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Supplementary material of paper “2-Manifold Tests for 3D Delaunay Triangulation-based Surface Reconstruction” Maxime Lhuillier

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1 Properties of c(T ∞ ) This section provides proofs of results (Theorem 1, Corollaries 1, 2, 3 and Lemma 4) whose statements are in Section 2 of the paper. The interior of a geometric simplex σ ⊂ R3 is σ\|∂σ| (the interior of point p is p, i.e. ∂p = ∅). Lemma 1 If σ ′ and σ are simplices in a simplicial complex and σ is not a face of σ ′ , (σ \ |∂σ|) ∩ σ ′ = ∅. Proof In a simplicial complex, σ ∩ σ ′ = ∅ or σ ∩ σ ′ is a face of σ. In the second case, σ ∩ σ ′ = σ implies σ ⊆ σ ′ (impossible). Therefore σ ∩ σ ′ is a face of σ which is not σ: σ ∩ σ ′ ⊆ |∂σ|. We obtain (σ \ |∂σ|) ∩ σ ′ = ∅. ⊓ ⊔ Lemma 2 If a set W ⊂ R3 includes a 3-ball, W cannot be included in a finite union of planes. Proof Let B be a 3-ball included in W with radius r > 0. Assume (reductio ad absurdum) that W ⊆ ∪ni=1 πi where πi is a plane in R3 and n is an integer. Thus B = ∪ni=1 (B ∩πi ). For every ǫ > 0, there is a cylinder Ci with radius r and height ǫ which includes 2-ball B ∩ πi . Since B ⊆ ∪ni=1 Ci , the volume of B is less than the sum of the volumes of the Ci s, i.e. 4πr3 /3 ≤ nπr2 ǫ. This is impossible if ǫ is small enough, e.g. ǫ = (2r)/(3n). ⊓ ⊔ Theorem 1 If c(T ) is a 3D Delaunay triangulation and τ is a vertex or an edge in c(T ), gτ is connected. Proof For all distinct tetrahedra ∆′0 and ∆′1 in Tτ∞ , we find a tetrahedron series ∆i ∈ Tτ∞ , i ∈ {0, 1, · · · k} Maxime Lhuillier Institut Pascal, 63171 Aubi` ere cedex, France Tel.: +33 (0)473407593 Fax: +33 (0)473407262 E-mail: [email protected]

Fig. 1 Notations of Theorem 1 proof (tetrahedra are represented by triangles). Left: p is a vertex of c(T ) \ c(∂T ). Middle: p is the center of an edge of c(T ). Right: p is a vertex of c(∂T ). 3-Ball B is centered at p. Point xi is in the intersection of B and tetrahedron interior ∆′i \ |∂∆′i | such that ∆′i ∈ Tτ . The gray area is |T \ Tτ |. The white triangles are in Tτ .

such that ∆i ∩ ∆i+1 is always a triangle, ∆0 = ∆′0 and ∆k = ∆′1 . The principle of the proof is the following: first assume that both ∆′i are finite and obtain the series in T by ray tracing along a line segment linking ∆′0 and ∆′1 . This segment should be “close enough” to τ to cross only tetrahedra ∆i ∈ Tτ , and it should “properly” intersect boundaries ∂∆i such that ∆i ∩ ∆i+1 is always a triangle. Here are the details. First, we show that there is a 3-ball B with center p such that p is a point in τ and B ∩ |T \ Tτ | = ∅. If τ is a vertex, we choose p = τ and obtain p ∈ / |T \ Tτ |. Furthermore R3 \ |T \ Tτ | is open in R3 (indeed, every geometric simplex is closed and T \ Tτ is a finite set): there is a 3-ball B centered at p such that B ⊆ R3 \ |T \ Tτ |, i.e. B ∩ |T \ Tτ | = ∅. If τ is an edge, let p be the center of τ . Now ∆ ∈ T and p ∈ ∆ imply (τ \ ∂τ ) ∩ ∆ 6= ∅, which in turn implies that τ is a face of ∆ (Lemma 1), i.e. ∆ ∈ Tτ . Thus p ∈ / |T \ Tτ |, and we obtain B as in the case where τ is a vertex. Assume that ∆′0 and ∆′1 are finite. Let B0 = B ∩ ′ (∆0 \|∂∆′0 |) and B1 = B∩(∆′1 \|∂∆′1 |). Second, we show that there are points x0 ∈ B0 and x1 ∈ B1 such that the

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line (x0 x1 ), which includes x0 and x1 , does not intersect the edges in c(Tτ ). Fig. 1 shows notations xi , B and ∆′i . We have Bi 6= ∅ (since B is centered at p which is in |∂∆′i |). Let x0 ∈ B0 . Thanks to Lemma 1, x0 is not in an edge of c(Tτ ). Let πj be a plane including x0 and the j-th edge in c(Tτ ). Since B1 is open in R3 (indeed, the interior of a tetrahedron is open in R3 [1]) and B1 6= ∅, B1 cannot be included in ∪j πj (Lemma 2). Now we have x1 ∈ B1 \ ∪j πj and (x0 x1 ) does not intersect the edges in c(Tτ ). Let x0 x1 be the line segment between x0 and x1 . Third, we show that x0 x1 ⊂ |Tτ |. Let x be a point in x0 x1 . Since |T | is convex (indeed, T is Delaunay) and xi ∈ |T |, there is ∆ ∈ T such that x ∈ ∆. Assume (reductio ad absurdum) that ∆ ∈ T \ Tτ . Since xi ∈ B and B is convex, x ∈ B. We obtain x ∈ B ∩ ∆ ⊆ B ∩ |T \ Tτ |, which contradicts B ∩ |T \ Tτ | = ∅. Thus x0 x1 ⊂ |Tτ |. Fourth, we show Theorem 1 assuming that ∆′0 and ′ ∆1 are finite. Let ∆ ∈ Tτ and I∆ = ∆ ∩ x0 x1 . Since ∆ is convex, I∆ is a line segment whose end-vertices are in |∂∆|∪{x0 , x1 }. Furthermore, the I∆ interior is included in the ∆ interior (since line (x0 x1 ) does not intersect the ∆ edges). Since two distinct tetrahedra in a simplicial complex have non intersecting interiors (Lemma 1) and x0 x1 ⊂ |Tτ |, the non empty segments I∆ form a partition of x0 x1 (two of them can only intersect at their end-vertices). Therefore we can order the ∆i ∈ Tτ intersected by x0 x1 from x0 to x1 such that x0 ∈ ∆0 , x1 ∈ ∆k , ∀i, ∆i ∩ ∆i+1 ∩ x0 x1 = I∆i ∩ I∆i+1 6= ∅. Now a ∆i triangle intersects a ∆i+1 triangle at a point which is not in a c(Tτ ) edge; we see that ∆i ∩ ∆i+1 is a triangle thanks to Lemma 1. We also have ∆′0 = ∆0 and ∆′1 = ∆k thanks to Lemma 1 and xi ∈ (∆′i \|∂∆′i |)∩∆i . Last, we show Theorem 1 assuming that ∆′0 is infinite. Thus τ ⊂ ∆′0 = v∞ abc. Since τ ∈ c(T ), v∞ is not a vertex of τ . We obtain τ ⊂ abc where abc ∈ ∂T . We have ∆′′0 ∈ T such that ∆′′0 ∩ ∆′0 = abc. Since ∆′′0 is finite and ∆′′0 ∈ Tτ , we replace ∆′0 by ∆′′0 in the previous part of the proof (step 4) and obtain a series of adjacent tetrahedra between ∆′0 and ∆′1 . The proof is the same if ∆′1 is infinite. ⊓ ⊔ Corollary 1 If τ is an edge in c(T ), gτ is a cycle. ∞ . A Proof We use notation τ = ab. Let abcd ∈ Tab ∞ tetrahedron in Tab which is adjacent to abcd is written abcd′ or abc′ d where d′ 6= d and c′ 6= c. Since every triangle is included in exactly two tetrahedra in c(T ∞ ), abcd′ exists and is unique (same as for abc′ d), and abcd′ 6= abc′ d (otherwise c ∈ {c′ , d}, which is impossible). Therefore every vertex in graph gτ exactly has two neighbors. Since gτ is a non empty finite graph which is connected (Theorem 1), gτ is a cycle. ⊓ ⊔

Maxime Lhuillier

Lemma 3 Assume that τ is a vertex or an edge in c(T ), O ⊆ T ∞ and ∂O ⊂ c(T ). Thus, τ ∈ c(∂O) iff Oτ 6= ∅ and Oτc 6= ∅. Proof If τ ∈ c(∂O), τ is in a triangle which is the intersection of a tetrahedron in Oτ and another in Oτc . Thus Oτ 6= ∅ and Oτc 6= ∅. Conversely, if Oτ 6= ∅ and Oτc 6= ∅, there is a tetrahedron series ∆i ∈ Tτ∞ between a tetrahedron in Oτ and another in Oτc such that ∆i ∩ ∆i+1 is always a triangle (Theorem 1). We thus have i such that ∆i ∈ Oτ , ∆i+1 ∈ Oτc , and ∆i ∩ ∆i+1 is a triangle including τ . We obtain τ ∈ c(∂O). ⊓ ⊔ Corollary 2 Let v be a vertex in c(T ). We have v ∈ c(∂T ) iff Tv∞ contains at least one infinite tetrahedron. Proof Since v ∈ c(T ), Tv 6= ∅. Note that Tvc is the set of infinite tetrahedra including v. We then use Lemma 3 with O = T and τ = v. ⊓ ⊔ Lemma 4 If v is a vertex in c(T ∞ ), we have ∂Tv∞ = {abc, abcv ∈ Tv∞ }. Proof Every triangle in ∂Tv∞ is included in a tetrahedron abcv in Tv∞ . Since the only tetrahedron in Tv∞ which includes abc is abcv, abc ∈ ∂Tv∞ . Since triangle abv is included in two tetrahedra in c(T ∞ ), we have abc′ v ∈ T ∞ such that c′ 6= c. Therefore abc′ v ∈ Tv∞ and abv ∈ / ∂Tv∞ . ⊓ ⊔ Corollary 3 Let v be a vertex in c(T ). The vertices and edges of c(∂Tv∞ ) form a connected graph. Proof Let a and a′ be two vertices in c(∂Tv∞ ). We have tetrahedra abcv and a′ b′ c′ v in c(Tv∞ ). If abcv = a′ b′ c′ v, a′ ∈ {a, b, c}. There is thus a path between a and a′ in triangle abc, whose edges are in c(∂Tv∞ ) (Lemma 4). Now we assume that abcv 6= a′ b′ c′ v. Since gv is connected (Theorem 1), we have a tetrahedron series ∆i ∈ Tv∞ , i ∈ {0, 1 · · · k} such that ∆i ∩ ∆i+1 is a triangle, ∆0 = abcv and ∆k = a′ b′ c′ v. Thanks to Lemma 4, ∆i = ai bi ci v where {ai , bi , ci , ai bi , bi ci , ci ai , } is included in c(∂Tv∞ ). Since ∆i ∩ ∆i+1 is a triangle ai bi v, which is also written ai+1 bi+1 v, triangles ai bi ci and ai+1 bi+1 ci+1 have common vertices in c(∂Tv∞ ). By chaining edges in c(∂Tv∞ ) coming from tetrahedra ∆0 , ∆1 , · · · ∆k , we obtain a path between a and a′ . ⊓ ⊔ 2 Proof for Theorems 1 and 2 Here we present a proof for Theorems 1 and 2 (known results) for the completeness of the paper. We use notations B2 = {x ∈ R2 , ||x|| < 1}, S1 = {x ∈ R2 , ||x|| = 1}. Here “regular” means “regular in |∂O|”, and “good”

Supplementary material of paper “2-Manifold Tests for 3D Delaunay Triangulation-based Surface Reconstruction”

Fig. 2 The three cases in Lemma 5. Left: p is in the interior of a triangle and B ∩ t1 is a 2-ball. Middle: p is in the interior of an edge and every B ∩ ti is a half-2-ball. Right: p is at a vertex. In all cases, B is a 3-ball centered at p.

means “∂O-good”. If X ⊆ R3 , X has the topology induced by R3 , i.e. the open sets of X are the sets V ∩ X where V is an open set in R3 . A 3-ball is an open set in R3 .

2.1 Convenient Neighborhoods Remember that a geometric simplex σ is closed in R3 , i.e. R3 \σ is open. A half-2-ball centered at point p is the intersection of a 3-ball centered at p and a half-plane whose border line contains p. Lemma 5 Let point p ∈ |∂O| and t1 · · · tn be the triangles in ∂O which include p. We have ǫ > 0 such that every 3-ball B centered at p with radius less than ǫ meets B ∩ |∂O| = B ∩ (∪ni=1 ti ) and the following conditions. If p is in t1 interior, n = 1 and B ∩ t1 is a 2-ball. If p is in the interior of a t1 edge, this edge is a face of every ti and every B ∩ ti is a half-2-ball. If p is not in the t1 interior or in a t1 edge interior, p is a t1 vertex. Proof Let U = |∂O\{t1 , · · · tn }|. Since p ∈ / U and R3 \U is open (indeed U is closed), we have a 3-ball B centered at p such that B ⊆ R3 \U . Thus B∩|∂O| = B∩(∪ni=1 ti ). We apply the following scheme for every ti . If p is in the interior of ti , we reduce the radius of B such that B ∩ ti is a 2-ball centered at p (the radius is less than the distance between p and all the edges of ti ). If p is in the interior of a ti edge, we reduce the radius of B such that B ∩ ti is a half-2-ball centered at p (the radius is less than the distance between p and the two edges of ti which do not contain p). Fig. 2 shows p, B and ti . If p is in t1 interior, p cannot be within another triangle of simplicial complex c(∂O) due to Lemma 1. Thus n = 1. If p is in the interior of one edge of t1 , p cannot be in the interior of triangle ti where i > 1 (Lemma 1). Thus p ∈ ti implies that p lies within one of the edges of ti . According to Lemma 1, this edge is the edge of t1 . ⊓ ⊔

Lemma 6 If point p is regular, we have 3-balls B as in Lemma 5 such that B ∩ |∂O| is homeomorphic to B2 .

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Proof Since p is regular in |∂O|, we have a homeomorphism ϕ and an open set U in R3 such that p ∈ U and ϕ(U ∩ |∂O|) = B2 . Here we would like to replace the general set U by a sufficiently small 3-ball B centered at p. The principle of the proof is the following: first use Lemma 5 to obtain B such that B ∩ |∂O| = B ∩ (∪i ti ) and B ⊆ U , second show that B ∩ (∪i ti ) is star-shaped with respect to p, third show that B ∩ (∪i ti ) is simply connected, and finally the Riemann Mapping Theorem asserts that ϕ(B ∩ |∂O|) is homeomorphic to B2 . First Lemma 5 provides ǫ > 0. Since p ∈ U and U is open in R3 , we have ǫ′ > 0 such that the p-centered 3-ball with radius ǫ′ is included in U . We replace ǫ by min{ǫ, ǫ′ }. Now every p-centered 3-ball B with radius less than ǫ simultaneously meets B ⊂ U , B ∩ |∂O| = B ∩ (∪i ti ) and ∀i, p ∈ B ∩ ti . Let X = B ∩ (∪i ti ). Secondly, we show that X is star-shaped with respect to p, i.e. x ∈ X implies that line segment px ⊆ X. There is triangle ti such that both x and p are in ti . Since B and ti are convex, px ⊆ B ∩ ti ⊆ X. Third we show that X is simply connected, i.e. X is path-connected and every continuous function from S1 to X can be extended to a continuous function from S1 ∪ B2 to X. Note that X is path-connected since it is star-shaped. Let f be a continuous function from S1 to X. We extend f in S1 ∪ B2 by g defined by g(x) = (1 − ||x||)p+||x||f (x/||x||) if x 6= 0 and g(0) = p. Since X is is star-shaped with respect to p, g(x) ∈ pf (x/||x||) ⊆ X. Furthermore, g is continuous at x 6= 0. It is also continuous at 0 since y 6= 0 implies ||g(y) − g(0)|| ≤ (||p|| + ||f ||∞ )||y|| (we have ||f ||∞ < +∞ since f is continuous in compact S1 ). Thanks to the homeomorphism ϕ, ϕ(B ∩ |∂O|) is also simply connected. Since B ∩ |∂O| is open in |∂O| (indeed B is open in R3 ) and ϕ is a homeomorphism, ϕ(B ∩ |∂O|) is open in B2 . Now ϕ(B ∩ |∂O|) is simply connected and open in R2 . Thus the Riemann Mapping Theorem implies that ϕ(B ∩ |∂O|) is homeomorphic to B2 . We see that B ∩|∂O| is homeomorphic to B2 thanks to ϕ. ⊓ ⊔

2.2 Proofs for Theorems 1 and 2 It should be remembered that A ⊂ Rn is not connected if A is the union of two (or more than two) disjoint non-empty open sets of Rn . Furthermore, the homeomorphism image of a connected set is connected, and B2 is also connected. If L is a list of triangles in a (abstract) simplicial complex, ∂L is the set of edges such that every edge is a face of exactly one triangle in L. Lemma 7 If O ⊂ T ∞ and ∂O ⊂ c(T ), ∂∂O = ∅.

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Maxime Lhuillier

Proof Let e be an edge in c(∂O). Thus Oe 6= ∅ and Oec 6= ∅. Since ∂O ⊂ c(T ), Corollary 1 implies that ge is a cycle of adjacent tetrahedra ∆0 · · · ∆n−1 ∆0 including e. Since Oe ∪Oec = {∆0 , · · · , ∆n−1 }, we have i such that ∆i ∈ Oe and ∆i+1 ∈ Oec and j 6= i such that ∆j ∈ Oec and ∆j+1 ∈ Oe (indices modulo n). Thus e is included in triangles ∆i ∩∆i+1 and ∆j ∩∆j+1 of ∂O. Since n ≥ 3, these triangles are different and we obtain ∂∂O = ∅. ⊓ ⊔ Fig. 3 Notations for the proof of Lemma 10.

Lemma 8 If triangle t is in ∂O and point p is in the interior of t, p is regular. Proof We use Lemma 5 and obtain a 3-ball B centered at p such that B ∩|∂O| = B ∩t and B ∩t is a 2-ball. ⊓ ⊔ Lemma 9 Let e be an edge in c(∂O) and point p be in the interior of e. Thus p is regular iff e is good. Proof First we assume that e is good and show that p is regular. There are exactly two triangles t1 and t2 in ∂O which include e. We use Lemma 5 and obtain a 3-ball B centered at p such that B ∩ |∂O| = B ∩ (t1 ∪ t2 ) and both B ∩ t1 and B ∩ t2 are half-2-balls joined by their common diameter in e. Thus B ∩ |∂O| is homeomorphic to a 2-ball. Conversely, assume that p is regular and show that e is good. Let n be the number of triangles ti ∈ ∂O which include e. Note that e ∈ c(∂O) implies n ≥ 1. Thanks to Lemma 6, there is a 3-ball B centered at p such that B ∩ |∂O| = B ∩ (∪ni=1 ti ) and every B ∩ ti is a half-2-ball and ϕ(B ∩(∪ni=1 ti )) = B2 where ϕ is a homeomorphism. Furthermore, n = 1 is impossible thanks to Lemma 7. Thus we define C2 = B ∩ (t1 ∪ t2 ). Let Ci = B ∩ (ti \ e) if i > 2. Partition {Ci } of B ∩ |∂O| is such that every Ci is homeomorphic to B2 . Thus {ϕ(Ci )} is a partition of B2 such that every ϕ(Ci ) is non-empty and open in R2 . Since the inequality n ≥ 3 contradicts the fact that B2 is connected, n = 2. ⊓ ⊔ Lemma 10 Let p be a vertex in c(∂O) such that p and the p-incident edges in c(∂O) are good. Thus p is regular. Proof Since p is good, the p-incident triangles in ∂O are t0 · · · tk−1 where ti = pvi v(i+1) mod k . Let t′i = ozi z(i+1) mod k be a triangle in R2 such that o = (0, 0) and zi = (cos(2πi/k), sin(2πi/k)). Let S ′ be the pure simplicial complex in R2 whose triangles are the t′i s. The proof is the following: first find homeomorphism ϕ such that ϕ(∪i t′i ) = ∪i ti , then find a neighborhood of p in |∂O| which is homeomorphic to B2 thanks to ϕ. We first define a function ϕ from |S ′ | to |∂O| by its values on vertices and linear interpolation in triangles t′i : ϕ(o) = p and ϕ(zi ) = vi . Function ϕ is continuous

and ϕ(t′i ) = ti . We have 3 ≤ k (otherwise a p-incident edge in c(∂O) is not good). Assume (reductio ad absurdum) that there are i and j such that i 6= j and vi = vj . Thus edge pvi is included in triangles ti−1 , ti , tj , tj−1 (indices modulo k). Since k ≥ 3, there are at least 3 different triangles in {ti−1 , ti , tj , tj−1 }. This is impossible since edge pvi is good. Since ϕ is injective on the vertex set of S ′ , ϕ is injective in |S ′ | [1] (ϕ is a simplicial map defined on S ′ ). Now ϕ is a piecewise linear function which is bijective between ∪i t′i and ∪i ti . Thus ϕ is homeomorphism. Second we obtain the neighborhood of p in |∂O| as follows: let V ∩ |∂O| where V is an open set in R3 and p ∈ V , furthermore it is homeomorphic to B2 since ϕ−1 (V ∩ |∂O|) is a 2-ball. Let B be a 3-ball centered at p such that B ∩|∂O| = B ∩(∪i ti ) (Lemma 5). Since ϕ is continuous at o, we have a 2-ball B ′ centered at o such that ϕ(B ′ ) ⊆ B ∩ (∪i ti ). Since ϕ is homeomorphism between ∪i t′i and ∪i ti , ϕ(B ′ ) is open in ∪i ti , i.e ϕ(B ′ ) = W ∩ (∪i ti ) where W is open in R3 . Let V = B ∩ W . Now V is an open set in R3 , p ∈ V and V ∩ |∂O| = W ∩ B ∩ (∪i ti ) = ϕ(B ′ ) ∩ B = ϕ(B ′ ). Fig. 3 shows notations ti , t′i , B, B ′ , o, p and ϕ(B ′ ).

⊓ ⊔

Lemma 11 Let p be a vertex in c(∂O) such that p is regular and the p-incident edges in c(∂O) are good. Thus p is good. Proof Since the p-incident edges are good, the adjacency graph of the p-incident triangles ti in ∂O is finite, non empty, and each of its vertex has exactly two edges. Thus this graph is a list of m disconnected cycles. Let Ui be the list of the triangles of the i-th cycle. Fig. 4 shows the Ui s. The principle of the proof is the following. Thanks to the good p-incident edges, we first show that |Ui | ∩ |Uj | = {p} (if i 6= j) and p is regular in every |Ui | (thanks to Lemma 10). Now every |Ui | has a p neighborhood homeomorphic to B2 , as has ∪i |Ui | (since p is regular in |∂O|). Then we show that B2 is homeomorphic to a union of m disjoint open sets associated to

Supplementary material of paper “2-Manifold Tests for 3D Delaunay Triangulation-based Surface Reconstruction”

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implies that every edge in c(∂O) is good (Lemma 9), and thus every vertex in c(∂O) is good (Lemma 11).

References

Fig. 4 Notations for the proof of Lemma 11.

the Ui s. Since B2 is connected, there is only one Ui (i.e. m = 1) and the result is finally obtained. First we show that i 6= j and x ∈ |Ui | ∩ |Uj | imply x = p. Since Ui ∩ Uj = ∅ and two distinct triangles cannot intersect in simplicial complex c(∂O) except at their edges (Lemma 1), x is in an edge. Since c(Ui ) and c(Uj ) have distinct edges (indeed the p-incident edges are good) which cannot intersect in their interiors (Lemma 1), x is a vertex. If x is not p, edge xp is in two c(Ui ) which is a contradiction. Second we show that p is regular in |Ui |. Since the adjacency graph of the triangles in Ui is a cycle, p is Ui -good and the p-incident edges in c(Ui ) are Ui -good. We obtain the result by using Lemma 10 and replacing ∂O by Ui . Thanks to Lemma 6 and since p is regular in |∂O|, there is a 3-ball B centered at p and a homeomorphism ϕ such that B ∩ |∂O| = B ∩ (∪i |Ui |) and ϕ(B ∩ (∪i |Ui |)) = B2 . Let Ci = B ∩ |Ui |. Since p is regular in |Ui |, we use Lemma 6 (replace ∂O by Ui ) to reduce the B radius such that every Ci is homeomorphic to B2 . Thus ϕ(Ci ) and ϕ(Ci \ {p}) are open in R2 . Since C1 , C2 \ {p} · · · Cm \ {p} is a partition of B ∩ (∪i |Ui |), B2 = ϕ(B ∩ (∪i |Ui |)) has a partition of non-empty open sets ϕ(C1 ), ϕ(C2 \ {p}) · · · ϕ(Cm \ {p}). Since B2 is connected, m = 1. Therefore U1 is the set of p-incident triangles in ∂O, and p is ∂O-good. ⊓ ⊔ Now we show Theorem 2 for vertex p in c(∂O). If p and the p-incident edges are good, p is regular (Lemma 10). Conversely, assume that vertex p is regular: there is a homeomorphism ϕ between B2 and |∂O|∩ U where U is an open set in R3 and p ∈ U . This implies that every p′ ∈ |∂O| ∩ U is also regular thanks to the same homeomorphism and neighborhood. If we choose p′ in the interior of a p-incident edge e in c(∂O), e is good due to Lemma 9 applied to p′ . Now p is good due to Lemma 11. We also show Theorem 1. If the edges and the vertices of c(∂O) are good, p ∈ |∂O| is regular if p is a vertex (Lemma 10), or if p is in an edge interior (Lemma 9), or if p is in a triangle interior (Lemma 8). Conversely, assume that every p ∈ |∂O| is regular. This

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