortest length in one cusp section. If N is Seifert fibered, we take a section F of the $S^1$ -bundle obtained from N by removing neighborhoods of the singular fibers. Using this F, we set a homology basis $(m,l)$ for each boundary torus $T \subset \partial N$ by taking m as the homology class of $\partial F \cap T$ and l as that of the fiber. Then, N is obtained from the $S^1$ -bundle over F by Dehn fillings along a tuple of slopes. We assume that the section F is chosen so that each filling slope $q/p$ satisfies $0 < q/p < 1$ by twisting along the $S^1$ -fiber. Then, a preferred slope on $T \subset \partial N$ is defined as the fiber of N or $\partial F \cap T$ .

Let T be a member of $\mathcal {T}$ . If T is a torus, we define $\delta _T$ as the maximum of the distance $\Delta (s, s')$ , where s and $s'$ are preferred slopes on $T = \partial N_1$ and $T = \partial N_2$ for the two blocks $N_1$ and $N_2$ adjacent to T, respectively. If T is a Klein bottle, its neighborhood W admits two fibrations, and we define $\delta _T$ as the maximum of $\Delta (s, s')$ on the torus $\partial W$ , where s is a preferred slope of the block adjacent to W at T and $s'$ is one of the slopes of the two fibers of W. The quantities $\delta _T$ depend on the choice of sections in the Seifert blocks of $M^i$ . By varying the sections, we minimize the total sum of the $\delta _T$ ’s for all $T \in \mathcal {T}$ . Under this restriction, for each i, we define $\Delta _T$ for $T \in \mathcal {T}$ in $M^i$ as the maximum of the $\delta _T$ ’s associated with T. Note that each $\delta _T$ has finite ambiguity due to the choice of sections in each Seifert block, but $\Delta _T$ is uniquely determined for T.

Now, since $|q_i| \to \infty $ and $|q^{\prime }_i| \to \infty $ , any member $T \in \mathcal {T}$ attaining the maximal $\Delta _T$ must lie in $M_0(s_i)$ and also in $M_0(s^{\prime }_i)$ within $M^i$ for sufficiently large i. (There may be several components in $\mathcal {T}$ attaining the same maximal value.) We observe that if $\Delta _T$ is maximal among those for the elements of $\mathcal {T}$ , then $\Delta _{h_i(T)}$ is also maximal. This follows from the definitions of $\Delta _T$ , together with the uniqueness of the Seifert block fibrations and the geometric decomposition.

By the observation above, we conclude that for sufficiently large i, either $h_i(M_0(s_i)) = M_0(s^{\prime }_i)$ , or $h_i(M_0(s_i)) = M_l$ and $h_i(M_l) = M_0(s^{\prime }_i)$ with $M_0 \cap M_l \ne \emptyset $ for some $l \in \{1, \dots , t\}$ .