Essential laminations and deformations of homotopy

equivalences, II: the structure of pullbacks

Mark Brittenham

University of Texas at Austin

§0

Introduction

It is a long-standing conjecture in 3-manifold theory that every homotopy equivalence f:M→N between closed, irreducible 3-manifolds M and N, with |π_{1}(M)| (=|π_{1}(N)|) = ∞, is homotopic to a homeomorphism. In [Wa], Waldhausen proved that this conjecture was true, if we assume that N contains a 2-sided incompressible surface S. The proof consists of first homotoping the map f so that the pullback surface f^{−1}(S) is an incompressible surface, and then splitting both manifolds open along these surfaces, and proceeding by induction on a hierarchy of N - a collection of incompressible surfaces in the successively split-open manifolds which end up splitting N into a 3-ball. The base case is the Alexander trick.

In [Br1], we began a program to extend this theorem to the case that N is laminar, i.e., N contains an essential lamination [G-O]. We showed that, given a homotopy equivalence f:M→N between non-Haken, (irreducible) 3-manifolds and a transversely-orientable essential lamination L⊆N, if the pullback lamination f^{−1}(L)⊆M is essential then we could homotope the map f to a homeomorphism. In this paper we study the ‘other half’ of Waldhausen’s proof - when can one deform a homotopy equivalence to give an essential pullback - by trying to understand what structure the pullback of an essential lamination has, in general, under a homotopy equivalence. We work, as in [Br1], under the hypothesis that M and N are non-Haken; otherwise, all that we shall use about the map f is that it has degree 1 (in particular, non-zero degree).

It has been conjectured that the method employed by Waldhausen - surgeries of the pullback achieved as homotopies of the map f - will be successful (with some modification) in creating an essential pullback in the present context. We begin this paper by showing that if the homotopy-through-surgery process will succeed in making the pullback essential, then, at the very beginning, the pullback must have had a fairly restrictive structure - each leaf of the pullback must map to its corresponding leaf in L with degree 0 or 1. In the remainder of the paper we attempt to show that leaves of a pullback do in fact have this property. We must admit at the outset that we do not succeed in proving this - but we will see that the

Typeset by AMS-TEX

Key words and phrases. essential lamination, homotopy equivalence.

Research supported in part by NSF grant # DMS−9203435

pullback has many of the same ‘tightly-wrappedness’ properties that an essential lamination in a non-Haken manifold has [Br2], which ‘almost’ implies that all leaves have degree 0 or 1.

§1

The homotopy through surgery process

Note: we assume, as in [B1], that the essential lamination L has every leaf dense in L, i.e., for every leaf L of L, \overline{L} = L. This necessary hypothesis can be easily arranged, by throwing away all but a minimal sublamination of any essential lamination L that we start with. We also assume that L contains no open sets in M, i.e., that transversely L looks like a nowhere-dense closed subset of an interval (typically, a Cantor set). This can be arranged, if necessary, by splitting L open along a finite number of leaves.

Given a homotopy equivalence f:M→N, where M and N are closed irreducible 3-manifolds, and an essential lamination L⊆N, we say that f is transverse to L if, for every point x∈f^{−1}(L), T_{f(x)}(L) + f_{∗}(T_{x}(M) = T_{f(x)}(N). This is most easily achieved by making f transverse to a branched surface B which carries L; then by imagining that L lies very close to B, f is immediately transverse to L. This method of achieving transversality is the one that we tend to think of as ‘standard’. However, we will occasionally make use of the more general point of view, especially when doing homotopies through surgery. Since for the most part we will be doing surgeries in order to eventually lead to a contradiction, these ‘excursions’ will from one point of view be unnecessary (hence our adoption of the more restrictive notion as ‘standard’).

Given a map f:M→N transverse to the lamination L⊆N, the pullback L′ = f^{−1}(L) is a lamination. If the map f is transverse to the branched surface B carrying L, then the pullback lamination L′ is carried by the pullback branched surface B′ = f^{−1}(B).

The main construction in our proofs will be that of \underline{surgering} the pullback lamination L′. This begins with a situation as in Figure 1a - a loop in leaf of L′, and a disk D, which is embedded in M, but not necessarily embedded in M|L′, and which compresses it. We assume that D is transverse to L′, so D∩L′ is a 1-dimensional lamination in M. Note that since L is essential, D∩L′ consists of (a finite number of parallel families of) closed loops - no loop of D∩L′ can have any holonomy around them. This is because they are null-homotopic in M, so their images under f are null-homotopic in N, so (by the essentiality of L) their images under f are null-homotopic in the leaves containing them, hence have no holonomy around them. So the original loops have no holonomy - holonomy is preserved under transverse maps. There can also be no non-compact leaves which \underline{don’t} limit on compact ones, since this would imply (by pushing forward under f) the existence of an end-compression for L.

We can then create a new lamination in M, by surgering L′ along this disk, by cutting L′ open along D, and sewing in a collection of disks to the resulting circles, working from innermost parallel family out (see Figure 1b). As with ‘ordinary’ surgery of a compact surface (see [He]), this operation can be achieved by a homotopy of f; the new lamination is the pullback of a map homotopic to (and which we will still call) f.

Figure 1

In many cases (e.g., with a Reeb component (Figure 2)), this surgery will create sphere leaves, or (necessarily inessential) leaves contained inside of embedded 3balls; but since we assume that both M and N are irreducible (so are both K(π, 1)’s), these spheres can be removed via a homotopy of f, as well, in the usual way(see [Wa].

Figure 2

§2 The degree of a leaf should be 0 or 1

In this section we prove:

Proposition 1. If a finite number of surgeries like those described above can make the pullback lamination L′ essential (modulo leaves lying in 3-balls), then every leaf of the pullback lamination maps to its corresponding leaf of L with degree either 0 or 1.

This follows fairly directly from the following results:

Lemma 2. Given f:M→N, L, and L′ = f^{−1}(L) as above, let L′_{0} be the result of surgering L′, achieved as a homotopy of the map f to the map g. Then for any leaf L′ of L′, mapping under f to the leaf L of L, if L′_{i}, i=1,2,… are the leaves of L′_{0} that L′ was surgered into, then

d(f|_{L′}) = Σ_{i=1}^{∞}d(g|_{L′_{i}}),

where d() refers to the degree of the map, thought of as a map into L.

Lemma 3. With the same initial conditions, if a leaf L′ of L′ is entirely contained in the 3-ball, then d(f|_{L′}) = 0.

Proof of Proposition 1: From [B1, Proposition 7], we know that when L′ is essential, every leaf of L′ maps to its corresponding leaf of L with degree 1. By our hypothesis, after a finite number of surgeries, for each leaf L of L, f^{−1}(L) consists of a single leaf mapping with degree 1, and a collection of leaves in 3-balls, which by Lemma 3 map with degree 0; in particular, the degrees of leaves are all non-negative. We can therefore inductively conclude, by undoing each surgery and appealing to Lemma 1, that the degree of each leaf at each stage is the sum of the degrees of leaves, all of which are \underline{non}-\underline{negative}. Consequently, all leaves of f^{−1}(L) map with non-negative degree. But for each leaf L of L, the degrees of each leaf of f^{−1}(L) must add up to the degree of f:M→N, which is 1, so we can conclude that one leaf of f^{−1}(L) maps with degree 1, and all others map with degree 0.

Proof of Lemma 2:

The general case is not much different, requiring just a bit more attention to detail. By cutting up the family of loops around the innermost one into a finite number of ‘thin’ pieces (using the fact that transversely they are nowhere dense), we can assume by deforming the map slightly that the family of loops map in ‘vertically’, i.e., the image of the annular region between innermost and outermost loop maps to the normal fence over the image of the innermost loop. If we then choose a null-homotopy of f of each loop to achieve our surgery, this null-homotopy will lift to the nearby leaves. By cutting our family of loops into finitely-many even thinner families, we can then assume that every null homotopy for our loops is the lift of the null homotopy for the innermost one. Now we replace, as in [He], a thin annular region in each leaf with two copies of these null-homotopies; this can be achieved by a homotopy of f. In this more genereal case, this could separate a leaf into infinitely-many leaves, but because the sewn-in disks all cancell out in any degree calculation, the total degree of the surgered leaf remains the same. A finite number of applications of this completes the proof.

Proof of Lemma 3: Since \overline{L′} is contained in the ball B^{3}, the inclusion-induced map π_{1}(L′)→π_{1}(M) is the zero map, and consequently (if f(L′)⊆L⊆L), the map f_{∗}π_{1}(L′)→π_{1}(L) must be the zero map (since composing with the inclusion-induced \underline{injective} map π_{1}(L)→π_{1}(M) gives the zero map).

If L is not simply-connected, then |π_{1}(L)|=∞. By the lifting criterion, there is a map f̃:L′→L̃ (where L̃ is the universal cover of L), so that

\begin{aligned}L′ & →^{f̃} & L̃ \\ & ↘^{f} & ↓p \\ & & L \end{aligned}

commutes. But now choose a regular value x for f:f^{−1}(L)→L (so x is also a regular value for f:L′→L). then f^{−1}(x) = f̃^{−1}(p^{−1}(x) = f̃^{−1}(∪x_{i}) = ∪f̃^{−1}(x_{i}), where {i}is an infinite index set. Since f is transverse to L, however, x is also a regular value for f:M→N, so f^{−1}(x) consists of finitely-many points. Consequently, most of the f̃^{−1}(x_{i}) must be empty, so d(f̃)=0. Therefore the signed sum over each f̃^{−1}(x_{i}) must also be 0, so the signed sum over f^{−1}(x) (for f:L′→L) is a sum of zeros, so d(f:L′→L) = 0.

If L \underline{is} simply-connected, we will argue by contradiction. If L′ maps with non-zero degree, then f(L′) = L, so f(\overline{L′}) is a (compact, hence) closed set in N containing L, hence contains \overline{L}=L. But L contains a non-simply connected leaf L_{0} (else N is the 3-torus [Ga1], hence Haken), so \overline{L′} contains leaves (in the 3-ball), namely f^{−1}(L_{0})∩\overline{L′}, which map to a non-simply-connected leaf, hence all map with degree 0.

§3

The structure of pullbacks, I: degree-zero vs. non-zero degree leaves

In this section we begin to establish some of the properties that a pullback lamination must always have. For this section, we assume only that the lamination L has every leaf dense (and, tacitly, that the map f has non-zero degree, so that \underline{some} leaf of f^{−1}(L)=L′ must map with non-zero degree). In many respects, the arguments mimic those of [B1, Section 2].

Proposition 4. If L_{0}⊆f^{−1}(L) is a sublamination, and z is a regular value of f in L, with I a (short open) arc though z transverse to L, then for every x∈I∩L, and every ϵ>0, there exists y∈I∩L with d(x,y)<ϵ and an open U⊆I containing y such that (for f^{−1}(I)=I_{1}∪I_{2}∪… ∪I_{n}) for each j=1,…,n, setting f^{−1}(U)∩I_{j}= U_{j}, either U_{j}∩f^{−1}(L)⊆L_{0} or U_{j}∩L_{0}=∅ .

Proof: f^{−1}(I)=I_{1}∪⋯ ∪I_{n}, each mapping homeomorphically to I under f. Now given x∈I, and ϵ>0, choose a y∈I with d(x, y) <ϵ such that f^{−1}(y)∩L_{0} consists of the fewest number of points; call them {y_{1},…,y_{k}}. Note that, for i ≥ k+1, since y_{i}∉L_{0}, and L_{0} is closed, all of the points in I_{i} sufficiently close to y_{i} are not in L_{0}; for some open U_{i} in I_{i} containing y_{i}, U_{i}∩L_{0}=∅ (sse Figure 3).

Figure 3

Claim: for i ≤ i ≤ k, all points of f^{−1}(L) sufficiently close to the y_{i} are in L_{0}.

For if not, then (passing to a subsequence and renumbering as necessary) there is a sequence of points z_{1},z_{2},… in I_{i}, converging to y_{i}, such that z_{j}∉L_{0} for every j. Renumber so that i=k. Then if we set x_{j}=f(z_{j})∈I, then the x_{j} converge to y, and so for large enough j, the x_{j}’s are within ϵ of x. But for large enough j, f^{−1}(x_{j})∩I_{i} is a point in U_{i}, for k+1≤ i ≤ n, so is not in L_{0}, while f^{−1}(x_{j})∩I_{k}=z_{k}, so is \underline{also} not in L_{0}. So f^{−1}(x_{j}) contains fewer points of L_{0} than f^{−1}(y) does, a contradiction.

Therefore, for 1≤ i ≤ k, there is an open interval U_{i} of I_{i} about y_{i} such that U_{i}∩f^{−1}(L)⊆L_{0}. Now set U=f(U_{1})∩⋯ ∩U_{n}.

Corollary 5. If L′ is a leaf of f^{−1}(L), and L_{0} is a \underline{proper} sublamination of \overline{L′}, then f(L_{0}) is \underline{nowhere} \underline{dense} in L (i.e., L\f(L_{0}) is open and dense in L). In particular, every leaf of L_{0} maps to L with degree zero.

Proof: By the proof of the proposition, for every regular value x of f in L , and ϵ>0, there is a regular value y of f in L, within ϵ of x along a transverse arc I, and an open interval U in I about y such that U_{j}∩f^{−1}(L)⊆L_{0} or U_{j}∩L_{0}=∅ But our hypothesis implies that for each j it must be that U_{j}∩L_{0}=∅; for otherwise given any point z∈U_{j}∩f^{−1}(L)⊆L_{0} , L′, which is \underline{not} in L_{0}, passes arbitrarily close to z, and hence intersects the open arc component U_{j} that contains z, a contradiction. Therefore U∩f(L_{0})=∅; in particular, y ∉ f(L_{0}). So A = L\f(L_{0}) is an open subset (since f(L_{0}) is the continuous image of a compact set) of L, and any (regular) point of L has a point of A arbitrarily close (in the transverse direction) to it (since x was arbitrary). So A is dense in L (since regular values are dense in L). In particular, A contains a regular value x for f in L. For any leaf L′_{0} of L_{0}, mapping under f to the leaf L_{0} of L, since L_{0} passes arbitrarily close to x (all leaves of L are dense in L), L_{0} contains a nearby regular value y in A (since A is open). Then f^{−1}(y)∩L_{0} = ∅, so f^{−1}(y)∩L′_{0} = ∅, so L′_{0} maps to L_{0} with degree zero.

Corollary 6. If L′_{0}, L′ are leaves of L′ mapping with non-zero degree, and L′_{0}⊆\overline{L′}, then \overline{L′_{0}}=\overline{L′}.

Let H denote the leaves of L having no holonomy; by [EMT], this is a dense subset of L. Let H′= f^{−1}(H); this is a (saturated) subset of the set of leaves of f^{−1}(L) = L′ having no holonomy, since a loop in H′ with non-trivial holonomy would be carried under f to one in H with non-trivial holonomy, since transverse pictures are preserved under f. H′ is also dense in L′, because transversely it looks the same as H in L (again, since transverse pictures are preserved under f).

Proposition 7. If L′⊆L′ maps to L⊆L under f, with non-zero degree, then (f(\overline{L′})=L, so) H_{L′}=\overline{L′}∩H′≠ ∅, every leaf of H_{L′} maps to its corresponding leaf either with degree 0 or with the \underline{same} (non-zero) degree d = d(\overline{L′}), and every leaf of \overline{L′} maps to its corresponding leaf with degree a \underline{non}-\underline{negative} multiple of d.

Proof: Pick a regular value x for f in a leaf of H, satisfying the conclusions of Proposition 4. That is, pick a regular value in a leaf of L satisfying the conditions, and then (since H is dense in L) pick a nearby point in H. Let f^{−1}(x)∩\overline{L′} = {x_{1},…,x_{k}}⊆H_{L′}. One of the leaves that these points lie in must map to its corresponding (H-) leaf with non-zero degree; otherwise, following the argument of [B1, Lemma 5], we can join these points together by arcs in their leaves, as the vertices of a tree, such that for each component of the tree the sum of the signs of its vertices is zero (i.e., the degree of the leaf containing it). Then by lifting these trees to nearby leaves (see Figure 4), since the resulting endpoints are all in the inverse image under f of the same point (because we are lifting from a leaf with no holonomy), we see that the degree of every nearby leaf (including L′) is a sum of 0’s, hence 0, a contradiction. So some H_{L′}-leaf L′_{0} has non-zero degree, so by Corollary 5, \overline{L′_{0}} = \overline{L′}. But now we can easily see, first, that every leaf of H_{L′} the \underline{same} degree; this is because each is dense in \overline{L′} (by Corollary 6), and so they pass arbitrarily close to one another. The argument of [B1, Lemma 5], applied verbatim, then gives the result. The final statement then follows by taking all of the points of {x_{1},…,x_{k}}⊆H_{L′} and stringing them together in their leaves (so the resulting trees each have vertex-sum either d or 0), and then lifting them to the leaves containing any other inverse image of a point close to x. The leaves have degree the sum of the degrees of the point inverses that lie in them, but since our trees lift to trees in these leaves (so any leaf containing one vertex contains them all), this sum is a sum of d’s and 0’s, hence is a non-negative multiple of d.

Figure 4

Proposition 8. If L′ maps with non-zero degree, then the set N_{L′} of leaves in \overline{L′} which map with non-zero degree is open in f^{−1}(L) = L′.

Proof: Let x be a regular value chosen as in Proposition 7, and let L′_{0} be a H_{L′}- leaf mapping with non-zero degree d. Let I be an arc, transverse to L, passing through x, and I_{1} the arc in the inverse image of I passing through a point x_{1}of f^{−1}(x) in L′_{0}. Assume that x=y has been chosen so that the number distinct leaves of H_{L′} meeting f^{−1}(y) is minimal among the points y lying in I∩H (otherwise, move to a different point!). Claim: All of the leaves of H_{L′} meeting f^{−1}(x) map with non-zero degree.

Proof of claim: As before, join together the points of f^{−1}(x)∩\overline{L′} contained in the same leaves, as the vertices of a tree; this tree lifts to nearby leaves, and the vertices of the lifted trees all lie in the inverse image of a single point lying near x.

If L′_{1} is a H_{L′}-leaf meeting f^{−1}(x) and mapping with degree 0, then, since \overline{L′_{0}}=\overline{L′}, L′_{0} limits on L′_{1}. But then when we lift the trees to L′_{0}(which corresponds to a point z near x), the vertices of the component contained in L′_{1}, since their degrees add up to 0, do not contain all of the points of f^{−1}(z)∩L′_{0}. So more than one component of the lifted tree lives in the same leaf, so the number of leaves of H_{L′} intersecting f^{−1}(z) is smaller, contradicting our choice of x.

This in turn allows us to conclude that \underline{every} leaf sufficiently close to L′_{0} maps with non-zero degree. All of the leaves of \overline{L′} meeting f^{−1}(x) map to the leaf containing x with the same degree d ≠0. Building our by now familiar trees joining the points of f^{−1}(x) together, and lifting these trees to any nearby point-inverse, we see that the degrees of these nearby leaves (in \overline{L′}) are positive multiples (each contains at least \underline{one} such tree) of d, and hence are all non-zero. Since any non-zero degree leaf L′_{1} in \overline{L′} limits on L′_{0}, L′_{1} is among these nearby leaves, so all of the leaves \underline{around} it have non-zero degree (and are contained in \overline{L′}). So the non-zero degree leaves of \overline{L′} form an open set in L′.

Corollary 9. If L′ maps to L with non-zero degree, then Z_{L′}, the set of leaves of \overline{L′} mapping to L with degree 0, is a closed set, forming a proper sublamination of \overline{L′} (so f(Z_{L′})⊆L is nowhere dense in L).

Proof: By Proposition 8, Z_{L′} = \overline{L′}\N_{L′} is a (relatively) closed set in \overline{L′}, hence is closed. So it is a sublamination of \overline{L′}. Because L′∉Z_{L′}, Z_{L′}is a \underline{proper} sublamination of \overline{L′}.

Corollary 10. A leaf L′_{0} of L′ mapping with degree zero cannot limit on a leaf L′ mapping with non-zero degree (i.e., \overline{L′_{0}} consists of degree-zero leaves).

Proof: Suppose it could; L′⊆\overline{L′_{0}}. We cannot have L′_{0}⊆\overline{L′}, since then L′_{0}⊆Z_{L′}, which is, by Corollary 9, a (\underline{closed}) sublamination consisting of degree-zero leaves, yet L′⊆\overline{L′_{0}}⊆Z_{L′}, a contradiction. Consequently, \overline{L′}is a \underline{proper} sublamination of \overline{L′_{0}}. But then Corollary 5 implies that every leaf of \overline{L′} (e.g, L′) maps with degree zero, a contradiction.

Corollary 11. The set Z of leaves of L′ which map under f with degree zero is a closed set, hence forms a sublamination of L′.

If L′ is a leaf mapping with non-zero degree (to the leaf L), then we know that f(\overline{L′})=L. But even more, since N_{L′} is open in L′, if we pick a regular value x for f in L, some inverse image of x is in L′, and since a small transverse arc I through x intersects every leaf of L (since every leaf is dense), and the component of the inverse image of I meeting L′ intersects only leaves of N_{L′}, this means that every leaf of L is the image of some (non-zero degree) leaf of N_{L′}; i.e., f(N_{L′})=L.

Corollary 12. For any two non-zero degree leaves L′_{1}, L′_{2}, either N_{L′_{1}}=N_{L′_{2}} or N_{L′_{1}}∩N_{L′_{2}}=∅ .

Proof: Suppose N_{L′_{1}}∩N_{L′_{2}}≠ ∅, so there is a leaf L′ contained in both (hence mapping with non-zero degree). Since L′⊆N_{L′_{i}}⊆\overline{L′_{i}} for i=1,2, N_{L′}⊆\overline{L′}⊆\overline{L′_{i}}, so N_{L′}⊆N_{L′_{i}} for i=1,2. If either of these containments is \underline{proper} (say, for i=1), then there is a (non-zero degree) leaf of N_{L′_{1}} not contained in N_{L′}, hence not contained in \overline{L′}. So \overline{L′} is a \underline{proper} sublamination of \overline{L′_{1}}, so, by Corollary 5, all of its leaves (including L′) maps with degree 0, a contradiction. Consequently, N_{L′_{1}}=N_{L′}= N_{L′_{2}}.

Corollary 13. The set N of leaves of L′ mapping with non-zero degree can be written as a disjoint union of finitely-many of the sets N_{L′_{1}},…,N_{L′_{k}}.

Proof: For each leaf L′ mapping with non-zero degree, f(N_{L′})=L. If we pick a regular value x for f in a leaf L of L, f^{−1}(x) consists of finitely-many points, and since every set N_{L′} has a leaf mapping (with non-zero degree) onto L, each set N_{L′} contains at least one of the points of f^{−1}(x). Consequently (since any two of the sets N_{L′} are either disjoint or identical) there are only finitely-many distinct sets of the form N_{L′}. Since any leaf L′⊆N is contained in one of these sets (the set N_{L′}, in fact!), the result follows.

§4

The structure of pullbacks, II: tightly-wrappedness

Now we add the hypothesis that our domain manifold, M, is irreducible and non-Haken. This allows us to mimic the arguments of [B2].

Proposition 14. If L′_{1} and L′_{2} are leaves of L′ mapping with non-zero degree, then \overline{L′_{1}}∩\overline{L′_{2}} ≠ ∅ .

Proof: If \overline{L′_{1}}∩\overline{L′_{2}} = ∅, then the two laminations have disjoint (fibered) neighborhoods, so they can be separated by a finite collection F of disjoint compact surfaces (e.g., the boundaries of one of the fibered neighborhoods); \overline{L′_{1}} and \overline{L′_{2}} are contained in distinct components of M\F. Because M is non-Haken, F cannot be incompressible, and, in fact, there must be a sequence of compressions of F (through surfaces F =F_{0}, F_{1}, …, F_{n}) so that F_{n} is a collection of disjoint 2-spheres; F_{n} = S_{1}∪⋯ ∪S_{k}. Because M is irreducible, each S_{i} bounds a 3-ball B_{i}.

Lemma 15. At most one component of M\F_{n} is \underline{not} contained in one of the 3-balls B_{i}.

Proof: This is a consequence of the fact that every surface in an irreducible, non-Haken 3-manifold M separates M. Every one of our 2-shperes S_{i} bounds a 3ball B_{i} on only one side (otherwise M=S^{3}, which can’t map, with non-zero degree, to a manifold having infinite fundamental group). This allows us to define a partial ordering on the S_{i}; S_{i}<S_{j} if B_{i}⊆B_{j}. If we take the maximal elements under this ordering, then they form a collection F′ of spheres bounding disjoint 3-balls. They also split M into one more component than the number of spheres, so exactly one component, M_{0}, \underline{isn’t} a 3-ball. M_{0} is a component of M\F_{n}, because any 2-sphere of F_{n}, in M_{0}, would not be maximal, so would be contained in one of the 3-ball components of M\F′, and hence would \underline{not} be in M_{0}. So every component of M\F_{n}, except M_{0}, is contained in one of the maximal 3-balls.

The two sublaminations \overline{L′_{1}} and \overline{L′_{2}} will meet our compressing disks; otherwise, by the lemma, one or the other is contained in a ball, so all of its leaves map with degree zero, a contradiction. But we can make L′ transverse to these disks; L′ then meets the disks in a collection of circles and arcs. More importantly, \overline{L′_{1}} and \overline{L′_{2}} meet the disks in a collection of circles only, because an arc of interscetion would require one of them to intersect our surface F. If we \underline{first} surger the \overline{L′_{i}} along these circles, we can then surger F without intersecting the (surgered) \overline{L′_{i}}. So after surgery (whose effect we will denote by s()), s(\overline{L′_{1}}) and s(\overline{L′_{2}}) are disjoint from the 2-spheres, so, again, one of them (s(\overline{L′_{1}}), say) is contained in a ball. So Lemma 2 implies that every leaf of s(\overline{L′_{1}}) maps with degree zero. But because L′_{1} maps with non-zero degree, Lemma 3 and induction on the number of surgeries involved in s implies that \underline{some} leaf of s(\overline{L′_{1}}) maps with non-zero degree, a contradiction. So \overline{L′_{1}}∩\overline{L′_{2}} ≠ ∅.

Proposition 16. If L′_{1}, …, L′_{n} are leaves of L′ mapping with non-zero degree, then \overline{L′_{1}}∩⋯ ∩\overline{L′_{n}} ≠ ∅.

Proof: This follows by induction on n. n=2 is Proposition 15. If n is the smallest number where we have leaves with \overline{L′_{1}}∩⋯ ∩\overline{L′_{n}} = ∅, then \overline{L′_{1}}∩⋯ ∩\overline{L′_{n−1}} = L′_{0} ≠ ∅, and L′_{0}∩\overline{L′_{n}} = ∅. But now if we apply the argument of Proposition 15, there is a sequence of surgeries s such that one of s(L′_{0}) and s(\overline{L′_{n}}) is contained in a ball. s(\overline{L′_{n}}) can’t be (it contains a leaf mapping with non-zero degree), so s(L′_{0}) is contained in a 3-ball. A further sequence of surgeries can therefore reduce it to a collection of 2-spheres; so we can assume that s(L′_{0}) is a collection of 2-spheres. By Reeb stability, these spheres have neighborhoods meeting only other sphere leaves, so no non-zero degree leaf can limit on any leaf of s(L′_{0}). But by Lemma 2, there is a non-zero degree leaf L′′_{i} in every s(L′_{i}), and \overline{L′′_{1}}∩⋯ ∩\overline{L′′_{n−1}}⊆s(\overline{L′_{1}})∩⋯ ∩s(\overline{L′_{n−1}}) = s(\overline{L′_{1}}∩⋯ ∩\overline{L′_{n−1}}) = s(L′_{0}), so \overline{L′′_{1}}∩⋯ ∩\overline{L′′_{n−1}} = ∅, since none of these leaves can actually limit on any leaf of s(L′_{0}). This is not actually a contradiction; we’ve reduced the number of leaves it takes to end up with the empty set, but we’ve done it for a \underline{different} pullback lamination, s(L′). However, we can then continue this same process, lumping together the first n-2 leaves. Eventually we will find a pullback lamination having two non-zero degree leaves with disjoint closures. But this would contradict Proposition 15.

Proposition 17. No sublamination of Z is essential.

Proof: Suppose L′_{0}⊆Z is an essential sublamination. Either L′_{0}∩\overline{N} =∅ or not. If L′_{0}∩\overline{N} =∅, surround L′_{0} by surfaces missing \overline{N} and surger them to spheres, bounding ball. s(L′_{0}) must end up contained in the balls, because s(\overline{N}) contains leaves mapping with non-zero degree. But if L′_{0} is essential, then s(L′_{0}) is isotopic (modulo sphere leaves) to L′_{0}, a contradiction; an essential lamination cannot be contained in a 3-ball.

But if L′_{0}∩\overline{N} ≠ ∅, then L′_{0}∩\overline{L′}=L′_{1}≠ ∅ for some leaf L′⊆N. But then (since L′_{1} is a \underline{proper} sublamination of \overline{L′}) f(L′_{1})≠L, By Corollary 5. Since L′_{1} is essential, every leaf π_{1}-injects into M, so every leaf π_{1}-injects to its corresponding leaf under f. Since every leaf maps with degree zero, [B3] says that every leaf of L′_{1} is a plane or annulus. If every leaf of L′_{1} is a plane, [Ga1] implies that M is a 3-torus, hence Haken, a contradiction. So some leaf L′_{1} of L′_{1} is an annulus, so [B3] L′_{1} must map under f onto a representative of an end ϵ of the corresponding leaf L_{1} of L. So f(L′_{1})⊇ f(\overline{L′_{1}})⊇ lim_{ϵ}(L_{1}), where lim_{ϵ}(L_{1}) is the limit set of the end ϵ of L_{1}, which is a non-empty sublamination of L, hence \underline{equals} L. So f(L′_{1})=L, a contradiction.

§5

Concluding remarks

It still remains to show that every leaf of L′ maps with degree either 0 or 1. Thus far this has eluded us, but it \underline{seems} that it should follow from the analysis we have presented here and the additional fact that our map f actually has degree 1. This means that if there are two inverse image leaves of a leaf L, each mapping with non-zero degree, then there are two such leaves with degrees of \underline{opposite} \underline{sign}. This in turn means that, ‘morally’, one maps orientation-preservingly, and the other orientation-reversingly. Yet they have, by Proposition 15, intersecting closures. This \underline{seems} impossible.

Conjecture. If f:M→N is a map of positive degree between non-Haken 3-manifolds, and L is an essential lamination in N, with pullback L′, then no leaf of L′ maps to its corresponding leaf with negative degree.

Of course, this has ‘seemed’ impossible (to me) for quite some time, without being \underline{provably} impossible!

§6

The (hypothetical) end

This last part is an attempt to give a very rough idea of where all these ideas might be leading. As such, we cannot make it particularly rigorous.

This conjecture would of course be wasted without some idea of how to use it to deform homotopy equivalences to homeomorphisms. There are (at least) two approaches:

(1) With the better understanding of the structure of pullbacks achieved, try to show that you \underline{can} surger the pullback to make it essential, or

(2) cheat.

The ‘cheat’ we have in mind is to try to mimic the argument of [Br1]. That is, we pick an essential loop γ in a leaf L of L, and look at f^{−1}(γ)⊆f^{−1}(L)=L′. The idea is that f^{−1}(γ) is a finite collection of loops γ_{1},…,γ_{n}. If L′ is the (conjectured) degree-one leaf mapping to L, containing the loops γ_{1},…,γ_{k}, say, then the sum of their degrees (thought of as mapping to γ) is one, and for all of the \underline{other} leaves mapping to L, the sum of degrees of loops in each leaf is zero. This means that we ought to be able to splice together the inverse images in each leaf, and then make the resulting degree-zero loops go away; they would, after all, map null-homotopically into the \underline{loop} γ. This would make the inverse image of the loop γ a single loop γ′, so f induces a map f:M\γ′→N\γ. This is precisely the setting of [Br1].

So we should try to show the same thing that was shown in [Br1]; namely that this map is injective on the level of π_{1}. This would then allow us to deform the homotopy equivalence f to a homeomorphism, as in [Br1].

It won’t be, in general, of course. But we can make some more conjectures:

Conjecture. If L′ is the (hypothetical) unique leaf of f^{−1}(L) mapping to L with degree one, then f^{−1}(γ)∩L′ consists of a loop mapping with degree one, together with loops mapping with degree zero. If L′_{0} is a leaf mapping to L with degree zero, then every loop of f^{−1}(γ)∩L′_{0} maps to γ with degree zero.

The first part of this is false; ‘folding’ across γ can create parallel loops mapping with degree 1 and -1. But if we change ‘a loop mapping with degree one’ to ‘ parallel loops mapping with degree ±1’, then this conjecture is perhaps not altogether farfetched.

Then we could ‘clean up’ everything, by splicing together loops, leaving a single degree-one loop γ′ in L′ \underline{without} altering the complement of this remaining loop (I think - this ‘cleaning up’ nonsense is kind of complicated). Then the question really becomes whether or not γ′ can cross infinitely many (really, really (whatever that means)) \underline{distinct} surgery disks. This problem (assuming we could get this far) at least seems easier than (1).

Much of the argument of [Br1], to establish the π_{1}-injectivity of this restricted map, actually goes through, except where the assertions that π_{1}(L′)→π_{1}(L) and π_{1}(L′\γ′)→π_{1}(L\γ) are π_{1}-injective are invoked. The first could probably be sidestepped by surgery, but the second one sounds a bit tougher. But it \underline{also} doesn’t seem like the second assertion should be quite as essential (pardon the pun) to the whole argument, since it is not really saying much (assuming the first can be sidestepped) about the complement of γ′ in M (if you think of γ′ as already being as ‘simple’ as possible).

This whole idea might make some people nervous - there are homotopy equivalences between 3-spheres, for example, which have a knot as the inverse image of the unknot, so the restricted map certainly cannot be π_{1}-injective. Nor does it seem easy to figure out how to deform the map, in those situations, so that it would be π_{1}-injective. But since we are looking at loops in a pullback lamination, we are dealing with (two levels of) pullbacks of codimension-one objects, which tend to be far more tractible. There is also a growing body of work showing that one can establish the sort of result we are after here, by finding the ‘right’ loop in a 3-manifold. This is the approach both of Casson and Jungreis’ proof of the Seifertfibered space conjecture [C-J], and Gabai’s recent work on deforming homotopy equivalences for hyperbolic 3-manifolds [Ga2].

BIBLIOGRAPHY

[B1] M. Brittenham, Essential laminations and deformations of homotopy equivalences: from essential pullback to homeomorphism, to appear in Topology and its Applications.

[B2] , Essential laminations in non-Haken 3-manifolds, Topology and its Applications.

[B3] , π_{1}-injective maps of open surfaces, preprint. [C-J] A. Casson and D. Jungreis, Convergence groups and Seifert-fibered 3-manifolds, preprint.

[Co] L. Conlon, Foliations of codimension one, lectures notes from Washington University, 1990.

[EMT] D.B.A. Epstein, K. Millett, and ?. Tischler, Leaves without holonomy.

[Ga1] D. Gabai, Foliations and 3-manifolds, Proc ICM Kyoto-1990 (1991), 609-619.

[Ga2] On the geometric and topological rigidity of hyperbolic 3-manifolds, preprint. [G-O] D. Gabai and U. Oertel, Essntial laminations in 3-manifolds, Annals of Math..

[He] 3-manifolds, Princeton University Press, 1976.

[Wa] F. Waldhausen, On irreducible 3-manifolds which are sufficiently large, Annals of Math 87 (1968), 56-88.

Department of Mathematics, University of Texas at Austin, Austin, TX 78712 E-mail address: britten@math.utexas.edu