EXTENSION OF INCOMPRESSIBLE SURFACES ON THE BOUNDARY OF 3-MANIFOLDS

Michael Freedman, Hugh Howards and Ying-Qing Wu

Abstract. An incompressible surface F on the boundary of a compact orientable 3manifold M is arc-extendible if there is an arc γ on ∂M − IntF such that F ∪ N(γ) is incompressible, where N(γ) is a regular neighborhood of γ in ∂M. Suppose for simplicity that M is irreducible, and F has no disk components. If M is a product F × I, or if ∂M − F is a set of annuli, then clearly F is not arc-extendible. The main theorem of this paper shows that these are the only obstructions for F to be arc-extendible.

Suppose F is a compact incompressible surface on the boundary of a compact, orientable, irreducible 3-manifold M. Let F′ be a component of ∂M − IntF. We say that F is arc-extendible (in F′) if there is a properly embedded arc γ in F′ such that F ∪ N(γ) is incompressible. In this case γ is called an extension arc of F. We study the problem of which incompressible surfaces on the boundary M are arcextendible. This is useful in, for example, finding a sequence of mutually nonparallel incompressible surfaces in a 3-manifold.

Denote by I the unit interval [0, 1]. We say that M is a product F × I if there is a homeomorphism φ : M ≅ F × I with φ(F) = F × 1. Note that in this case F′ = ∂M − IntF, and F is not arc-extendible. A surface F is diskless if it has no disk component. An incompressible surface with a disk component is always arc-extendible, unless the disk lies on a sphere component of ∂M. Thus to avoid trivial cases, we will only consider arc-extension of diskless surfaces.

Theorem 1. Let F be a diskless, compact, incompressible surface on the boundary of a compact, orientable, irreducible 3-manifold M, and let F′ be a non-annular component of ∂M − IntF. Then either F is arc-extendible in F′, or M is a product F × I.

The proof of the theorem involve some deep results about incompressible surfaces related to Dehn surgery and 2-handle additions. It breaks down into three cases. The

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1991 Mathematics Subject Classification. Primary 57N10..

Freedman was partially supported by an NSF grant. Wu’s research at MSRI was supported in part by NSF grant #DMS 9022140.

case that F′ is a thrice punctured sphere is treated in Theorem 4, which shows that if the surface obtained by gluing F and F′ along one of the boundary curve of F′ is compressible for all the three boundary curves of F′, then M must be a product. The second case is that F′ is parallel into F (see below for definition). A similar result as above holds in this case. Theorem 9 shows that in the remaining case there is an arc γ intersecting some circle C in F′ at one point, so that all but at most three Dehn twists of γ along C are extension arcs of F. Moreover, in this case the extension arc γ of F can be chosen to have endpoints on any prescribed components of ∂F′. See Theorem 10 below.

Note that the irreducibility of M is irrelevant to the compressibility of surfaces on ∂M. However, this does make the conclusion of the theorem simpler. If we drop this assumption from the theorem, the conclusion should be changed to “Either F is arc-extendible in F′, or there is a component F_{0} of F, and a homeomorphism φ : M ≅ F_{0} × I#M′ for some M′, such that φ(F_{0}) = F_{0} × 1, and φ(F′) = F_{0} × 0 ∪ ∂F_{0} × I.”

Given a simple closed curve α on a surface S on the boundary of M, we use M[α] to denote the manifold obtained by adding a 2-handle to M along the curve α. More explicitly, M[α] is the union of M and a D^{2} × I, with the annulus (∂D^{2}) × I glued to a regular neighborhood N(α) of α on ∂M. Use S[α] to denote the surface in M[α] corresponding to S, i.e. S[α] = (S − N(α)) ∪ (D^{2} × ∂I). The following two lemmas are very useful in dealing with incompressible surfaces. Various versions of Lemma 2 have been proved by Przytycki [Pr], Johannson [Jo], Jaco [Ja], and Scharlemann [Sch]. The lemma as stated is due to Casson and Gordon [CG].

Lemma 2. (The Handle Addition Lemma [CG].) Let α be a simple closed curve on a surface S on the boundary of an orientable irreducible 3-manifold M, such that S is compressible and S − α is incompressible. Then S[α] is incompressible in M[α], and M[α] is irreducible.

Lemma 3. (The Generalized Handle Addition Lemma.) Let S be a surface on the boundary of an orientable 3-manifold M, let γ be a 1-manifold on S, and let α be a circle on S disjoint from γ. Suppose S − γ is compressible and S − (γ ∪ α) is incompressible. If D is a compressing disk of S[α] in M[α], then there is a compressing disk D′ of S − α in M such that ∂D′ ∩ γ ⊂ ∂D ∩ γ.

Proof. This is essentially [Wu2, Theorem 1]. The theorem there stated that ∂D′ ∩ γ has no more points than ∂D ∩ γ, but the proof there gives the stronger conclusion that ∂D′ ∩ γ ⊂ ∂D ∩ γ. □

We first study the case that the surface F′ in Theorem 1 is a thrice punctured sphere. Let α_{1}, α_{2}, α_{3} be the boundary curves of F′. Since F′ is a component of ∂M − IntF, we have α_{i} ⊂ ∂F for i = 1, 2, 3. Note that if IntF ∪ IntF′ ∪ α_{i} is incompressible for some i, then for any essential arc γ on F′ with ∂γ ⊂ α_{i}, the surface F ∪ N(γ) is incompressible. Hence the following theorem proves Theorem 1 in the case that F′ is a twice punctured disk. However, it should be noticed that a similar statement is false if we drop the assumption that F′ is a sphere with three holes.

Theorem 4. Let F be a diskless compact incompressible surface on the boundary of a compact, orientable, irreducible 3-manifold M, and let F′ be a component of ∂M − IntF which is a punctured sphere with ∂F′ = α_{1} ∪ α_{2} ∪ α_{3}. If IntF ∪ IntF′ ∪ α_{i} is compressible for i = 1, 2, 3, then M is a product F × I.

Proof. We fix some notation. Write F̂ = F ∪ F′. Denote by F̂_{i} the surface obtained by gluing IntF and IntF′ along α_{i}, i.e. F̂_{i} = IntF ∪ IntF′ ∪ α_{i}. Similarly, write F̂_{ij} = IntF ∪ IntF′ ∪ α_{i} ∪ α_{j}.

First notice that F′ is incompressible. This is because each simple closed curve on F′ is isotopic to one of the α_{i} ⊂ F, and because F is incompressible and diskless. Since IntF ∩ IntF′ = ∅, the surface IntF ∪ IntF′ is incompressible.

Let M′ be a maximal compression body of ∂M in M. Then a surface on the boundary of M is compressible in M if and only if it is compressible in M′. Notice that if M ≠ M′, then M′ is never a product F × I, so if the theorem is true for M′, it is true for M. Hence after replacing M by M′ if necessary, we may assume without loss of generality that M is a compression body.

We claim that the curves α_{1}, α_{2}, α_{3} are mutually nonparallel on F̂, that is, no component of F is an annulus with both boundary components on F′. If two curves α_{1}, α_{2}, say, are parallel on F̂, then the surface IntF ∪ IntF′ = F̂ − α_{1} ∪ α_{2} ∪ α_{3} is incompressible if and only if F̂_{1} = F̂ − α_{2} ∪ α_{3} is incompressible. However, by assumption F̂_{1} is compressible, and we have shown that IntF ∪IntF′ is incompressible. Hence the claim follows.

Since F̂_{i} is compressible, and F̂_{i} − α_{i} = IntF ∪ IntF′ is incompressible, we can apply the Handle Addition Lemma (Lemma 2) to F̂_{i} and α_{i} to conclude that after adding a 2-handle along α_{i}, the surface F̂_{i}[α_{i}] is incompressible in M[α_{i}], and M[α_{i}] is irreducible.

Consider the surface F̂[α_{1}]. Notice that after adding the 2-handle, the surface F′ becomes an annulus on F̂[α_{1}] with boundary α_{2} ∪ α_{3}, so the two curves α_{2}, α_{3} are parallel on F̂[α_{1}]. Thus, F̂_{1}[α_{1}] = F̂[α_{1}] − α_{2} ∪ α_{3} being incompressible in M[α_{1}] implies that F̂[α_{1}] − α_{2} is incompressible in M[α_{1}]. With the above notation, this says that F̂_{13}[α_{1}] is incompressible in M[α_{1}].

By assumption F̂_{3} is compressible in M. Let D be a compressing disk of F̂_{3} in M. Then ∂D is disjoint from α_{1} ∪ α_{2}, because ∂D ⊂ F̂_{3}. Also, ∂D is not isotopic to α_{1} in F̂_{13}, otherwise α_{1} would bound a disk in M, contradicting the assumption that F is diskless and incompressible. We have shown that F̂_{13}[α_{1}] is incompressible in M[α_{1}], so D is not a compressing disk of F̂_{13}[α_{1}] in M[α_{1}], and hence ∂D must bound a disk in F̂_{13}[α_{1}]. This is true if and only if ∂D is coplanar to α_{1} on F̂_{13}, that is, either ∂D is parallel to α_{1}, or it bounds a once punctured torus T in F̂_{13} which contains α_{1} as a nonseparating curve. The first possibility has been ruled out, so the second must be true. Let T̂ be the torus T ∪D. Since we have assumed above that M is a compression body, either (i) T̂ is parallel to a boundary component of M, or (ii) T̂ bounds a solid torus.

If T̂ is parallel to a boundary component T_{0} of M, then after adding the 2-handle, the surface T̂[α_{1}] becomes a sphere which separates T_{0} from F̂[α_{1}], hence is a reducing sphere of M[α_{1}], which contradicts the irreducibility of M[α_{1}]. Similarly, if T̂ bounds a solid torus V but α_{1} is not a longitude of V, then after adding the 2-handle the manifold would have a lens space or S^{2} × S^{1} summand, which again contradicts the irreducibility of M[α_{1}]. (Note that M[α_{1}] cannot be a lens space because it has some boundary components.)

We have now shown that there is a compressing disk D of F̂_{3} in M which cuts the manifold into two pieces, one of which is a solid torus V which contains α_{1} as a longitude, but is disjoint from α_{2}. Let D_{1} be a meridian disk of V. Then ∂D_{1} ∩α_{1} is a single point, and ∂D_{1} is disjoint from α_{2} because ∂V is disjoint from α_{2}. Notice that ∂D_{1} is not coplanar to α_{2}, for if ∂D_{1} were parallel to α_{2} then α_{2} would also intersect α_{1}, and if ∂D_{1} would bound a once punctured torus containing α_{2} then ∂D_{1} would be a separating curve on ∂M, so it would intersect α_{1} in an even number of points, either case leading to a contradiction. Thus, after adding a 2-handle to M along α_{2}, the disk D_{1} remains a compressing disk of F̂[α_{2}]. Since the two curves α_{1} and α_{3} are parallel in F̂[α_{2}], and since D_{1} intersects α_{1} in a single point, we can isotope D_{1} to another disk D_{2} in M[α_{2}] so that it intersects each of α_{1} and α_{3} in a single point. We are looking for such a disk in M; however D_{2} is not necessary the one because it may intersect the attached 2-handle.

Recall that the surface F̂_{2} is compressible, but the surface F̂_{2} −α_{2} = IntF ∪IntF′ is incompressible. Hence we can apply the Generalized Handle Addition Lemma (Lemma 3, with S = F̂, γ = α_{1} ∪ α_{3}, and α = α_{2}) to conclude that there is also a compressing disk D_{3} of F̂ in M, such that ∂D_{3} is disjoint from α_{2}, and ∂D_{3} ∩ (α_{1} ∪ α_{3}) is a subset of ∂D_{2} ∩ (α_{1} ∪ α_{3}).

The set ∂D_{3} ∩ (α_{1} ∪ α_{3}) is nonempty, otherwise, since ∂D_{3} is also disjoint from α_{2}, D_{3} would be a compressing disk of IntF ∪ IntF′, contradicting the incompressibility of IntF ∪IntF′. Since α_{1} ∪α_{2} ∪α_{3} is separating on F̂, the curve ∂D_{3} can not intersect α_{1} ∪ α_{2} ∪ α_{3} at a single point. It follows that ∂D_{3} ∩ (α_{1} ∪ α_{3}) = ∂D_{2} ∩ (α_{1} ∪ α_{3}), that is, ∂D_{3} intersects each of α_{1}, α_{3} in a single point. Such a disk is called a bigon.

Denote by D_{13} the bigon D_{3} above. Interchanging the rules of α_{1} and α_{2} in the above argument, we get another compressing disk D_{23} of F̂ in M, which is disjoint from α_{1}, and intersects each of α_{2}, α_{3} in a single point. By a simple innermost circle outermost arc argument, we can isotope D_{13} so that it is disjoint from D_{23}, and still has the same number of intersection points with each α_{i}. Cutting M along D_{13} ∪ D_{23}, we get a submanifold M′ of M, in which the surface F′ becomes a disk F̃′ ⊂ F′, and the surface F becomes a surface F̃ ⊂ F. It is clear that one boundary component C of F̃ bounds a disk on ∂M′, namely the union of F̃′ and the two copies of D_{13} ∪ D_{23}. Since F is incompressible, this curve C bounds a disk in F, so F̃ must be a disk. These disks together form a sphere boundary component of M′. Since M is irreducible, M′ must be a 3-ball, so it is a product F̃ × I. Gluing back along D_{13} and D_{23}, we see that M is a product F × I. This completes the proof of Theorem 4. □

Below, F, F′ and M will be as in Theorem 1. Using Theorem 4 we may assume that F′ is not a thrice punctured sphere. A curve C′ on F′ is ∂-nonseparating if (i) C′ is not parallel to a boundary curve on F′, and (ii) there is a proper arc γ in F′ intersecting C′ in a single point. A sub-surface G′ of F′ is parallel into F if there is a product G′ × I ⊂ M such that G′ = G′ × 0, and G′ × 1 ⊂ F. Similarly, a curve C′ on F′ is parallel into F if there is an embedded annulus A ⊂ M with ∂A = C′ ∪ C, where C ⊂ F.

Lemma 5. If F′ is compressible, then there is a ∂-nonseparating curve C′ on F′ which is not parallel into F.

Proof. Let D be a compressing disk of F′. If ∂D is non-separating on F′, let C′ be a curve in F′ that intersects ∂D in one point. Then C′ is nonseparating, hence ∂- nonseparating on F′. We want to show that C′ is not parallel into F. Otherwise, let A be an annulus with ∂A = C′ ∪ C, where C ⊂ F. Then A ∩ D is a proper 1-manifold on D. But ∂(A ∩ D) = (∂A) ∩ ∂D is a single point, which is absurd. Hence C′ is the curve required.

Now assume that ∂D is separating on F′, cutting F′ into F′_{1} and F′_{2}. Choose a simple loop C_{i} on F′_{i} as follows. If F′_{i} is nonplanar, then there are a pair of nonseparating curves intersecting each other in one point, at least one of which is not null-homologous in M. Choose this one as C_{i}. If F′_{i} is planar, choose C_{i} to be isotopic to a boundary curve of F′. Note that since F is incompressible and diskless, C_{i} is not null-homotopic in M. Also notice that in both cases there is a properly embedded arc γ on one of the F′_{i} which intersects C_{1} ∪ C_{2} in one point.

Now choose a band B = I × I on F′ such that B ∩ ∂D = I × \frac{1}{2}, B ∩ C_{1} = I × 0, B ∩ C_{2} = I × 1, and B is disjoint from the arc γ above. Such band exists because γ is a nonseparating arc on F′_{i}. Let C′ be the band sum of C_{1} and C_{2}, that is, C′ = (C_{1} ∪ C_{2} − I × {0, 1}) ∪ ({0, 1} × I). Then C′ intersects γ in one point. Since C′ intersects ∂D essentially in two points, it is not parallel to any boundary component on F′. Therefore C′ is ∂-nonseparating.

We want to show that C′ is not parallel into F. Using the property that C_{i} are not null-homotopic in M, one can show by an innermost circle argument that C′ is not null-homotopic in M. Now suppose that there is an annulus A in M with ∂A = C′∪C, where C ⊂ F. Since C′ is not null-homotopic in M, A is incompressible in M. By surgery along an innermost circle of D ∩ A one can eliminate all circle intersections of A ∩ D. Since ∂(A ∩ D) consists of two points, A ∩ D is a single arc, which has endpoints on the same component of ∂A, hence it cuts off a disk D′ from A. Assume without loss of generality that D′ ∩ F′ is on F′_{1}. Let D′′ be the disk on D bounded by (A ∩ D) ∪ (B ∩ D), and let B_{1} = B ∩ F′_{1}. Then D′ ∪ D′′ ∪ B_{1} is a disk with boundary C_{1}, which contradicts the fact that C_{1} is not null-homotopic in M. Therefore, C′ is not parallel into F. □

Lemma 6. Suppose F′ is incompressible, and is not a thrice punctured sphere. Then either (i) there is a ∂-nonseparating curve C′ on F′ which is not parallel into F, or (ii) F′ is parallel into F.

Proof. Since F′ is not a thrice punctured sphere, one can easily find a ∂-nonseparating curve α_{0} on F′. Assume that (i) is not true, so all ∂-nonseparating curves are parallel into F. We want to show that F′ is parallel into F.

Since α_{0} is parallel into F, the annulus N(α_{0}) is also parallel into F. It is an incompressible annulus because α_{0} is essential on F′ and F′ is incompressible. Among all connected incompressible surfaces in IntF′ which contain α_{0} and are parallel into F, choose G′ such that the complexity (χ(G′), |∂G′|) is minimal in the lexical-graphic order, where χ(G′) is the Euler characteristic of G′, and |∂G′| is the number of boundary components of G′. All incompressible sub-surfaces of F′ have Euler characteristics bounded below by χ(F′), hence such G′ does exist.

If all boundary components of G′ are parallel to some boundary components on F′, then either G′ is contained in a collar of ∂F′, or F′ − IntG′ = ∂F′ × I. The first case does not happen because G′ contains the ∂-nonseparating curve α_{0}, which by definition is not parallel to any boundary curve on F′. In the second case F′ is isotopic to G′, so it is parallel into F, and we are done. Hence we may assume that some boundary curve β of G′ is not parallel to any boundary curve on F′.

We want to find a ∂-nonseparating curve α′ which intersects β essentially in one or two points. If β is nonseparating on F′, choose α′ to be any curve on F′ that intersects β in a single point. Then α′ is nonseparating, hence ∂-nonseparating on F′.

If β separates F′ into F′_{1} and F′_{2}, choose an essential arc α′_{i} on F′_{i} with ∂α′_{1} = ∂α′_{2} ⊂ β. Moreover, if F′_{i} is nonplanar, choose α′_{i} to be nonseparating on F′_{i}. Then α′ = α′_{1} ∪ α′_{2} is ∂-nonseparating, and intersects β essentially in two points, as required.

Isotope α′ so that it intersects ∂G′ minimally. The geometric intersection number between α′ and β is 1 or 2, so α′ ∩ ∂G′ ≠ ∅. Since α′ is ∂-nonseparating, by our assumption above it is parallel into F, so there is an annulus A with ∂A = α′ ∪ α, where α ⊂ F. Isotope A rel α′ so that it intersects (∂G′) × I minimally. Since G′ is incompressible, (∂G′)×I consists of incompressible annuli in M, hence A∩((∂G′)×I) has no trivial circles. Since F and F′ are also incompressible, one can show that A ∩ ((∂G′) × I) has no trivial arcs on A either. Therefore A ∩ ((∂G′) × I) consists of vertical arcs (α′ ∩∂G′)×I. These arcs cut A into several squares α′_{i} ×I, where each α′_{i} is the closure of a component of α′ −∂G′. Choose i so that α′_{i} lies outside of G′. Let H be the component of F′ − IntG′ that contains α′_{i}. Then G′′ = G′ ∪ N(α′_{i}) is a surface parallel into F, and χ(G′′) = χ(G′) − 1. The arc α′_{i} is essential on H, so the only case that some boundary component γ of G′′ bounds a disk on F′ is when H is an annulus, and γ is the boundary of the disk obtained by cutting H along α′_{i}. Since F and F′ are incompressible and M is irreducible, both ends of the annulus γ × I ⊂ G′′ × I ⊂ M bound disks on F ∪F′, which together with γ ×I bounds a 3-ball in M. It follows that G′∪H is parallel into F. Since G′∪H has the same Euler characteristic as G′ but fewer number of boundary components, this contradicts the choice of G′. Therefore ∂G′′ consists of essential curves on F′. Since F′ is incompressible, G′′ is also incompressible. Since χ(G′′) < χ(G′), this again contradicts the choice of G′. □

Given a simple closed curve α and a proper arc γ on F′, denote by τ_{α}^{n}γ the curve obtained from γ by Dehn twist n times along α, and by N(τ_{α}^{n}γ) a regular neighborhood of τ_{α}^{n}γ on ∂M. Suppose T is a fixed torus boundary component of a 3-manifold M. Denote by M(r) the manifold obtained by Dehn filling on T along a slope r on T, that is M(r) is obtained by gluing a solid torus V to M along T so that the curve r on T bounds a meridian disk on V. Denote by Δ(r_{1}, r_{2}) the minimal geometric intersection number between two slopes r_{1}, r_{2}. The following two theorems will be used in the proof of Theorem 9, which proves Theorem 1 in the case that F′ contains a ∂-nonseparating curve which is not parallel into F.

Lemma 7. ([Wu2], Theorem 1) Let T be a torus component on the boundary of a 3-manifold M, and let S be an incompressible surface on ∂M − T. Suppose there is no incompressible annulus in M with one boundary component on each of S and T. If S is compressible in M(r_{1}) and M(r_{2}), then Δ(r_{1}, r_{2}) ≤ 1. In particular, S is incompressible in all but at most three M(r). □

Lemma 8. ([CGLS], Theorem 2.4.3) Let T, S, M be as in Lemma 7, and assume further that M is irreducible. Suppose that there is an incompressible annulus A in M with one boundary component on S and the other a curve r_{0} on T. Then either S is a torus and M = S × I, or S remains incompressible in all M(r) with Δ(r, r_{0}) > 1. □

Theorem 9. Let α be a ∂-nonseparating curve on F′ which is not parallel into F, and let γ be a proper arc on F′ intersecting α in one point. Then F_{n} = F ∪ N(τ_{α}^{n}γ) is incompressible for all but at most three consecutive n’s.

Proof. Let K be the knot obtained by pushing α slightly into M. There is an embedded annulus A_{0} in M with ∂A_{0} = α∪K. Consider the manifold M_{K} = M −IntN(K), where N(K) is a regular neighborhood of K in M. Let T be the torus ∂N(K), and let (m, l) be the meridian-longitude pair on T such that l = A_{0} ∩ T. Denote by M_{K}(p/q) the manifold obtained by Dehn filling on T along the slope pm + ql. The Dehn twist τ_{α}^{−n} on F′ extends to a Dehn twist of M_{K} along the annulus A = A_{0} ∩ M_{K}, which sends the meridian slope m of T to the slope m − nl, so it extends to a homeomorphism φ_{n} : M = M_{K}(1/0) ≅ M_{K}(−1/n), which maps the curve τ_{α}^{n}γ to the curve γ, and hence the surface F_{n} to the surface F_{0} = F ∪ N(γ). It follows that φ_{n} is a homeomorphism of pairs

φ_{n} : (M, F_{n}) → (M_{K}(−1/n), F_{0}).

Therefore to prove the theorem we need only show that for all but at most three consecutive integers n, the surface F_{0} is incompressible in M_{K}(−1/n).

CLAIM 1. T = ∂N(K) is incompressible in M_{K}, and M_{K} is irreducible.

If D is a compressing disk of T in M_{K}, then ∂D must intersect the meridian m of K in one point, because otherwise after the trivial Dehn filling, M = M_{K}(1/0) would contain a lens space or S^{2} × S^{1} summand, contradicting the irreducibility of M. It follows that K, and hence α, bounds a disk in M. In this case α is parallel to a trivial curve on F, which contradicts the assumption that α is not parallel into F. Similarly, if M_{K} is reducible, then since M is irreducible, K is contained in a ball in M, so α would be null-homotopic. Using Dehn’s Lemma, we see that α bounds a disk in M, hence is parallel to a trivial circle in F, contradicting the assumption that α is not parallel into F.

CLAIM 2. F_{0} is incompressible in M_{K}.

Recall that A denotes the annulus A_{0} ∩ M_{K}. Since α intersects γ in a single point, A ∩ F_{0} is a single arc C on the boundary curve α of A. Let D be a compressing disk of F_{0}, chosen so that |D ∩ A|, the number of components in D ∩ A, is minimal. After disk swapping along disks on A bounded by innermost circles, we can assume that no component of D ∩ A is a trivial circle on A. Since T is incompressible by Claim 1, the annulus A is also incompressible, so D ∩ A contains no essential circle component on A either. Hence D ∩ A consists of arcs only. If some arc e of D ∩ A is parallel to a sub-arc on C = A ∩ F_{0}, then after boundary compressing D along a disk Δ cut off by an outermost such arc we will get two disks D_{1}, D_{2} with boundary on F_{0}, at least one of which has boundary an essential curve on F_{0}, hence is a compressing disk of F_{0}. Since |D_{i} ∩ A| < |D ∩ A|, this contradicts the minimality of |D ∩ A|. Therefore, all arcs of D ∩ A are essential relative to C, in the sense that it is not parallel to an arc on C. See Figure 1(a). Notice that |D ∩ A| ≠ 0, otherwise D would be a compressing disk of F, contradicting the incompressibility of F.

Consider an outermost disk Δ on D, as shown in Figure 1(b). Then ∂Δ consists of two arcs e_{1}, e_{2}, where e_{1} is an arc on A which is essential relative to C, and e_{2} is an arc on F_{0} with interior disjoint from C. Thus e_{2} ∩N(γ) consists of two arcs e′_{2}, e′′_{2}. Let t_{1} be the subarc of C connecting the two ends of e′_{2} ∪e′′_{2} on C, and let t_{2} be the subarc on ∂N(γ) connecting the other two ends of e′_{2} ∪e′′_{2}. Then e′_{2} ∪t_{1} ∪e′′_{2} ∪t_{2} bounds a disk Δ′ on N(γ). Now A′ = Δ ∪ Δ′ is an annulus in M, with one boundary component e_{1} ∪ t_{1} an essential circle on A, which is parallel to α, and the other component e_{2} ∪ t_{2} a curve on F. This contradicts the assumption that α is not parallel into F.

Figure 1

CLAIM 3. There is no incompressible annulus P in M_{K} with one boundary component C_{1} on F_{0} and the other component C_{2} a curve on T which is disjoint from l = A ∩ T.

The proof is similar to that of Claim 2. Choose P so that |P ∩A| is minimal. Using the fact that P is incompressible, one can show as above that P ∩ A has no trivial circle component. Note that since C_{2} is disjoint from l, P ∩ A has no arc component with endpoints on l = A∩T. If P ∩A had some essential circle component,choose such a component t which is closest to l on A. By cutting and pasting along the annulus on A bounded by t ∪ l, one would get another incompressible annulus P′ which has fewer intersection components with A. As in the proof of Claim 2 one can eliminate all arc components of P ∩ A which are inessential relative to C = A ∩ F_{0}. Hence P ∩ A consists of arcs with ends on C and are essential relative to C, as shown in Figure 1(a). Also, since P is disjoint from l, P ∩ A are inessential arcs on P. Now one can use a disk Δ cut off by an outermost arc on P, proceed as in the proof of Claim 2 to get an annulus with one boundary on α and the other on F, and get a contradiction. Finally, if P ∩ A = ∅ then P extends to an annulus with one boundary on α and the other on F, contradicting the assumption that α is not parallel into F. This completes the proof of Claim 3.

We now continue with the proof of Theorem 9. We have shown that F_{0} is incompressible in M_{K}. If there is no essential annulus in M_{K} with one boundary component on each of F_{0} and T, then by Lemma 7 we know that F_{0} is incompressible in M_{K}(r) for all but at most three slopes r with mutual intersection number 1. In particular, it is incompressible in M_{K}(−1/n) for all but at most two consecutive n’s, so the theorem follows. Now suppose there is an essential annulus P in M_{K} with one boundary component on F_{0} and the other a curve r_{0} on T. Since F_{0} is not a closed surface, it is not a torus. Hence by Lemma 8, F_{0} remains incompressible in M_{K}(−1/n) unless Δ(−1/n, r_{0}) ≤ 1. By Claim 3, r_{0} is not the longitude slope 0/1, therefore, Δ(−1/n, r_{0}) ≤ 1 holds for at most three consecutive integers n. This completes the proof of Theorem 9. □

Proof of Theorem 1. By Theorem 4, Lemmas 5 and 6, and Theorem 9, we can now assume that F′ is incompressible and is parallel into F. We want to show that either F is arc-extendible in F′, or M is a product F × I. As in the proof of Theorem 4, we may assume without loss of generality that M is a compression body, so all closed incompressible surfaces of M are boundary parallel. Let α_{1}, …, α_{k} be the boundary curves of F′. Let F′ × I be a product in M such that F′ = F′ × 0 and F′ × 1 ⊂ F. Write α_{i}^{1} = α_{i} × 1, which is a curve on F isotopic to α_{i} in M.

We have assumed above that F′ is incompressible in M, so IntF ∪ IntF′ is incompressible in M. Write F̂_{i} = IntF ∪ IntF′ ∪ α_{i}. If F̂_{i} is incompressible for some i, then F ∪ N(γ) is incompressible for any essential arc γ in F′ with endpoints on α_{i}, and we are done. (Such an arc exists because F′ is not an annulus or disk.) So assume that F̂_{i} is compressible for all i. By the Handle Addition Lemma (Lemma 2), after adding a 2-handle to M along α_{i}, the surface F̂_{i}[α_{i}] is incompressible, and M[α_{i}] is irreducible. Notice that since F′ is incompressible, the curve α_{i}^{1} = α_{i} × 1 in F is essential on F. But after adding the 2-handle, α_{i}^{1} bounds a disk in M[α_{i}], so it must also bound a disk on F̂_{i}[α_{i}] because F̂_{i}[α_{i}] is incompressible. By definition F̂_{i}[α_{i}] is obtained from (IntF ∪IntF′)−IntN(α_{i}) by capping off the two copies of α_{i} with disks, hence α_{i}^{1} ∪α_{i} bounds an annulus A_{i} on F_{i}. Denote by A′_{i} the annulus α_{i} × I ⊂ F′ × I ⊂ M. Then T_{i} = A_{i} ∪ A′_{i} is a torus in M. Since we have assumed above that M is a compression body, either T_{i} bounds a solid torus V_{i}, or it is parallel to some torus component of ∂M. However, since M[α_{i}] is irreducible, one can show as in the proof of Theorem 4 that V_{i} is a solid torus, and α_{i} is a longitude of V_{i}. This is true for all i. It is now easy to see that M is a product F × I. □

The following theorem supplements Theorem 1. It says that in most case there are extension arcs with endpoints on any prescribed boundary compponents of F′.

Theorem 10. Let F, F′, M be as in Theorem 1. Suppose M is not a product F × I, and suppose F′ is not parallel into F and is not a thrice punctured sphere. Then it contains an extension arc γ of F with endpoints on any prescribed components of ∂F′.

Proof. If F′ is nonplanar, then by the proof of Lemmas 5 and 6, there is a ∂- nonseparating circle α (denoted by C′ there) on F′ which is not parallel into F, and is actually nonseparating on F′. Hence given any boundary components ∂_{1}, ∂_{2} of F′, (possibly ∂_{1} = ∂_{2}), there is an arc γ with endpoints on ∂_{1} and ∂_{2}, intersecting α in one point. By Theorem 9, for all but at most three integers n, the arc γ_{n} = τ_{α}^{n}γ is an extension arc of F.

Now suppose F′ is planar with |∂F′| ≥ 4. First assume that ∂_{1}, ∂_{2} are distinct boundary components of F′. By the proof of Lemmas 5 and 6, the curve α is a band sum of two boundary components of F′. From the proofs one can see that we can always choose α to be the band sum of ∂_{1} and ∂_{3}, with ∂_{3} ≠ ∂_{1}, ∂_{2}. Hence there is an arc γ from ∂_{1} to ∂_{2} intersecting α in one point. We can then apply Theorem 9 to get an extension arc γ_{n} with one endpoint on each of ∂_{1} and ∂_{2}.

We now proceed to find an extension arc in F′ with boundary on the same component ∂_{1} of ∂F′. By the proof of Lemmas 5 and 6, we can choose the curve α above to be the band sum of of ∂_{2} and ∂_{3}, with ∂_{1} ≠ ∂_{2}, ∂_{3}. Recall that α is not parallel into F. Choose an arc γ as follows. Let ∂′_{2} be a curve on F′ parallel to ∂_{2}, let γ′ be an arc connecting ∂′_{2} to ∂_{1} intersecting α in one point, and let Q be the sub-surface N(γ′ ∪ ∂′_{2}) of F′. Then γ is the closure of the arc component of ∂Q ∩ IntF′, that is, γ is the arc component of the frontier of Q in F′, see Figure 2 below. Consider the surface F_{0} = F ∪ N(γ), and observe that F_{0} is isotopic to the surface F ∪ Q. After Dehn twist along α, it is isotopic to the surface F ∪ N(τ_{α}^{n}γ); hence to show that all but at most three τ_{α}^{n}γ are extension arcs of F in F′, we need only show that F ∪ Q is incompressible after all but at most three Dehn twist along α. Since F ∪Q intersects α in a single arc, the argument in the proof of Theorem 9 is still valid, with the following easy modifications. We use the notations in that proof.

Figure 2

The proof of Claim 2 needs the following modifications. (i) The arc e_{2} on the boundary of the outermost disk Δ may be on Q. In this case, notice that the other arc e_{1} on ∂D is isotopic to an arc α_{1} on α, and e_{2} ∪ α_{1} is isotopic in F′ to the curve ∂_{3}, so the fact that e_{1} ∪ e_{2} bounds a disk Δ would imply that ∂_{3} bounds a disk. Since ∂_{3} is also on ∂F, this contradicts the fact that F is incompressible and diskless. (ii) The compressing disk D of F ∪ Q could be disjoint from the annulus A. But since F is incompressible, this would imply that ∂D lies on Q, hence is isotopic to ∂_{2}, which would imply that ∂_{2} bounds a disk, again contradicting the assumption that F is incompressible and diskless.

The proof of Claim 3 applies to show that the annulus P there can be modified to be disjoint from the annulus A. Then notice that the component of ∂P on F ∪ Q is either in F, or in Q and hence parallel to ∂_{2}. Since ∂_{2} ⊂ F, in either case P can be extended to an annulus with one boundary component on α and the other on F, which contradicts the assumption that α is not parallel into F.

The rest part of the proof of Theorem 9 follows verbatim to show that F ∪ Q is incompressible after all but at most three Dehn twist along α. □

Remark. Theorem 10 is not true if F′ is a thrice punctured sphere.

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Michael Freedman Department of Mathematics, UC San Diego, La Jolla, CA 92093

   E-mail address: freedman@math.ucsd.edu

Hugh Howards Department of Mathematics, UC San Diego, La Jolla, CA 92093

   E-mail address: howards@math.ucsd.edu

Ying-Qing Wu Department of Mathematics, University of Iowa, Iowa City, IA 52242; and

MSRI, 1000 Centennial Drive, Berkeley, CA 94720-5070

   E-mail address: wu@math.uiowa.edu