Proof. Fix a regular neighbourhood $N_\Gamma$ of $|\Gamma|$ in $\mathcal{S}$ disjoint from $\partial B \cup \partial B'$. The subspace $\overline{N_\Gamma} := \mathcal{S} \setminus \operatorname{int}(N_\Gamma)$ is also a PL 3-manifold [Reference Rourke and Sanderson15, Corollary 3.14]. Moreover, $\overline B := \mathcal{S} \setminus \operatorname{int} (B) \subseteq \operatorname{int}(\overline{N_\Gamma})$ is a 3-ball (similarly for $\overline{B'} := \mathcal{S} \setminus \operatorname{int}(B')$) since closures of complements of PL-embedded n-balls in n-spheres are n-balls [Reference Rourke and Sanderson15, Corollary 3.13].

Since a PL homeomorphism between the boundaries of two balls extends to a PL homeomorphism of their interiors [Reference Rourke and Sanderson15, Lemma 1.10], we may extend $\Phi_\partial$ to an orientation-preserving PL homeomorphism $\Phi_{\overline B} \colon \overline{B} \to \overline{B'}$. Apply Theorem 3.1 to produce a PL ambient isotopy of $\overline{N_\Gamma}$ from the inclusion $\overline{B} \hookrightarrow \overline{N_\Gamma}$ to the composition $\overline B \overset{\Phi_{\overline B}}{\longrightarrow} \overline{B'} \hookrightarrow \overline{N_\Gamma}$. As this ambient isotopy keeps $\partial N_\Gamma$ fixed, the homeomorphism $\Phi_{\overline{N_\Gamma}} \colon \overline{N_\Gamma} \to \overline{N_\Gamma}$ extends to $\mathcal{S}$ by setting it to the identity on $N_\Gamma$. This extension $\Phi_\mathcal{S} \colon \mathcal{S} \to\mathcal{S}$, when restricted to B, is a PL homeomorphism $\Phi_B \colon B \to B'$ satisfying the conclusion of the lemma.

Proof. By Lemma 3.3 there is an orientation-preserving PL homeomorphism $\Phi_1 : B_1 \to B_1'$ with $\Phi_1\vert_{\vert \Gamma_1 \vert} = \mathrm{id}$. Similarly, let $\Phi_2 \colon B_2 \to B_2'$ be an orientation-preserving PL homeomorphism fixing $|\Gamma_2|$ and whose restriction to $\partial B_2$ is $f' \circ \Phi_1|_{\partial B_1} \circ f^{-1}$. The maps $\Phi_i$ assemble to a PL homeomorphism $\Phi \colon B_1 \sqcup_f B_2 \to B_1' \sqcup_{f'} B_2'$ giving the required isomorphism between $\Gamma_1 \sqcup_f \Gamma_2$ and $\Gamma_1\sqcup_{f'}\Gamma_2$.

Proof. Let $B_i \subset \mathcal{S}_i$ be the enclosing balls from which the disjoint union was formed, with attaching map $f \colon \partial B_1 \to \partial B_2$. To construct a PL homeomorphism $\Phi \colon \mathcal{S}_1 \to B_1 \cup_f B_2$ with $\Phi\vert_{\vert \Gamma_1 \vert} = \mathrm{id}$, take Φ as the identity on B ₁ and on $\overline{B_1}:= \mathcal{S}_1 \setminus \operatorname{int}(B_1)$ choose any extension $\overline{B_1} \to B_2$ of f [Reference Rourke and Sanderson15, Lemma 1.10].

Proof. Form the disjoint union $\Gamma_1 \sqcup_f \Gamma_2$ by using a suitable PL homeomorphism $f \colon \partial B_1 \to \partial B_2$ between the boundaries of enclosing balls for $\Gamma_1, \Gamma_2$. Writing $B_i':= \Phi_i(B_i)$ and defining $f' \colon \partial B_1' \to \partial B_2'$ as $f' := \Phi_2|_{\partial B_2} \circ f\circ \Phi_1^{-1}|_{\partial B_1'}$, we can form the disjoint union $\Gamma_1' \sqcup_{f'} \Gamma_2'$. The restrictions $\Phi_i|_{B_i}$ then assemble to the desired isomorphism $\Phi_1 \sqcup \Phi_2$.

Proof. Lemma 3.5 implies the first claim, and commutativity follows by using the same enclosing balls for both disjoint unions, and mutually inverse attaching maps.

Associativity is illustrated in Figure 3.1. Let $\mathcal{S}_i$ be the ambient sphere of $\Gamma_i$ and B ₁, B ₃ enclosing balls for $\Gamma_1, \Gamma_3$, respectively, and let B ₂₁, B ₂₃ be enclosing balls for Γ₂ such that $\operatorname{int}(B_{21})\cup \operatorname{int}(B_{23}) = \mathcal{S}_2$ (that is, $\mathcal{S}_2 \setminus \operatorname{int}(B_{21}) \cap \mathcal{S}_2 \setminus \operatorname{int}(B_{23}) = \emptyset$). Choosing attaching maps $f_1\colon \partial B_1 \to \partial B_{21}$, $f_3 \colon \partial B_{23} \to \partial B_3$, it follows that $B_1 \cup_{f_1} (B_{21} \cap B_{23})$ is an enclosing ball for $\Gamma_1 \sqcup_{f_1} \Gamma_2$, and $(B_{21} \cap B_{23}) \cup_{f_3} B_3$ is an enclosing ball for $\Gamma_2 \sqcup_{f_3} \Gamma_3$. The spatial graphs $(\Gamma_1 \sqcup_{f_1} \Gamma_2) \sqcup_{f_3} \Gamma_3$ and $\Gamma_1 \sqcup_{f_1} (\Gamma_2 \sqcup_{f_3} \Gamma_3)$ are thus the same.

Figure 3.1. The proof of associativity of the disjoint union.

Proof. Use Lemma 3.5 to regard each $\Gamma_i$ as a sub-graph of $\Gamma_1 \sqcup \Gamma_2$, and take B_i as an enclosing ball for $\Gamma_i$. If $f \colon S \to S$ is the identity map, then $\Gamma = \Gamma_1 \sqcup_f \Gamma_2$.

Proof. Since $\Lambda \ne \boldsymbol{0}$, certainly $|\Lambda|$ cannot be contained in both B_i. Denote by $\Lambda_i$ the sub-graph of Λ whose vertices and edges are contained in B_i. By Lemma 3.8, we see S decomposes Λ as $\Lambda_1 \sqcup \Lambda_2$. Since Λ is non-split, one of the summands, say Λ₁, is empty. By the first part of Proposition 3.7, it follows that $\Lambda_2 = \Lambda$.

Proof. Write $\Gamma_1:= \bigsqcup_{i \in I_1} \Lambda_i$ and $\Gamma_2:=\bigsqcup_{i \in I_2} \Lambda_i$. We induct on $\# I_1$.

If $I_1 = \emptyset$ then $\Gamma_1 = \boldsymbol{0} = \Gamma_2$, whence $I_2 = \emptyset$ and there is nothing left to show. If $I_1 = \{i_1\}$, then $\Gamma_1 = \Lambda_{i_1}$ is a piece. Hence Γ₂ is also a piece and therefore $I_2=\{i_2\}$. We thus set $f(i_1) := i_2$.

If I ₁ has more than one element, choose any partition into non-empty subsets $I_1 = I_1 ^+ \sqcup I_1^-$. Let S ₁ be a 2-sphere decomposing Γ₁ as $\bigl(\bigsqcup_{i \in I_1^+} \Lambda_i\bigr) \sqcup \bigl(\bigsqcup_{i \in I_1^-} \Lambda_i\bigr)$, and write $\Gamma_1^+ := \bigsqcup_{i \in I_1^+} \Lambda_i$ and $\Gamma_1^- := \bigsqcup_{i \in I_1^-} \Lambda_i$. Now $S_2 := \Phi(S_1)$ is a 2-sphere in the ambient sphere of Γ₂ disjoint from $|\Gamma_2|$ whose sides correspond to $\Gamma_1^+$ and $\Gamma_1^-$. By Lemma 3.10, each $|\Lambda_i|$ is contained in either the ‘+’-side or the ‘−’-side of S ₂. Partition I ₂ accordingly as $I_2 = I_2^+ \sqcup I_2^-$, and write $\Gamma_2^\pm := \bigsqcup_{i \in I_2^\pm} \Lambda_i$. Since Φ maps the support $|\Gamma_1^{\pm}|$ into $|\Gamma_2^{\pm}|$, we conclude that Φ doubles as a pair of isomorphisms of sub-graphs $\Phi^\pm \colon \Gamma_1^\pm \to \Gamma_2^\pm$. Both $I_1^\pm$ have fewer elements than I ₁, so by induction we obtain bijections $f^\pm \colon I_1^\pm \to I_2^\pm$, which assemble to the required $f \colon I_1 \to I_2$.

Proof. Consider the double $\mathcal D_N(M)$ of M along N, which is a union of two copies of M glued along the identity map on N, one of the copies with its orientation reversed. Using the fact that N is closed in $\partial M$ one sees that $(\mathcal D_N(M), N)$ is a PL manifold pair, and its boundary is $(\mathcal D_{\partial N} (\partial M \setminus \operatorname{int}(N)), \partial N)$. Doubling also B along D yields an unknotted ball pair $(\mathcal D_D(B),D)$.

Now, the maps $\iota_1, \iota_2$ extend to orientation-preserving PL embeddings:

\begin{equation*}\mathcal D(\iota_1), \mathcal D(\iota_2)\colon(\mathcal D_D(B), D) \to (\operatorname{int}(\mathcal D_N(M)), \operatorname{int}(N)).\end{equation*}

Theorem 4.4 yields a PL ambient isotopy of $(\mathcal D_N(M), N)$ relative $\mathcal D_{\partial N} (\partial M \setminus \operatorname{int}(N))$ that carries $\mathcal D(\iota_1)$ to $\mathcal D(\iota_2)$. A connectivity argument shows that it restricts to an isotopy from ι ₁ to ι ₂ relative $\partial M \setminus \operatorname{int}(N)$.

Proof. We may assume that $A_0 = C \times [0,1]{} \subset \mathbb{R}^{n}$ for some PL circle $C\subset \mathbb{R}^{n-1}$, with $\gamma_0 = C\times \{0\}$ and $\delta_0 = C \times \{1\}$, because the cone $v (C \times [0,1]) \to vA_0$ preserves cones at v for every PL homeomorphism $C \times [0,1]{} \to A_0$.

Choose a finite set of points in $\gamma \subset v(C \times \{0\})$ subdividing γ into straight line segments [Reference Rourke and Sanderson15, Theorem 2.2]. Pushing these points radially into $\gamma_0 = C \times \{0\}$ and projecting onto C yields a finite set of points in C (note that since $v\gamma$ is a cone, no two points of γ get pushed to the same point of γ ₀). Doing the same with δ yields a second finite subset of C. Finally, choose a third finite subset of C subdividing C itself into straight line segments. We denote by $p_1, \ldots, p_k$ the points in the union of these three subsets, ordered cyclically around C (with indices $1, \ldots, k$ modulo k). Now push $(p_1, 0), \ldots, (p_k,0) \in \gamma_0$ radially into γ to obtain points $p_1^\gamma, \ldots, p_k^\gamma$. Similarly, pushing $(p_1, 1), \ldots, (p_k, 1)$ radially yields $p_1^\delta, \ldots, p_k^\delta$.

Since the p_i subdivide C into straight line segments, we see that for each $i \in \mathbb{Z} / k$, the points $(p_i, 0), (p_{i+1} , 0), (p_i, 1), (p_{i+1}, 1)$ are the vertices of a rectangle R_i contained in A ₀. In particular, the cone $vR_i\subset vA_0$ is convex.

For each $i \in \mathbb{Z}/k$, denote by $T_i^\gamma$ the triangle spanned by the points $p_i^\gamma, p_{i+1}^\gamma, p_i^\delta$, and by $T_i^\delta$ the one spanned by $p_i^\delta, p_{i+1}^\delta, p_{i+1}^\gamma$. By the previous observation, these triangles are contained in vR_i. The union $A:=\bigcup_{i \in \mathbb{Z} / k} (T_i^\gamma \cup T_i^\delta)$ is then a PL annulus embedded in vA ₀, with $\partial A = \gamma \cup \delta$. It is also clear that each point of A lies in a unique ray from v through a point in A. The cone condition on vA follows.

Proof of Proposition 4.2

Write $\overline{B} := \mathcal{S} \setminus \operatorname{int}(B)$ and $\overline{B'}:=\mathcal{S} \setminus \operatorname{int}(B')$, and choose any extension of $\Phi_\partial$ to a PL homeomorphism $\Phi_{\overline B}\colon \overline B \to \overline{B'}$. We will find an extension $\Phi_\mathcal{S} \colon \mathcal{S} \to \mathcal{S}$ of $\Phi_{\overline{B}}$ that fixes $|\Gamma|$, and whose restriction $\Phi_B$ to B therefore satisfies the conclusion of the lemma. The intricate construction of $\Phi_\mathcal{S}$, for which we need some notation, is illustrated in Figure 4.2.

Figure 4.2. The 3-balls $\overline B$ and $\overline{B'}$.

First, choose a star neighbourhood N ₀ for v in the pair $(\mathcal{S}, \overline{B'} \cup |\Gamma|)$. More explicitly, N ₀ is a 3-ball such that $(\overline{B'} \cup |\Gamma|) \cap N_0$ is a cone with base its intersection with $\partial N_0$, and vertex v [Reference Rourke and Sanderson15, p. 50]. In particular, $D_0:= \overline{B'} \cap \partial N_0$ is a disc and $\overline{B'} \cap N_0$ is the cone vD ₀ with base D ₀ and vertex v.

We then pick a smaller star neighbourhood $N_v \subset \operatorname{int}(N_0)$ of v in $(\mathcal{S}, \overline{B'} \cup |\Gamma|)$, such that N_v is also a star neighbourhood of v in $(\mathcal{S}, \overline B \cup |\Gamma|)$, and $\overline B \cap N_v$ is mapped conically by $\Phi_{\overline B}$ into $\operatorname{int}(N_0)$. Denoting by D the disc $\overline B \cap \partial N_v$, so $\overline B \cap N_v$ is a cone vD with base D and vertex v, this means that $\Phi_{\overline B}(vD)$ is a cone vD ^ʹ with base the disc $D':= \Phi_{\overline B}(D)$ and vertex v, and that $\Phi_{\overline B}|_{vD}\colon vD \to vD'$ is the cone of $\Phi_{\overline B}|_D\colon D \to D'$. The existence of such N_v follows from the definitions of PL map and polyhedron, say, by taking N_v to be a sufficiently small ϵ-neighbourhood of v. We denote by $\overline{N_v}$ the 3-ball $\mathcal{S} \setminus \operatorname{int}(N_v)$.

To apply the disc theorem at the boundary, we need to move $\overline{B'}$ to the more convenient configuration given by the following claim and illustrated in Figure 4.3.

Claim. There exists an orientation-preserving PL homeomorphism $\Psi\colon \mathcal{S} \to \mathcal{S}$ fixing $|\Gamma|$ and such that:

Figure 4.3. The 3-ball $\overline{B'}$ and its Ψ-image $\widetilde B$.

• • Ψ maps the pair $(\Phi_{\overline B}(\overline B \cap \overline{N_v}), D')$ into the pair $(\overline{N_v}, \partial N_v)$, and

• • writing $\widetilde D := \Psi(D')$, the map Ψ is given on vD ^ʹ as the cone $v D' \to v\widetilde D$ of the PL homeomorphism $D' \to \widetilde D$.

Assuming the claim for the moment, let us see how to use the resulting Ψ to construct the desired extension $\Phi_\mathcal{S}$ of $\Phi_{\overline B}$.

Let $\widetilde B$ be the 3-ball $\Psi(\overline{B'})$ and choose a regular neighbourhood $N_\Gamma$ of $|\Gamma| \cap \overline{N_v}$ in $\overline{N_v}$, small enough to be disjoint from $\overline B$ and $\widetilde B$. Denote by M the closure of $\overline{N_v} \setminus N_\Gamma$ in $\overline{N_v}$, and consider the closed codimension-0 submanifold $N:= \partial N_v \cap M$ of $\partial M$. By construction of Ψ, its restriction to $\Phi_{\overline B}(\overline{B}\cap \overline {N_v})$ is a PL homeomorphism of pairs $(\Phi_{\overline B}(\overline B \cap \overline{N_v}), D') \to (\widetilde B \cap \overline{N_v}, \widetilde D)$. We may thus apply Corollary 4.5 to the inclusion $(\overline B \cap \overline{N_v},D) \hookrightarrow (M,N)$ and the composition:

\begin{equation*}(\overline B \cap \overline{N_v}, D) \overset{\Phi_{\overline B}}{\longrightarrow} (\Phi_{\overline B}(\overline B \cap \overline{N_v}), D') \overset{\Psi}{\longrightarrow} (\widetilde B \cap \overline N_v, \widetilde D) \hookrightarrow (M,N).\end{equation*}

This is illustrated in Figure 4.4.

Figure 4.4. Applying the Disc Theorem at the boundary to ambiently isotope $(\overline B \cap \overline{N_v}, D)$ onto $(\widetilde B \cap \overline{N_v}, \widetilde D)$ within (M, N).

The final PL homeomorphism $\widetilde \Phi_M \colon M \to M$ of the resulting PL isotopy of M extends the composition $\Psi|_{\Phi_{\overline B}(\overline B \cap \overline {N_v})} \circ \Phi_{\overline B}|_{\overline B \cap \overline{N_v}}$ and fixes $\partial M \setminus \operatorname{int}(N) = \partial M \cap N_\Gamma$. We may thus extend $\widetilde \Phi_M$ to a PL homeomorphism $\widetilde \Phi_{\overline{N_v}} \colon \overline{N_v} \to \overline{N_v}$ by setting it to be the identity on $N_\Gamma$. In particular, $\widetilde \Phi_{\overline{N_v}}$ fixes $|\Gamma| \cap \overline{N_v}$. Finally, extend $\widetilde \Phi_{\overline{N_v}}$ to a PL homeomorphism $\widetilde \Phi_\mathcal{S} \colon \mathcal{S}\to \mathcal{S}$ by defining it on $N_v = v(\partial N_v)$ as the cone of the already prescribed PL homeomorphism $\partial N_v \to \partial N_v$.

The restriction $\widetilde \Phi_\mathcal{S}|_{\overline B}$ is now the composition $\Psi|_{\overline B}\circ\Phi_{\overline B}$: indeed, we have seen that the two maps agree on $\overline B \cap \overline{N_v}$, and on $\overline B \cap N_v = vD$ both are defined as the cone of $D \overset{\Phi_{\overline B}} \to D' \overset{\Psi} \to \widetilde D$. Moreover, $\widetilde \Phi_\mathcal{S}$ fixes $|\Gamma|$. Hence, the map $\Phi_\mathcal{S}:= \Psi^{-1}\circ \widetilde \Phi_\mathcal{S}$ extends $\Phi_{\overline B}$ and fixes $|\Gamma|$, as desired.

Proof of the Claim

The argument is illustrated in Figure 4.5. Choose a collar for $\partial D_0$ in $B' \cap \partial N_0$, that is, a PL embedding $c\colon \partial D_0 \times [0,1]{} \to B' \cap \partial N_0$ such that $c(-, 0)$ is the identity on $\partial D_0$, and $c(\partial D_0 \times [0,1[)$ is an open neighbourhood of $\partial D_0$ in $B' \cap \partial N_0$. We may also assume that the image A ₀ of c is disjoint from $|\Gamma|$; see [Reference Rourke and Sanderson15, p. 24] for more on collars.

Figure 4.5. The construction of Ψ.

Let $D_0^+$ be the ‘enlarged disc’ $D_0 \cup A_0$, and consider the 3-ball $vD_0^+$, which is a cone with base $D_0^+$ and vertex v. We shall define Ψ as the identity on $\mathcal{S} \setminus \operatorname{int}(v D_0^+)$ and then suitably extend the identity on $\partial (vD_0^+)$ to all of $vD_0^+$.

Denote by $\widetilde D^+$ the disc $vD_0^+ \cap \partial N_v$ and consider the PL circles $\partial D'$ and $\partial \widetilde D^+$, each lying in the cone of a distinct component of $\partial A_0$. Each of these circles is the base of a cone with vertex v, so we can use Lemma 4.6 to find an annulus A with $\partial A = \partial D' \cup \partial \widetilde D^+$, such that vA is a cone with base A and vertex v. Denote by $D'^+$ the disc $D' \cup A$ and note that by construction, $\partial D'^+= \partial \widetilde D^+$.

To define Ψ inside $v D_0^+$, we first choose any extension of the identity map $\partial D'^+ \to \partial \widetilde D^+$ to a PL homeomorphism $D'^+ \to \widetilde D^+$. Since both $vD'^+$ and $v\widetilde D^+$ are cones at v, we can define Ψ on $vD'^+$ as the cone of the above PL homeomorphism $D'^+ \to \widetilde D^+$. This is consistent with the definition of Ψ as the identity on $\partial(vD_0^+)$.

It remains only to define Ψ on $vD_0^+ \setminus vD'^+$, whose closure in $\mathcal{S}$ is a 3-ball C (because it is the complement in $vD_0^+$ of an open regular neighbourhood of a boundary point). Denoting by $\widetilde C$ the closure in $\mathcal{S}$ of $vD_0^+ \setminus v\widetilde D^+$, which is a 3-ball, this amounts to choosing a PL homeomorphism $C \to \widetilde C$ extending the prescribed map $\partial C \to \partial \widetilde C$. We choose any extension, and this completes the construction of Ψ. It is straightforward to verify that all required conditions on Ψ are satisfied.

Proof. The argument can be copied almost word-by-word from the proof of Proposition 3.4, with the role of Lemma 3.3 now played by Proposition 4.2.

Proof. Identical to the proof of Proposition 3.7 save for the following modifications:

• • The first item relies on vertex summands being sub-graphs (Lemma 4.8), rather than disjoint union summands being sub-graphs (Lemma 3.5).

• • To prove associativity in the case $v_{21} = v_{23}$, the requirement on enclosing balls is that $\operatorname{int}(B_1)\cup \operatorname{int}(B_2) = \mathcal{S} \setminus\{v_{21}\}$ (equivalently, $\overline{B_{21}} \cap \overline{B_{23}} = \{v_{21}\}$).

Proof. We again induct on $\# J$. When J has at most one element, no choices are made in defining $[\mathcal T]$, so there is nothing to show.

Suppose then that J contains two elements $j_1 \neq j_2$. For each $k \in \{1,2\}$, denote by $[\mathcal T ]_k$ the realization of $\mathcal T$ constructed by splitting T at j_k. Moreover, let $L_k \subset L$ be the set of edges incident with j_k, and consider, for each $l \in L_k$, the tree of spatial graphs $\mathcal T_l := (T_l, I_l, J_l, L_l, (\Gamma_i)_{i \in I_l}, (v(l'))_{l' \in L_l})$ defined as before.

Now, there is exactly one edge $l_1 \in L_1$ such that $T_{l_1}$ contains the vertex j ₂, and one edge $l_2 \in L_2$ such that $T_{l_2}$ contains j ₁. Consider the tree $\dot T:=T_{l_1} \cap T_{l_2}$. We have a tree of spatial graphs:

\begin{equation*} \dot{\mathcal T} := (\dot T, \dot I , \dot J, \dot L, (\Gamma_i)_{i \in \dot I}, (v(l'))_{l' \in \dot L}), \end{equation*}

where $\dot I := I_{l_1} \cap I_{l_2}, \dot J := J_{l_1} \cap J_{l_2}$, and $\dot L := L_{l_1} \cap L_{l_2}$; see Figure 4.8.

Figure 4.8. The tree T, the sub-trees $T_{l_1}, T_{l_2}$ and their intersection $\dot T$. Large vertices represent elements of I, and small ones elements of J.

By induction, the realizations $[\mathcal T_{l_k}]$ are well-defined. One readily checks that:

\begin{align*} [\mathcal T_{l_1}]{} &= [\dot{\mathcal T}]{} \, _{v(l_2)}\!{\bullet}_{v_2}\, \underset {l \in L_2 \setminus \{l_2\}} \bigstar ([ \mathcal T_l],v(l)),\\ [\mathcal T_{l_2}]{} &= [\dot{\mathcal T}]{} \, _{v(l_1)}\!{\bullet}_{v_1}\, \underset {l \in L_1 \setminus \{l_1\}} \bigstar ([ \mathcal T_l],v(l)), \end{align*}

where v_k is the result of identifying the vertices v(l) with $l \in L_k \setminus \{l_k\}$. We finish the proof by applying the ‘$v_{21} \neq v_{23}$’ case of associativity from Proposition 4.10:

\begin{align*} [\mathcal T]_1 & = [\mathcal T _{l_1}]{} \, _{v(l_1)}\!{\bullet}_{v_1}\, \bigg(\underset {l \in L_1 \setminus \{l_1\}} \bigstar ([ \mathcal T_l],v(l))\bigg)\\ & = \left( \bigg(\underset {l \in L_2 \setminus \{l_2\}} \bigstar ([ \mathcal T_l],v(l))\bigg) \, _{v_2}\!{\bullet}_{v(l_2)}\, [\dot{\mathcal T}]{} \right) \, _{v(l_1)}\!{\bullet}_{v_1}\, \bigg( \underset {l \in L_1 \setminus \{l_1\}} \bigstar ([ \mathcal T_l],v(l))\bigg)\\ & = \bigg(\underset {l \in L_2 \setminus \{l_2\}} \bigstar ([ \mathcal T_l],v(l))\bigg) \, _{v_2}\!{\bullet}_{v(l_2)}\, \left( [\dot{\mathcal T}]{} \, _{v(l_1)}\!{\bullet}_{v_1}\, \bigg( \underset {l \in L_1 \setminus \{l_1\}} \bigstar ([ \mathcal T_l],v(l)) \bigg) \right)\\ & = \bigg(\underset {l \in L_2 \setminus \{l_2\}} \bigstar ([ \mathcal T_l],v(l))\bigg) \, _{v_2}\!{\bullet}_{v(l_2)}\, [\mathcal T_{l_2}]{} = [\mathcal T]_2. \end{align*}

Proof. If a vertex summand, say Γ₁, is split, let $\mathcal{S}_1$ be its ambient sphere and let $S\subset \mathcal{S}_1$ be a splitting sphere. Choose an enclosing ball B ₁ for $(\Gamma_1, v_1)$ containing S in its interior. Then if we use B ₁ to form the vertex sum, S will be contained in the ambient sphere of $\Gamma_1 \, _{v_1}\!{\bullet}_{v_2}\, \Gamma_2$, with both sides of S intersecting $|\Gamma_1 \, _{v_1}\!{\bullet}_{v_2}\, \Gamma_2|$. Hence S is a splitting sphere for $\Gamma_1 \, _{v_1}\!{\bullet}_{v_2}\, \Gamma_2$ by Lemma 3.8.

Conversely, suppose S is a splitting sphere for $\Gamma_1 \, _{v_1}\!{\bullet}_{v_2}\, \Gamma_2$. The component of $\mathcal{S} \setminus S$ not containing the vertex $v_1= v_2$ has non-empty intersection with the support of a summand, say Γ₁. Then both components of $\mathcal{S} \setminus S$ intersect $|\Gamma_1|$ and so, regarding Γ₁ as a sub-graph of $\Gamma_1 \, _{v_1}\!{\bullet}_{v_2}\, \Gamma_2$, we see S is a splitting sphere for Γ₁.

Proof. Since Λ is a piece, the case where $S \cap |\Lambda| = \emptyset$ follows from Lemma 3.10. If $S \cap |\Lambda|$ consists of one vertex v of Λ, then since $\Lambda \not \cong \boldsymbol{1}$, certainly $|\Lambda|$ is not contained in both B_i. By Lemma 4.17, S decomposes Λ as $\Lambda_1 \, _{v}\!{\bullet}_{v}\, \Lambda_2$. As Λ has no cut vertices, one of the summands, say Γ₁, is isomorphic to 1. This means $\Lambda = \Lambda_2$.

Proof. To see that each v(j) is cut: by definition of a tree of spatial graphs, j has degree at least 2, so if $L_0 \subseteq L$ is the set of edges incident to j, one has a non-trivial partition $L_0 = L_1 \sqcup L_2$. For each $k\in\{1,2\}$, choose $l_k \in L_k$. Then $\Lambda_{i(l_k)}$, being a block, has an edge, hence $[\mathcal T_{l_k}]$ also has an edge. Thus each vertex summand in:

\begin{equation*}[\mathcal T]{} = \biggl( \underset {l \in L_1}\bigstar([\mathcal T_l], v(l)) \biggr) \, _{v(l_1)}\!{\bullet}_{v(l_2)}\, \biggl( \underset {l \in L_2} \bigstar([\mathcal T_l], v(l)) \biggr),\end{equation*}

has an edge and so is not isomorphic to 1. Thus $v(j) = v(l_1)=v(l_2)$ is cut. Looking at the vertex set of $[\mathcal T]$, as given by the description of $\langle[\mathcal T]\rangle$, it is clear that the assignment $j \mapsto v(j)$ is injective.

Conversely, suppose v is a vertex of $[\mathcal T]$ that does not result from such an identification, and consider a PL-embedded 2-sphere S in the ambient sphere $\mathcal{S}$ of $[\mathcal T]$ intersecting $|[\mathcal T]|$ precisely at v. Say S decomposes $[\mathcal T]$ as $\Gamma_1 \, _{v}\!{\bullet}_{v}\, \Gamma_2$; we argue that some $\Gamma_i$ is 1. The on v implies that all edges of $[\mathcal T]$ incident to v come from the same block $\Lambda_i$. Using Lemma 4.13 to regard $\Lambda_i$ as a sub-graph of $[\mathcal T]$, we see from Lemma 4.19 that all edges of $[\mathcal T]$ incident to v are in one of the $\Gamma_i$, say in Γ₁. Hence v is an isolated vertex of Γ₂. But since $[\mathcal T]$ is non-split by Corollary 4.16, Γ₂ is non-split by Lemma 4.15, which implies $\Gamma_2 \cong \boldsymbol{1}$.

Proof. We induct on the number of edges in Γ to produce a tree of blocks $\mathcal T = (T, I, J, L, (\Lambda_i)_{i \in I}, (v(l))_{l \in L})$ realizing Γ. If Γ has no edges, then $\Gamma = \boldsymbol{0}$ and we take T to be the empty tree.

Now suppose Γ has edges. If Γ is a block, take T to be a tree with a single vertex $i \in I$ and set $\Lambda_{i} := \Gamma$. Otherwise Γ can be expressed as $\Gamma_1 \, _{v}\!{\bullet}_{v}\, \Gamma_2$ with each $\Gamma_k$ not isomorphic to 1, and also non-split by Lemma 4.15. Hence both $\Gamma_k$ have at least one edge, and thus fewer edges than Γ, so the induction hypothesis applies.

Let $\mathcal T_k = (T_k, I_k, J_k, L_k, (\Lambda_i)_{i \in I_k}, (v(l))_{l \in L_k})$ be a tree of blocks for $\Gamma_k$. We construct T as in Figure 4.10 from modified versions $T_k'$ of the T_k, according to the following two cases:

• • If v is not a cut vertex of $\Gamma_k$, so by Proposition 4.20 there is no edge $l_k \in L_k$ with $v(l_k) = v$, construct $T_k'$ from T_k by adding a new vertex j_k and a new edge l_k connecting j_k to the vertex $i_k \in I$ whose block $\Lambda_{i_k}$ contains v. We also write

  \begin{equation*}J_k':= J_k \sqcup \{j_k\},\quad L'_k:= L_k \sqcup \{l_k \}, \quad L^0_k:=\{l_k\},\end{equation*}

  and set $v(l_k):= v$. (In this case, $T_k'$ with its vertex set partitioned as $I_k \sqcup J_k'$ is not admissible as the tree in a tree of spatial graphs, since j_k is a leaf.)

• • If v is a cut vertex of $\Gamma_k$, then there is a corresponding $j_k \in J_k$, whose set of incident edges we denote by $L_k^0$. In $\Gamma_k$, the vertices v(l) with $l \in L_k^0$ are identified into the vertex v. Set $T_k':= T_k$, $J_k':= J_k$, $L_k':=L_k$.

We define

\begin{equation*}T:= T_1' \, _{j_1}\!{\bullet}_{j_2}\, T_2', \quad I:= I_1 \sqcup I_2,\quad J:= (J_1' \sqcup J_2')/j_1 \sim j_2,\quad L:= L_1' \sqcup L_2',\end{equation*}

Figure 4.10. Constructing T from T ₁ and T ₂. Large (resp. small) vertices represent elements of I ₁, I ₂ (resp. of J ₁, J ₂). Elements of I ₁, I ₂ whose corresponding blocks contain v are indicated. Here, v is not a cut vertex in Γ₁, but it is in Γ₂, where it corresponds to j ₂.

and this turns $\mathcal T$ into a tree of spatial graphs whose realization is Γ:

\begin{align*} [\mathcal T]{} & = \underset {\substack{l \in L\\ j(l)=j_1=j_2}} \bigstar ([\mathcal T_l], v(l))\\ & = \bigg( \underset {l \in L_1^0} \bigstar ([\mathcal T_l], v(l))\bigg) \, _{v}\!{\bullet}_{v}\, \bigg(\underset {l \in L_2^0} \bigstar ([\mathcal T_l], v(l))\bigg) \\ & = [\mathcal T_1]{} \, _{v}\!{\bullet}_{v}\, [\mathcal T_2]{} = \Gamma_1 \, _{v}\!{\bullet}_{v}\, \Gamma_2. \end{align*}

Proof. We induct on $\# I_1$. If $I_1 = \emptyset$, then $[\mathcal T_1]{} = \boldsymbol{0} = [\mathcal T_2]$, so T ₂ is the empty tree and there is nothing to show. If I ₁ is comprised of a single element i ₁, and hence $J_1 = \emptyset$, then $[\mathcal T_1]{} = \Lambda_{i_1}$ is a block, so $[\mathcal T_2]$ is a block. In particular, $[\mathcal T_2]$ has no cut vertices and so by Proposition 4.20 we conclude $J_2 = \emptyset$. Hence I ₂ contains exactly one element i ₂, with $[\mathcal T_2]{} = \Lambda_{i_2}$, and we set $f(i_1):=i_2$.

Assume now that I ₁ contains at least two elements, so $J_1 \neq \emptyset$. Choose $j_1 \in J_1$, write $v_1:=v(j_1)$, and let S ₁ be a cut sphere for $[\mathcal T_1]$ decomposing it as $[\mathcal T_1]{} = \Gamma_1^+ \, _{v_1}\!{\bullet}_{v_1}\, \Gamma_1^-$, so $\Gamma_1^+, \Gamma_1^-$ are pieces non-isomorphic to 1. Similarly, $v_2:= \Phi(v_1)$ is a cut vertex for $[\mathcal T_2]$, so let $j_2 \in J_2$ be the corresponding element. The sphere $S_2 := \Phi(S_1)$ is now a cut sphere for $[\mathcal T_2]$ decomposing it as $[\mathcal T_2]{} = \Gamma_2^+ \, _{v_2}\!{\bullet}_{v_2}\, \Gamma_2^-$, with Φ giving isomorphisms of sub-graphs $\Phi^\epsilon \colon \Gamma_1^\epsilon \to \Gamma_2^\epsilon$, for each $\epsilon \in \{+,-\}$.

Let $k \in \{1,2\}$. Our goal is to extract from $\mathcal T_k$ a description of the $\Gamma_k^+, \Gamma_k^-$ as realizations of trees of blocks, to which we then apply the induction hypothesis. The procedure is analogous for all four spatial graphs, so fix $\epsilon \in \{+, -\}$.

Denote by $L_k^0 \subseteq L_k$ the set of edges incident to j_k. By Lemma 4.19, for each $i \in I_k$, the block $\Lambda_i$ is a sub-graph of exactly one among $\Gamma_k^+, \Gamma_k^-$. Consider the partition $L_k^0 = L_k^{0+} \sqcup L_k^{0-}$, where $l\in L_k^0$ is in $L_k^{0\epsilon}$ if $\Lambda_{i(l)}$ is a sub-graph of $\Gamma_k ^\epsilon$. Then $[\mathcal T_k]$ decomposes as:

\begin{equation*} [\mathcal T_k]{} = \bigg( \underset {l \in L_k^{0+}} \bigstar ([(\mathcal T_k)_l], v(l))\bigg) \, _{v_k}\!{\bullet}_{v_k}\, \bigg(\underset {l \in L_k^{0-}} \bigstar ([(\mathcal T_k)_l], v(l))\bigg).\end{equation*}

We claim that this is the same as the decomposition given by S_k, that is:

\begin{equation*}\Gamma_k^\epsilon = \bigstar_{l \in L_k^{0\epsilon}} ([(\mathcal T_k)_l], v(l)).\end{equation*}

To see this, first notice that since the $\Lambda_i$ are non-split, each $[(\mathcal T_k)_l]$ is also non-split by Corollary 4.16. Now, for every $l \in L_k^0$, it follows from Proposition 4.20 that v(l) is not a cut vertex of $[(\mathcal T_k)_l]$. Therefore, S_k decomposes $[(\mathcal T_k)_l]$ as a trivial vertex sum $[(\mathcal T_k)_l]\, _{v(l)}\!{\bullet}_{}\,\boldsymbol{1}$. In other words, $|[(\mathcal T_k)_l]|$ is entirely contained in one side of S_k, which must be the same as $|\Lambda_{i(l)}|$. Thus, each $\bigstar_{l \in L_k^{0\epsilon}} ([(\mathcal T_k)_l], v(l))$ is a sub-graph of $\Gamma_k^\epsilon$, whence the above description of the $\Gamma_k^\epsilon$ holds.

Next, we write down an explicit tree of blocks $\mathcal T_k^\epsilon$ for $\bigstar_{l \in L_k^{0\epsilon}} ([(\mathcal T_k)_l], v(l))$. If $L_k^{0 \epsilon}$ has only one element $l_k^\epsilon$, put $\mathcal T_k^\epsilon := (\mathcal T_k)_{l_k^\epsilon}$. Otherwise, recover the notation introduced when defining the realization of a tree of spatial graphs:

\begin{equation*}(\mathcal T_k)_l = ((T_k)_l, (I_k)_l, (J_k)_l, (L_k)_l, (\Lambda_i)_{i \in (I_k)_l}, (v(l'))_{l' \in (L_k)_l} ),\end{equation*}

and set $\mathcal T_k^\epsilon := (T_k^\epsilon, I_k^\epsilon, J_k^\epsilon, L_k^\epsilon, (\Lambda_i)_{i \in I_k^\epsilon},(v(l))_{l \in L_k^\epsilon})$ to be the tree of blocks comprised of the branches of $\mathcal T_k$ at j_k that stem from edges in $L_k^{0\epsilon}$. Explicitly, $T_k^\epsilon$ is the sub-tree of T_k with vertex and edge sets given by:

\begin{equation*}I_k^\epsilon := \bigsqcup_{l \in L_k^{0\epsilon}} (I_k)_l, \quad J_k^\epsilon:= \{j_k \} \sqcup \bigsqcup_{l \in L_k^{0\epsilon}} (J_k)_l, \quad L_k^\epsilon:= {L_k^{0\epsilon}} \sqcup \bigsqcup_{l \in L_k^{0\epsilon}} (J_k)_l.\end{equation*}

Observe that in the first case $[\mathcal T_k^\epsilon]$ does not have v_k as a cut vertex, and in the second case it does, with j_k being the corresponding element of $J_k^\epsilon$.

It is now clear that, in either case, $[\mathcal T_k^\epsilon]{} = \bigstar_{l \in L_k^{0\epsilon}} ([(\mathcal T_k)_l], v(l)) = \Gamma_i^\epsilon$. By induction hypothesis, the isomorphisms $\Phi^\epsilon \colon [\mathcal T_1^\epsilon]{} \to [\mathcal T_2^\epsilon]$ yield tree isomorphisms $f^\epsilon \colon T_1^\epsilon \to T_2 ^\epsilon$, which we now assemble to the desired $f \colon T_1 \to T_2$. On each sub-tree $T_1^\epsilon$ of T ₁, we want to set $f = f^\epsilon$ but have to ensure that $f^+$ and $f^-$ agree where they overlap, and we must also define f on vertices and edges of T ₁ that are not in any of the $T_1^\epsilon$.

Fix $\epsilon \in \{+,-\}$ for this paragraph. The isomorphism $\Phi^\epsilon$ ensures that v ₁ is a cut vertex of $[\mathcal T_1^\epsilon]$ if and only if v ₂ is a cut vertex of $[\mathcal T_2^\epsilon]$. If this is the case, then for both $k \in \{1,2\}$, the vertex j_k of T_k is in $T_k^\epsilon$, along with the edges in $L_k^{0 \epsilon}$. Moreover, in this situation we have in $\mathcal T_2^\epsilon$ that $v(j_2) = v_2 = \Phi^\epsilon (v_1) = \Phi^\epsilon (v(j_1)) = v(f^\epsilon(j_1))$, whence it follows by injectivity of $j \mapsto v(j)$ that $j_2 = f^\epsilon(j_1)$. On the other hand, if one (hence both) v_k is not cut in $[\mathcal T_k^\epsilon]$, then the corresponding j_k and the unique edge $l_k^\epsilon$ in $L_k^{0\epsilon}$ are not in $T_k^\epsilon$. In this situation, $i(l_k^\epsilon)$ is the only element of $I_k^\epsilon$ whose corresponding block $\Lambda_{i(l_k^\epsilon)}$ contains v_k as a vertex.

At this point there are three cases to consider:

• • If for some (hence both) $k\in \{1,2\}$ the vertex v_k is cut in $[\mathcal T_k^+]$ and $[\mathcal T_k^-]$, then the sub-trees $T_k^+, T_k^-$ jointly cover T_k, and they overlap precisely at j_k. As $f^+(j_1) = j_2 = f^-(j_1)$, we can glue together the f ^ϵ into the desired $f \colon T_1 \to T_2$.

• • Suppose, for both $k \in \{1,2\}$, the vertex v_k is cut in $[\mathcal T_k^+]$ but not in $[\mathcal T_k^-]$ (the reverse situation being analogous). Then $T_k^+$ and $T_k^-$ do not overlap, and jointly they cover all of T_k except for the edge $l_k^-$ described above. In this case, extend the definition of $f^+, f^-$ to all of T ₁ by setting $f(l_1^-) := l_2^-$. This respects the endpoints of the edge: we have seen that $f^-(j_1) = j_2$, and the characterization of $i(l_k^\epsilon)$ given above, together with the fact that $\Phi^-(v_1)=v_2$, shows that $f^-(i(l_1^-)) = i(l_2^-)$.

• • If for both $k \in \{1,2\}$ the vertex v_k is not cut in $[\mathcal T_k^+]$ nor in $[\mathcal T_k^-]$, then $T_k^+, T_k^-$ are disjoint and cover all of T_k except for the vertex j_k and its only two incident edges $l_k^+, l_k^-$. We extend $f^+, f^-$ by putting $f(j_1):=j_2$ and, for each $\epsilon \in \{+,-\}$, setting $f(l_1^\epsilon):=l_2^\epsilon$. This respects the incidence of each $l_1^\epsilon$ at the endpoint j ₁, and for the other endpoint we argue as in the previous case.

Having defined the isomorphism $f: T_1 \to T_2$, almost all properties stated in the proposition are inherited from the f ^ϵ. We are only left to check that, in the second and third cases above, the definition of f on the new edge(s) $l_1^\epsilon$ satisfies $\Phi(v(l_1^\epsilon)) = v(f(l_1^\epsilon))$. And indeed it does: $\Phi(v(l_1^\epsilon)) = \Phi(v_1) = v_2 = v(l_2^\epsilon) = v(f(l_1^\epsilon))$.

Proof. Let $N_{V,k}$ and $N_{E,k}$ be the regular neighbourhoods from the construction of $X_k^\circ$. Similarly, use the corresponding notation $N_{v,k}, N_{e,k}$ for the components corresponding to single vertices and edges, as well as $X_{V,k}$ for $\mathcal{S}\setminus\operatorname{int}(N_{V,k})$.

By the Regular Neighbourhood Theorem for pairs [Reference Rourke and Sanderson15, Theorem 4.11] there is a PL isotopy of $\mathcal{S}$ carrying $N_{V,1}$ to $N_{V,2}$ and fixing $|\Gamma|$. This induces an orientation-preserving PL homeomorphism $\Phi_V\colon N_{V,1}\to N_{V,2}$. As $|\Gamma|$ is fixed during the isotopy, $\Phi_V$ maps $N_{v,1}$ to $N_{v,2}$ for each vertex v of Γ and fixes the edges as well.

Since PL homeomorphisms take regular neighbourhoods to regular neighbourhoods, $\Phi_V(N_{E,1})$ and $N_{E,2}$ are regular neighbourhoods of $|\Gamma|\cap X_{V,2}$ in $X_{V,2}$. Using the Regular Neighbourhood Theorem again, we find a PL isotopy of $X_{V,2}$ that carries $\Phi_V(N_{E,1})$ to $N_{E,2}$. This isotopy carries $\Phi_V(X_1^\circ)$ to $X_2^\circ$. This shows that the two marked exteriors are in fact PL-homeomorphic via an orientation-preserving PL homeomorphism $\Phi\colon X_1^\circ\to X_2^\circ$. As these isotopies fix $|\Gamma|$, each vertex/edge region of the boundary gets mapped to the corresponding vertex/edge region, thus mapping $P_1^\circ$ to $P_2^\circ$. As Φ is orientation-preserving and the orientation of the junctures is determined by the orientation of X_k, Φ also preserves the orientation of the $P_k^\circ$.

Proof. Recall that the orientation of $P_\Gamma^\circ$ determines which regions of $\partial X_\Gamma^\circ$ are vertex regions. Thus, preserving the orientation of $P_k^\circ$ is equivalent to mapping vertex regions to vertex regions (and edge regions to edge regions).

Denote the regular neighbourhoods used in the construction of X_k by $N_{V,k}$ and $N_{E,k}$. The discs $N_{E,k}\cap \partial N_{V,k}$ are bounded by the junctures in $P_k^\circ$, and each disc intersects $|\Gamma_k|$ in exactly one point in its interior. Since discs are PL-homeomorphic to cones over any of their interior points, we can use the cone construction to extend $\Phi_X$ to a PL homeomorphism $\Phi_{X^+}\colon X_1^\circ\cup (N_{E,1}\cap \partial N_{V,1})\to X_2^\circ \cup (N_{E,2}\cap \partial N_{V,2})$ mapping the intersection points of $|\Gamma_k|$ with $N_{E,k}\cap \partial N_{V,k}$ to each other.

Note that each $(N_{e,k},e\cap X_{V,k})$ is an unknotted ball pair. As any PL homeomorphism of the boundary of an unknotted ball pair extends to the interior [Reference Rourke and Sanderson15, Theorem 4.4], we can extend $\Phi_{X^+}$ to a PL homeomorphism $\Phi_{X_V}\colon X_{V,1}\to X_{V,2}$ that maps $|\Gamma_1|\cap X_{V,1}$ to $|\Gamma_2|\cap X_{V,2}$.

Since each $N_{v,k}$ is PL-homeomorphic to a cone with base a 2-sphere containing $R_{v,k}$ and cone point v, such that $|\Gamma|\cap N_{v,k}$ corresponds to the cone over $|\Gamma|\cap \partial N_{v,k}$, we may cone the already defined map on each $\partial N_{v,1}$. This finishes the extension of $\Phi_X$ to the ambient sphere of Γ₁.

Proof. By Proposition 6.2, the oriented marked exteriors $(X_k^\circ,P_k^\circ)$ used to construct the $(X_k,P_k)$ are homeomorphic as manifolds with boundary pattern via a homeomorphism $\Phi\colon (X_1^\circ,P_1^\circ)\to (X_2^\circ, P_2^\circ)$ respecting vertex and edge regions of the boundary, as well as the orientation of each factor. By the Regular Neighbourhood Theorem [Reference Rourke and Sanderson15, Theorem 3.24] there is a PL isotopy H_A of $\partial X_2$ fixing $P_2^\circ$ and pushing the Φ-image of the chosen regular neighbourhood of $P_1^\circ$ onto the chosen regular neighbourhood of $P_2^\circ$. Denote the final homeomorphism of this isotopy by Ψ.

For each juncture $\gamma\subset P_1^\circ$, apply the Disc Theorem for pairs (Theorem 4.4) to get a PL isotopy H_γ of $A_{\Phi(\gamma)}$ that fixes $\partial A_{\Phi(\gamma)}$, preserves the juncture $\Phi(\gamma)$, and isotopes the postcomposition with $\Psi\circ \Phi$ of the embedding of the model disc at γ to the embedding of the model disc at $\Phi(\gamma)$. Using all the PL isotopies H_γ and extending to the whole boundary $\partial X_2$ as the identity, we obtain a PL isotopy H_D of $\partial X_2$ carrying $\Psi\circ\Phi(P_1)$ to P ₂.

The isotopy of $\partial X_2$ obtained by the concatenation of H_A and H_D extends to a PL isotopy of X ₂ [Reference Rourke and Sanderson15, Proposition 3.22(ii)]. Its final homeomorphism, when precomposed with Φ, yields a homeomorphism $(X_1,P_1)\to (X_2,P_2)$ respecting the vertex and edge regions.

Proof. For $k\in\{1,2\}$ consider the boundary patterns $P_k^\circ\subset P_k$ consisting only of the junctures. As an unoriented manifold, $P_k^\circ$ can be intrinsically characterized as the union of all embedded circles in P_k that are not the only circle in their component of $\partial X_k$. Hence, $\Phi_X$ maps $P_1^\circ$ to $P_2^\circ$.

Observe that $\Phi_X$ preserves vertex regions and edge regions: indeed, components of $\partial X_k$ with a single component of P_k correspond to isolated vertices and, on all other connected components, vertex regions are those receiving arcs from 3 distinct points of each juncture, while edge regions only receive arcs from at most two distinct points. As the number of arcs extended into the vertex region encodes the orientation of each juncture, $\Phi_X$ has to preserve the orientation of $P_k^\circ$ as well as the orientation of the triangles at isolated vertices. As the orientation of $X_k^\circ$ is determined by the orientation of the junctures and which side is the vertex region, $\Phi_X$ also preserves the orientation of the $X_k^\circ$.

Overall, we conclude that $\Phi_X$ is a PL homeomorphism $(X_1^\circ,P_1^\circ)\to (X_2^\circ,P_2^\circ)$ preserving the orientation of each factor. Thus, Proposition 6.3 can be applied to conclude that, up to decorations, $\Phi_X$ extends to an isomorphism $\Phi\colon \Gamma_1\to\Gamma_2$. From the way the decorations got encoded into the boundary patterns, it is clear that Φ respects decorations.

Proof. ($\Longrightarrow$) Suppose S is a splitting sphere for Γ. Building $X_\Gamma$ out of small enough regular neighbourhoods N_V and N_E as to avoid S, we see S is a reducing 2-sphere for $X_\Gamma$ since no component of $X_\Gamma \setminus S$ is an open 3-ball, as both have non-empty boundary.

($\Longleftarrow$) Denote by $\mathcal{S}$ the ambient 3-sphere of Γ, assume S is a reducing sphere for a marked exterior $X_\Gamma \subset \mathcal{S}$, and let $B_1, B_2 \subset \mathcal{S}$ be the 3-balls into which S splits $\mathcal{S}$. If for some $i\in \{1,2\}$ the intersection $|\Gamma| \cap B_i$ were empty, then we would have $B_i \subset X_\Gamma$, in contradiction with S being a reducing sphere. Hence, if S decomposes Γ as $\Gamma_1 \sqcup \Gamma_2$, then none of the $\Gamma_i$ is empty.

Proof. Let $\Gamma = (\mathcal{S}, V, E)$, let v be a cut vertex for Γ, and S a cut sphere through v. We construct a marked exterior $(X_\Gamma, P_\Gamma)$ using a small enough regular neighbourhood N_V of V so that N_V is in fact a regular neighbourhood of V in $(\mathcal{S}, |\Gamma| \cup S)$, and also, we use N_E small enough to be disjoint from S. Additionally, we ensure that $P_\Gamma$ is built from disc embeddings with image small enough to be disjoint from S, so that $S \cap P_\Gamma = \emptyset$. As $S \cap N_v$ is a regular neighbourhood of $\{v\}$ in S, it is a disc. The other side $D := S\cap X_\Gamma$ is thus a clean properly embedded disc in $(X_\Gamma, P_\Gamma)$.

We claim that D is a reducing disc for $(X_\Gamma, P_\Gamma)$. To see this, consider the two balls $B_1, B_2$ into which S separates $\mathcal{S}$. The curve $\partial D$ separates the component C of $\partial X_\Gamma$ containing the vertex region R_v into the two regions $C_i:= C \cap B_i$, for $i \in \{1,2\}$. We need to show that no C_i is a clean disc. Since S is a cut sphere, there is at least one edge e_i incident to v on each B_i. As the corresponding component $N_{e_i}$ of N_E is disjoint from S, we have $N_{e_i} \subset B_i$, and in particular $R_{e_i} \subset C_i$. The juncture between $R_{e_i}$ and R_v is thus contained in C_i, whence C_i is not clean.

Proof of Proposition 6.13

We need the following:

Claim. For every PL embedded arc $I\subset \operatorname{int}(M)$ with endpoints $v,u$, and for every $u' \in I\setminus \{v\}$, there is a PL isotopy of M fixing v and carrying I to the sub-arc of I with endpoints $v, u'$.

Before justifying this claim we use it to prove the proposition. By homogeneity of manifolds [Reference Rourke and Sanderson15, Lemma 3.33], there is a PL isotopy of M carrying v ₁ to v ₂, so we may assume $v_1=v_2=:v$. Choose a star neighbourhood N_v of v in the pair (M, I), and denote by $u_1', u_2'$, respectively, the points of intersection of the $(n-1)$-sphere $\partial N_v$ with each arc $I_1, I_2$. Using the above claim on both arcs reduces the problem to showing that there is a PL isotopy of M carrying the straight line segment $[v,u_1']$ onto $[v, u_2']$, with v being carried to itself.

Since $\partial N_v$ is connected, again by homogeneity of manifolds, there is a PL isotopy of $\partial N_v$ carrying $u_1'$ to $u_2'$. By coning at v, this isotopy extends to N, taking $[v,u_1']$ to $[v, u_2']$ as required. To extend it to all of M, we use the general fact every PL isotopy of the boundary of a manifold (in this case $M \setminus \operatorname{int}(N_v)$) extends to the interior [Reference Rourke and Sanderson15, Proposition 3.22(ii)].

Proof of the Claim

It suffices to show that the subspace $Q\subset I\setminus\{v\}$ of points u ^ʹ for which the claim holds is non-empty, open and closed in $I\setminus\{v\}$. Clearly $u \in Q$.

Let us verify that Q is open, beginning with u. A regular neighbourhood N_u of $\{u\}$ in the pair (M, I) is PL-homeomorphic to the standard n-ball $[-1,1]^n$, with $N_u \cap I$ corresponding to the straight line segment from 0 to a point p ₀ in $\partial([-1,1]^n)$. Let q ₀ be in the interior of this line segment. Since $[-1,1]^n$ is a cone with base $\partial([-1,1]^n)$ over any of its interior points, the formula $tp \mapsto (1-t) q_0 + tp$ with $p \in \partial([-1,1]^n)$ defines a PL homeomorphism of $[-1,1]^n$ fixing the boundary and taking $[0, p_0]$ to $[q_0,p_0]$. By Alexander’s trick [Reference Rourke and Sanderson15, Proposition 3.22(i)], such a map is PL-isotopic to the identity on $[-1,1]^n$ keeping the boundary fixed. This isotopy can then be transferred to a PL isotopy of N_u and extended as the constant isotopy in all of M. This shows that the point $q \in N$ corresponding to q ₀ is in Q, and so Q contains the half-open interval $\operatorname{int}(N_u) \cap I$.

To verify the openness condition at points $u' \in Q \setminus \{u\}$, proceed similarly: choose a regular neighbourhood $N_{u'}$ of u ^ʹ in (M, I), and model $(N_{u'}, N_{u'}\cap I)$ as the standard ball pair $([-1,1]^n, [-1,1]\times \{0\}^{n-1})$. The previous construction shows that $\operatorname{int}(N_{u'})\cap I \subset Q$. The same argument shows Q is closed in $I \setminus \{v\}$.

Proof. Let D be a reducing disc for a marked exterior $(X_\Gamma, P_\Gamma)$ in the ambient sphere $\mathcal{S}$ of Γ. Since $\partial D$ is disjoint from $P_\Gamma$, it is contained in a vertex region R_v or an edge region R_e of $\partial X_\Gamma$.

Let us first treat the case where $\partial D \subset \partial R_v$. Consider the 3-ball N_v containing v, which is a component of the vertex set neighbourhood N_V used in constructing $(X_\Gamma, P_\Gamma)$. Since N_v is a regular neighbourhood of $\{v\}$ in $(\mathcal{S}, |\Gamma|)$, the pair $(N_v, N_v \cap |\Gamma|)$ is PL-homeomorphic to a cone of $(\partial N_v, \partial N_v\cap|\Gamma|)$, with v corresponding to the cone point. Let $D_v\subset N_v$ be the disc corresponding to the cone of $\partial D$, and consider the 2-sphere $S := D \cup D_v$, which intersects $|\Gamma|$ precisely at v. We claim that S is a cut sphere.