Hyperbolic Geometry and Homotopic Homeomorphisms of Surfaces
Abstract.
The EpsteinBaer theory of curve isotopies is basic to the remarkable theorem that homotopic homeomorphisms of surfaces are isotopic. The groundbreaking work of R. Baer was carried out on closed, orientable surfaces and extended by D. B. A. Epstein to arbitrary surfaces, compact or not, with or without boundary and orientable or not. We give a new method of deducing the theorem about homotopic homeomorphisms from the results about homotopic curves via the hyperbolic geometry of surfaces. This works on all but 13 surfaces where ad hoc proofs are needed.
Key words and phrases:
hyperbolic surface, homeomorphism, homotopy, isotopy, Poincaré disk, circle at infinity, limit set, half plane, end1991 Mathematics Subject Classification:
Primary 37E301. Introduction
On arbitrary connected surfaces , compact or not, orientable or not and with or without boundary, D. B. A. Epstein proves that (with some qualifications) properly homotopic homeomorphisms of surfaces are isotopic [4]. For this, he extends a result of R. Baer [1] to prove that homotopic, essential, simple, 2sided closed curves on are ambient isotopic. To accomodate the case that , he further proves that properly imbedded arcs which are homotopic mod the endpoints are ambient isotopic mod the endpoints. These are Theorems 2.1 and 3.1 in [4], the proofs of which in the PL category are quite elegant. Their truth in the topological category follows from results in [4, Appendix] about approximating topological curves and homeomorphisms by their PL counterparts in dimension two. We restate these results as Theorem 4.3 and Theorem 4.4 and refer the reader to [4] for the proofs.
The method in [4] of deducing from these two results the theorem about homotopic homeomorphisms is a bit involved and requires that the homotopy be proper if has any noncompact components. In this note we propose a new method of deducing the theorem (in fact, a stronger theorem) from the same two results using a hyperbolic metric on with certain reasonable properties (which we will call a standard metric). There are exactly exceptional surfaces, which do not admit a standard metric (see Section 6). Our proof, therefore, works for all surfaces except the “unlucky” and nowhere requires the homotopy to be proper.
In the following definition and throughout the paper we use the open unit disk as our model of the hyperbolic plane and denote the circle at infinity by .
Definition 1.1.
A hyperbolic half plane is a hyperbolic surface isometric to the subsurface of which is the union of a geodesic and one of the components of .
Remark 1.
Generally, we just refer to these as half planes. Many results fail for hyperbolic metrics with imbedded half planes on noncompact surfaces. Any noncompact surface can be given a complete hyperbolic metric with geodesic boundary and imbedded half planes (see the example below).
Definition 1.2.
A hyperbolic metric on a surface is called “standard” if it is complete, makes geodesic, and admits no isometrically imbedded half planes. We call a surface equipped with such a standard metric a standard hyperbolic surface. If a surface is homeomorphic to a standard hyperbolic surface, it will simply be called a standard surface.
In this paper we prove for standard hyperbolic surfaces several results that are wellknown for complete, hyperbolic surfaces with geodesic boundary and finite area. We believe that our proofs of two of these results (Theorems 3.1 and 4.1) are new, even for compact surfaces.
Example 1.3.
We give an easy example showing how half planes can occur in a noncompact surface with infinite genus. In Figure 1 we depict a 2ended surface of infinite genus and a shaded subsurface homeomorphic to . Excise the interior of and on the remaining surface put a complete hyperbolic metric making the one boundary component a geodesic. Now glue on a hyperbolic half plane by an isometry along the boundary to obtain a complete hyperbolic metric on with an isometrically imbedded hyperbolic half plane , as pictured in Figure 2.
Throughout this paper we assume that every hyperbolic surface (standard or not) is complete with geodesic boundary. We will use the term “half plane” for “isometrically imbedded hyperbolic half plane”.
2. Limit Points and the Ideal Boundary
We will let denote the open unit disk with the Poincaré metric. The closed unit disk will be denoted by , where is the unit circle, called the circle at infinity. If is a complete hyperbolic surface without boundary, the universal covering space is . If and is geodesic, the double has a canonical hyperbolic metric which is the double of the metric on and . If one fixes a lift of , this serves as the universal covering space. The group of deck transformations is the restriction to of the subgroup of deck transformations for which leaves invariant.
We will write for the closure of in . We will set and call this the ideal boundary of . Thus is a compact subset of and is equal to the unit circle precisely when .
Many results are more easily proven for surfaces without boundary. One often proves a result for by proving the corresponding result for and remarking that its truth there implies its truth in . For this, the following lemma will be needed.
Lemma 2.1.
If , then is a standard hyperbolic surface if and only if is a standard hyperbolic surface.
Proof.
The “if” direction is trivial. For the converse, suppose has a standard hyperbolic metric. Write and consider the tiling of by lifts and . Let denote the set of ideal endpoints of , either a finite set for all , or countably infinite for all . Since and (hence ) has no half planes, it is elementary that is a countable, dense subset of .
Let be given. If is countably infinite, then all but finitely many of the components of are sets with Euclidean diameter . (We use the standard Euclidean metric on the open unit disk.) Similarly, whether or not the ’s are infinite, the components of have Euclidean diameter for sufficiently large.
Suppose that the doubled metric on is not standard. Thus, contains a half plane with lift having a nondegenerate arc as its ideal boundary. The projection of onto by the covering map is a homeomorphism. Let be an interior point of . Since is countable and dense in , we find a sequence of points converging to . If infinitely many of the ’s lie in a single , then for some integer , the component of having as an ideal endpoint lies entirely in . Even if not, it remains true that there is an integer such that the component of having as an ideal endpoint lies entirely in . Since the ’s tile , the half plane bounded by contains infinitely many ’s which are lifts of . Consequently, there are covering transformations permuting these ’s nontrivially, hence identifying distinct points of under the projection to . This contradiction implies that is standard. ∎
Definition 2.2.
The limit points of are the accumulation points in of the set for fixed . The union of these points is the limit set of .
The following is well known and elementary.
Lemma 2.3.
The limit set of is independent of .
Let be the set of fixed points of the (extensions to of the) nontrivial deck transformations of . Then .
Definition 2.4.
We will call an end of a simply connected end if it has a simply connected neighborhood in .
Simply connected ends can occur only if . For instance, an isolated simply connected end has a neighborhood homeomorphic to . But the simply connected ends can form a very complicated subset of the endset, even a Cantor set of such ends being possible.
Theorem 2.5.
If is a standard hyperbolic surface with no simply connected ends, then is dense in the ideal boundary and .
Proof.
If is nonorientable, its orientation cover is intermediate between and , hence we can assume that is orientable. Now suppose the contrary, i.e. that there is a point not approached by points in . If , our surface is simply connected and has been excluded by hypothesis. Let be the maximal open interval in containing the point , letting denote the endpoints of . We consider the cases and .
If , then all deck transformations would be parabolics fixing this point. In this case, the covering group is infinite cyclic and . If , this is the open annulus with one end a cusp and the other a “flaring” annular end (an infinite hyperbolic trumpet). This surface contains an imbedded half plane and has been excluded by hypothesis. If , then has one cusp and, since the boundary is geodesic, at least one simply connected end, also excluded by hypothesis.
Suppose then that . Let be the geodesic with endpoints . Since lies in the ideal boundary of , so do and , hence . Let be the portion of bounded by , and set . Since contains , . Remark that, if is a deck transformation, either , or and g is hyperbolic with axis . Otherwise, .
We consider two cases: (a) is nondegenerate and is not the axis of a deck transformation, or (b) is the axis of a deck transformation .
In case (a), the above remark assures us that no deck transformation identifies distinct points of or of . Therefore, under the covering projection, projects to s a geodesic homeomorphic to the reals which cuts off the image of under the covering projection, which is a homeomorphism on . If , the image of is a half plane, contradicting the fact that contains no half planes. If , the image of contains at least one simply connected end of , again a contradiction..
In case (b), the geodesic projects to an essential closed curve which cuts off the image of in . Thus, if , projects to a neighborhood of a flaring annular end of , contrary to hypothesis. If , projects to a neighborhood of at least one simply connected end of , contrary to hypothesis.
Thus, , so we have also proven that . ∎
As this proof reveals, simply connected ends cause essentially the same obstruction to this theorem as imbedded half planes.
For the sake of completeness we include the standard proof of the following.
Corollary 2.6.
If is a standard hyperbolic surface with no simply connected ends and , then the orbit of under the group of covering transformations is dense in .
Proof.
We will show that the orbit clusters at every point of the dense set . Let and a covering transformation transformation fixing . Choose in the orbit of not fixed by . Since is either hyperbolic or parabolic, applying the positive and negative iterates of to produces a subset of the orbit of clustering at . ∎
Corollary 2.7.
If is a standard hyperbolic surface with no simply connected ends and , then the ideal boundary is a Cantor set.
Proof.
By Corollary 2.6, every point of is a limit point of . If is not a Cantor set, it contains a closed, nondegenerate interval. Let , with endpoints , be a maximal such interval. Let be fixed by a deck transformation . Clearly, so which is a contradiction. ∎
Remark 2.
Theorem 2.5 and its corollaries are standard for complete hyperbolic surfaces of finite area.
Remark 3.
Remark that, even if has simply connected ends, its double does not and so the conclusion of Theorem 2.5 holds for . In applications of this theorem in what follows, we will always be working either in or in if .
3. Extensions of Homeomorphisms to the Ideal Boundary
Theorem 3.1.
If is a standard hyperbolic surface and is a homeomorphism, then any lift extends canonically to a homeomorphism .
Remark 4.
If is not standard, that is contains imbedded half planes, then there are homeomorphisms that do not extend.
If , then sends geodesics in to geodesics in and therefore, if the assertion holds for the double , it holds for . Thus, we may assume that . If is nonorientable, admits orientation reversing lifts as well as orientation preserving ones. If is orientable, then is orientation preserving (respectively, reversing) if and only if all of its lifts are orientation preserving (respectively, reversing).
Lemma 3.2.
Any lift induces a homeomorphism which is orientation preserving or reversing according as is orientation preserving or reversing.
Proof.
Recall that is the set of fixed points of nontrivial deck transformations and, since , is dense in (Theorem 2.5). Since consists of endpoints of the lifts of essential closed curves, it is clear that induces a bijection . This bijection preserves or reverses the cyclic order induced on by the orientation of according as is orientation preserving or reversing. In either case, it is well known that the fact that is dense in implies that extends to a homeomorphism, again denoted by with the asserted orientation properties. ∎
It is now clear how to define . We set and . Since is compact, this bijection will be a homeomorphism precisely if it is continuous at each point .
Definition 3.3.
If is a curve such that some (hence every) lift has two distinct, well defined ideal endpoints on , is called a pseudogeodesic. We denote the extension of the lift to by and call it a completed lift of .
Remark that a completed lift is actually the metric completion of relative to the standard Euclidean metric on .
It is well known that a loop will be a pseudogeodesic if and only if it is essential and does not bound a cusp. Remark that the endpoints of completed lifts depend only on the free homotopy class of .
We regularly write for , where .
Definition 3.4.
Remark 5.
Because the lifts of a closed pseudogeodesic have distinct endpoints, a closed pseudogeodesic is homotopic to a closed geodesic. In particular, closed pseudogeodesics are not parabolic curves.
Corollary 3.5.
The homeomorphism carries bijectively onto .
Indeed, carries the lifts of pseudogeodesic loops exactly onto lifts of pseudogeodesic loops, so the assertion is a consequence of (the proof of) Lemma 3.2.
Definition 3.6.
Let be the set of pseudogeodesics in such that some, hence every, completed lift has both endpoints in .
Lemma 3.7.
if , then .
Proof.
Fix a lift and let be the homeomorphism given by Corollary 3.5. If has an endpoint , we will show that limits on the point . Applying this to both endpoints of , , will prove the lemma.
Let be a closed geodesic with a lift such that one endpoint of is . The existence of such a closed geodesic follows from Definition 3.3 and the fact that . Let be a closed geodesic intersecting transversely and let be one of these intersection points. Let be the lifts of in , indexed so that monotonically along . Let be the lift of through . Each has endpoints bounding a subarc containing the point . The sequences and monotonically as . We claim that . Otherwise, the geodesics accumulate locally uniformly (in the hyperbolic metric) on the geodesic with endpoints . This would mean that the closed geodesic in accumulates locally uniformly on a distinct geodesic. This is impossible. The sets bounded by form a fundamental system of closed neighborhoods of .
Applying to this picture gives a family of pseudogeodesics with endpoints , meeting the curve in points , such that and monotonically (by the fact that is a homeomorphism that either preserves or reverses orientation). Let be the subset of bounded by the arc and the arc of with endpoints containing the point . Since is a homeomorphism, is a nest with empty intersection. Also and and it follows that the form a descending nest of closed neighborhoods of . If is an open neighborhood of and no , then, by compactness of , there exists a point for all . Since and , it follows that . Thus, . Since is a homeomorphism, it follows that there exists an such that . Since the form a fundamental system of neighborhoods of , it follows that for some . Thus, which is a contradiction so the form a fundamental system of neighborhoods of . Since has endpoint , a neighborhood of one end of lies in . Therefore . It follows that has endpoint . ∎
Remark 6.
Crucial in the previous proof is the existence of a closed geodesic intersecting the closed geodesic transversely. The existence of such a can be seen as follows. If self intersects, let . Otherwise is a simple closed geodesic. If it does not intersect another closed geodesic, then is peripheral. Since and is standard, would have to bound a cusp on one side. But in this case is not a geodesic, contrary to hypothesis.
Since is a union of orbits in of the group of (extended) covering transformations and since , we can apply Corollary 2.6 to obtain the following.
Lemma 3.8.
The set is dense in .
Proof of Theorem 3.1.
We want to show that is continuous at . This is clear if , so we assume . Let be an open neighborhood of . Since is dense in , we can choose a compact arc such that . Let be any curve in with endpoints and and the projection of to . Then so by Lemma 3.7, .
Therefore the curve, , which is an extended lift of , has endpoints . Then the subset of bounded by and is a closed neighborhood of in and . This proves continuity at arbitrary . ∎
Corollary 3.9.
If is a pseudogeodesic in and is a homeomorphism, then is a pseudogeodesic.
Remark 7.
Our proof of Theorem 3.1 includes the case in which is compact, hence gives a fundamentally different proof in that case from the ones given by Casson and Bleiler [3, Lemma 3.7] and Handel and Thurston [5, Corollary 1.2]. These proofs make use of compactness, whereas we do not. The analogous result holds for higher dimensional, compact, hyperbolic manifolds [2, Proposition C.1.2], [7, Theorem 11.6.2], where compactness is only used to guarantee that is a pseudoisometry [7, p. 555]. In [3], compactness is only used to guarantee that is uniformly continuous.
4. The Basic Isotopy Theorem.
The following is well known when is compact.
Theorem 4.1.
If is a standard hyperbolic surface and is a homeomorphism, then is isotopic to the identity if and only if it has a lift to such that is the identity.
The “only if” part of this theorem is elementary. In fact, working in the double , if is isotopic to the identity, it is obvious that is the identity on and since is dense in then and this remains true for .
Corollary 4.2.
If is a standard hyperbolic surface and are homeomorphisms, then is isotopic to if and only if there are lifts to such that agree on the ideal boundary .
Indeed, set .
We cannot find a proof of Theorem 4.1 in the literature for surfaces of infinite Euler characteristic, so we give a detailed sketch here. We do not need orientability nor empty boundary. Basic to our proof are the following two EpsteinBaer theorems. They are, respectively, Theorem 2.1 and Theorem 3.1 in [4]. While the proofs are carried out in the PL category, it is shown in the Appendix of [4] that these and other results in the paper remain true in the topological category. There is no hyperbolic metric assumed on in these two theorems.
Theorem 4.3 (EpsteinBaer).
Let be freely homotopic, imbedded, sided, essential circles. Then there is a compactly supported homeomorphism and an ambient isotopy with and , compactly supported in , such that .
Theorem 4.4 (Epstein).
Let be properly imbedded arcs with the same endpoints which are homotopic modulo the endpoints. Then there is a compactly supported homeomorphism and an ambient isotopy with and , compactly supported in and carrying by , such that .
We will generally call itself a compactly supported ambient isotopy. As noted earlier, we sometimes abuse terminology by identifying a curve with its image. Thus, instead of writing , we might write (or ).
4.1. Preliminaries
The following is elementary and well known.
Lemma 4.5.
Orientation preserving homeomorphisms and are isotopic to the identity.
Corollary 4.6.
The homeomorphism of Theorem 4.1 admits an ambient isotopy , supported near , such that restricts to the identity on .
Proof.
Since fixes pointwise, it is clear that preserves the components of and is orientation preserving on each. In a collar neighborhood of each boundary component, one extends the isotopy of Lemma 4.5 to an ambient isotopy supported in the collar. ∎
From now on, therefore, we will assume that .
Lemma 4.7.
If admits a lift such that , then it admits a unique such lift.
Proof.
Let be lifts of so that . Then is a covering transformation which is the identity on , hence . ∎
Remark 8.
Hereafter, will denote this unique lift. By abuse, its extension to will also be denoted by . Remark that, if is varied by an ambient isotopy , then also admits a lift . Indeed, itself has such a lift, since it preserves the free homotopy classes of loops, and so .
Corollary 4.8.
The canonical lift commutes with all covering transformations.
Proof.
Indeed, if is a covering transformation, is a lift of which fixes pointwise. By Lemma 4.7, . ∎
The following is a key lemma.
Lemma 4.9.
Let be an essential imbedded circle such that . Then there is an ambient isotopy , supported in an arbitrarily small neighborhood of , such that is the identity on any lift of .
Proof.
Since and since preserves the endpoints of in , . If we set , this parametrizes and so parametrizes as (up to an additive integer). The fact that then implies that is just translation of by an integer . Fix a normal neighborhood of , with . Do this so that is a lift of an annular neighborhood of . If is a component of , make a 1sided normal neighborhood . Now translation on by is isotopic to the identity through translations and this easily defines an ambient isotopy , supported in , such that is the identity along . As the notation suggests, we have taken care that is a lift of an ambient isotopy supported in which leaves (as a point set) invariant. If is another lift of , it is the image of under a deck transformation . The parametrization of corresponds to that of under up to translation by an integer. By Corollary 4.8 and Remark 8, commutes with . Then for any ,
Thus, fixes pointwise. ∎
Corollary 4.10.
There is an ambient isotopy , supported in any preassigned neighborhood of , such that .
Proof.
For each compact component of , apply Lemma 4.9. The lifts of noncompact components project onetoone onto and the assertion is trivial. ∎
From now on, therefore, we will assume both that and .
4.2. The proof of Theorem 4.1
We will concentrate on the case in which is noncompact, remarking at the end on how our methods adapt easily to the compact case.
There is an exhaustion of by compact, connected surfaces , with , . We can require that the finitely many connected components of are neighborhoods of ends (that is, are noncompact with compact frontier), the frontier of each being a finite, disjoint union of properly imbedded arcs and 2sided, essential circles. Let the collection of all of these arcs and circles be denoted by . Note that partitions into a family of compact submanifolds with boundary composed of elements of and arcs and/or circles in . Enumerate these submanifolds as , in such a way that , are the components of , are the components of , etc.
Lemma 4.11.
There is an ambient isotopy such that and , for each and each lift .
Proof.
If is a properly imbedded arc in , then is properly imbedded with the same endpoints as , since . Since , the lifts and have the same endpoints. It follows that and are homotopic modulo their endpoints. By Theorem 4.4, there is a compactly supported ambient isotopy , keeping pointwise fixed, such that . Note that perturbs only finitely many images of elements of . Note that each lift has endpoints in , fixed by . It is evident, then, that .
Similarly, if is a circle imbedded in , any lift is an imbedded copy of in having well defined ideal endpoints in . Since fixes these endpoints, the lift of has these same ideal endpoints and is freely homotopic to in . By Theorem 4.3, there is an ambient isotopy , compactly supported in , such that . Again, perturbs only finitely many images of elements of . An application of Lemma 4.9 shows that the ambient isotopy can be modified so that .
Let be contained in . Construct as above. If , we must choose so . The trick is to temporarily cut apart along (the image of) and apply the above argument in the resulting component containing . Suppose that is the full list of elements of in . Then continuing in this way, we find , compactly supported, the identity on , such that restricts to the inclusion map on and restricts to the inclusion map on , for each lift . Now let be the component of containing . Work in to produce , a compactly supported isotopy on , the identity on , such that restricts to the inclusion map on and restricts to the inclusion on , for each lift . It is important to note that can be viewed as a compactly supported isotopy on which is the identity on throughout the isotopy. Proceeding in this way, produce an isotopy such that and are the inclusions, for each and each lift . The isotopy is well defined since, for each , all but finitely many are the identity isotopy on . ∎
Hereafter, we replace by , assuming that is the identity on and that is the identity on the entire lift of this set.
Proof of Theorem 4.1.
Each is a compact surface with nonempty boundary. As such it can be viewed as the result of attaching finitely many bands (twisted and/or untwisted) to a disk (see [6, pp. 4345]). Thus, there are finitely many disjoint, properly imbedded arcs in which decompose this surface into a disk. The properly imbedded arcs and have the same endpoints since fixes pointwise. Furthermore, since fixes pointwise, they have lifts in with common endpoints. This implies that they are homotopic in by a homotopy keeping their endpoints fixed. Applying Theorem 4.4 in the usual way, allows us to assume that fixes pointwise, . Since cutting apart along these arcs gives a disk and the homeomorphism induced by is the identity on , we apply Alexander’s trick to find an isotopy of to the identity which is constant on . Regluing gives an isotopy of to the identity which is constant on the boundary. Carrying this out for each , we obtain isotopies that fit together to an isotopy on since they leave all boundary components pointwise fixed. ∎
Remark 9.
While the isotopy in the above proof is constant on , this is only after altering the original by an ambient isotopy that is not generally constant on . Thus the isotopy in Theorem 4.1 is not generally constant on the boundary. A simple example is a Dehn twist on the closed annulus.
Remark 10.
If is compact, the above proof easily adapts. Indeed, if , use our methods to isotope to be the identity on and so that is the identity on . Now introduce the arcs as above and use Alexander’s trick to complete the isotopy. If , find a simple, closed, 2sided curve that does not disconnect (). First do the isotopy of that makes it the identity on and makes the identity on all lifts of . Now cut apart along and apply Alexander’s trick. Since this last isotopy is constant on the boundary, we reglue to obtain the desired isotopy on .
5. Homotopic Homeomorphisms
In [4], Epstein states and proves the following theorem. Here we say that a homotopy is rel if the homotopy is through maps that carry into itself.
Theorem 5.1 (EpsteinBaer).
Let be a surface and a homeomorphism homotopic to the identity. If is the open disk, the closed disk, the open annulus, or the closed annulus require that be orientation preserving. Suppose that either of the following two conditions is satisfied:

The surface has all boundary components compact;

The homotopy is rel and proper.
Then is isotopic to the identity.
Corollary 5.2.
Let be a surface and homotopic homeomorphisms. If is the open disk, the closed disk, the open annulus, or the closed annulus require either that and both preserve orientation or both reverse it. If either condition or of Theorem 5.1 is satisfied, then is isotopic to .
Proof.
If is the homotopy, then is a homotopy of to and the hypotheses of Theorem 5.1 are satisfied. Therefore is isotopic to the identity. ∎
Using the hyperbolic methods of this paper, we will prove these results for standard surfaces, but with substantially weaker hypotheses.
A weaker condition than (1) is that has no simply connected ends. It is simply connected ends that cause difficulties rather than noncompact boundary components. If is isotopic to the identity, then must fix all ends but there are many examples of standard surfaces (simply connected or not) admitting homeomorphisms that are homotopic to the identity but do not fix some simply connected ends. Here are the simplest ones.
Example 5.3.
Let be an ideal gon, a surface with simply connected ends. Since is contractible, all maps are in the same homotopy class. There is a homeomorphism which permutes the ends cyclically. Then is homotopic to the identity, but clearly not isotopic to the identity.
Moreover, there are examples of homeomorphisms that are homotopic to the identity and fix all ends but are not isotopic to the identity. Part (3) of Theorem 5.6 shows that the following are the only possible surfaces that admit such examples.
Example 5.4.
If the surface is the connected sum along a loop of any surface (other than the 2sphere or closed disk or open disk) and a noncompact simply connected surface (other than the open disk) and if is the homeomorphism given by a nontrivial Dehn twist in an annular neighborhood of , then is homotopic to the identity (deform the simply connected summand into its interior, untwist, and deform back) but not isotopic to the identity. Note that fixes each simply connected end.
Recall that is the set of fixed points of nontrivial deck transformations.
Lemma 5.5.
Let be a standard surface and a homeomorphism that is homotopic to the identity. Then admits a lift such that fixes pointwise.
Proof.
Let , , be a homotopy such that and . By the homotopy lifting property, there is a homotopy such that and is a lift of . Let be an essential closed loop in , any lift. If is a pseudogeodesic, the ideal endpoints of are distinct. Otherwise they coincide. In any event, these endpoints make up as and its lifts vary. Set , a homotopy of to through lifts of , . Since has empty interior, the ideal endpoints of are constant in . Consequently, fixes these endpoints, hence fixes pointwise. ∎
Theorem 5.6.
Let be a standard surface and a homeomorphism which is homotopic to the identity. Suppose that any one of the following conditions is satisfied:

has no simply connected ends;

the homotopy is rel ;

is any surface except those in Example 5.4 and the homeomorphism fixes each simply connected end.
Then is isotopic to the identity.
Proof.
Assume condition (1). By Lemma 5.5 and Theorem 2.5, there is a lift such that fixes pointwise. By Theorem 4.1, is isotopic to the identity.
Assume condition (2). The double has no simply connected ends and both and the homotopy double because the homotopy is rel . Apply the argument for condition (1) to conclude that there is a lift such that fixes pointwise. Therefore leaves the universal cover invariant and fixes pointwise. Again, Theorem 4.1 implies that is isotopic to the identity.
Assume condition (3). If , then is simply connected, the projection map from to is oneone, and the unique lift of fixes every end of . Thus fixes and Theorem 4.1 implies that is isotopic to the identity. Therefore, suppose and let be the lift given by Lemma 5.5. Assume there is a point and let be the component containing . Then are fixed by .
If then is the only point of and is the fixed point of a parabolic deck transformation . Then for a lift of and a suitable choice of , is a lift of which fixes every lift of each simply connected end while still fixing . Again, Theorem 4.1 implies that is isotopic to the identity.
Next suppose . Let be the geodesic with endpoints and . Let be the closed set cut off by on the side determined by . Since , the only way a covering transformation can identify distinct points in is for that transformation to have as its axis. Let be the covering transformation, the powers of which give every covering transformation with axis . If is orientation preserving, then is of the form , where is noncompact and simply connected and is an open disk. The boundary circle is a closed geodesic. If , then and as in the case where has only one point, one can choose a covering of such that fixes all the lifts of the ends in while still fixing and . Again, Theorem 4.1 implies that is isotopic to the identity. Otherwise, separates and is a connected sum violating the condition on in (3). If is orientation reversing, the reader will easily check that is the connected sum of a projective plane and a noncompact, simply connected surface, violating the condition on in (3).
Therefore carries homeomorphically onto a complete submanifold having only simply connected ends and cut off by the noncompact geodesic which is the projection of . All the ends of are each fixed by . This implies that fixes the corresponding ends of upstairs, hence that fixes pointwise. Indeed, leaves invariant each component of , hence fixes the ideal endpoints of these arcs. Since has no half planes, these endpoints are dense in . Thus, fixes pointwise and Theorem 4.1 implies that is isotopic to the identity. Theorem 5.6 is proven. ∎
For standard surfaces, Theorem 5.6 is substantially stronger than Theorem 5.1. Indeed, (1) of Theorem 5.1 implies (1) of Theorem 5.6, but not conversely, and (2) of Theorem 5.1 implies (2) of Theorem 5.6, but not conversely. Part (3) of Theorem 5.6 shows that Example 5.4 gives the only possible examples of surfaces admiting homeomorphisms that are homotopic to the identity and fix all simply connected ends but are not isotopic to the identity.
Corollary 5.7.
The proof is essentially the same as that of Corollary 5.2.
The restriction in Theorem 5.6 to standard surfaces might cause concern. As the following shows, this restriction is very mild.
Theorem 5.8.
Up to homeomorphism there are exactly nonstandard surfaces. They are: the open disk, the closed disk, the open annulus, the half open annulus, the closed annulus, the open Möbius band, the closed Möbius band, the half plane , the doubly infinite strip , the sphere, the projective plane, the torus, the Klein bottle.
For these 13 surfaces, there are ad hoc proofs of Theorem 5.1 (requiring that be orientation preserving in four cases) which can be gleaned from [4]. Some are sufficiently elementary to be posed as exercises.
The next section will be devoted to the proof of Theorem 5.8.
6. The Nonstandard Surfaces
The following is quite elementary.
Lemma 6.1.
The surfaces listed in Theorem 5.8 are nonstandard.
For instance, the closed disk has a hyperbolic metric, but the boundary cannot be geodesic. The open disk has a complete hyperbolic metric, necessarily isometric to the Poincaré metric. This has imbedded half planes. The open annulus has a complete hyperbolic metric and either one end is a cusp and the other a hyperbolic trumpet, or both ends are trumpets. In any case there are imbedded half planes. The open Möbius strip has the open annulus as its orientation cover. The nontrivial deck transformation is an isometry of (relative to the lift of any complete hyperbolic metric on ) which interchanges the ends. Thus both ends are trumpets and there is a half plane in which is carried to a disjoint half plane by the deck transformation. Thus contains a half plane. Checking out the rest can be left to the reader.
We need the notion of Euler characteristic for possibly noncompact surfaces. For any space with finite dimensional real homology, the Euler characteristic is the alternating sum of the betti numbers . For a closed, orientable surface , this is . For every other surface, . If is infinite, we set . In particular, nonnegative Euler characteristic is finite.
Recall that, if is a surface with handles, crosscaps, compact boundary components, annular ends (all finite integers) and no other boundary components or ends, then the Euler characteristic of is given by the following formula:
We fix the meaning of these letters. Remark that the values of and are not individually well defined by . For example, if , we can also take and (cf. [6, p. 26, Lemma 7.1]). In fact, 3 crosscaps equals 1 handle and 1 crosscap.
Remark 11.
In particular, all of this applies to complete hyperbolic surfaces with finite area and compact geodesic boundary, where one also has the formula
computed by integrating the constant curvature . In such a surface, the annular ends are cusps. (See [3, pp. 31  37] for a discussion of such surfaces).
Definition 6.2.
A complete hyperbolic surface with geodesic boundary and finite area is called a generalized pair of pants if and . We will refer to a cusp of a generalized pair of pants as a boundary component of length .
The following lemma is well known.
Lemma 6.3.
Given a triple of numbers , , there exists a generalized pair of pants whose three boundary components have length .
Proposition 6.4.
Given a surface with finite Euler characteristic and with boundary closed curves and endset consisting of annular ends, and given positive real numbers , there exists a complete hyperbolic metric on such that is a geodesic of length , , and the annular ends are cusps. In particular, this metric is standard.
Proof.
Cut the handles of apart along curves and the crosscaps apart along curves to yield a surface homeomorphic to a sphere with boundary components and punctures. Since it follows that . Thus and has a generalized pair of pants decomposition. By Lemma 6.3, each generalized pair of pants can be given a hyperbolic metric so that the annular ends are cusps and the boundary lengths are such that the regluing can be done to give a hyperbolic metric on with the having the correct lengths. ∎
The following lemma is classical. See CassonBleiler [3, p. 37].
Lemma 6.5.
If is a surface with finite Euler characteristic and noncompact boundary components then,
Proposition 6.6.
If is a nonstandard surface with finite Euler characteristic and finitely many boundary components, then it is homeomorphic to one of the surfaces in Lemma 6.1.
Proof.
Let , as above, have compact boundary and finite Euler characteristic. If , then . The possibilities are:

and , so is the sphere or torus.

and , so is the projective plane or Klein bottle.

and , so is the closed disk or closed annulus.

and , so is the open disk or open annulus.

and , so is the closed Möbius strip.

and , so is the open Möbius strip.

and , so is the half open annulus.
Thus, any surface with finite Euler characteristic and only compact boundary components, other than the surfaces just listed, has negative Euler characteristic and so, by Proposition 6.4, has a standard hyperbolic metric. By Lemma 6.1, the above 11 are nonstandard.
Let have finite Euler characteristic and finitely many boundary components, not all of which are compact. Since is nonstandard, so is (Lemma 2.1). By Lemma 6.5, has finite Euler characteristic. Since the boundary is empty, must be in the above list of surfaces. Since cannot be compact and cannot have boundary, is either the open disk, the open annulus or the open Möbius strip. The only surface whose double is the open disk is the half plane. The doubly infinite strip is the only surface with noncompact boundary whose double is the open annulus, and no surface has double equal to the open Möbius strip. This completes the proof. ∎
Lemma 6.7.
If is a surface with infinitely many boundary components, then has at least one end that is not annular.
Proof.
Suppose that is a surface such that every end of is annular. Then has finitely many annular ends and there exists a compact surface such that the finitely many components of are open annuli. Thus, every one of the finitely many components of is either of the form or . Thus, has finitely many boundary components. The lemma now follows by contradiction. ∎
Proposition 6.8.
If is a surface with infinitely many boundary components and/or infinite Euler characteristic, then is standard.
Proof.
If has nonempty boundary, then is standard if and only if is standard (Lemma 2.1). Also, under our hypothesis, has at least one end that is not annular (Lemma 6.7). Hence we can limit our attention to surfaces without boundary with at least one end that is not annular.
Thus, there is an exhaustion of by connected surfaces with finite Euler characteristic such that the finitely many connected components of are neighborhoods of nonannular ends, the frontier of each being a simple closed curve. Let the collection of all of these simple closed curves be denoted by . Note that partitions into a countable family of submanifolds with finite Euler characteristic and with boundary composed of elements of . We can assume the exhaustion is chosen so that all the have negative Euler characteristic. Then by Proposition 6.4, each has a standard hyperbolic metric such that every simple closed curve in is a geodesic of length . Thus, these hyperbolic metrics can be fitted together to give a hyperbolic metric on . Evidently, an imbedded half plane would have to meet closed geodesics in , for infinitely many indices . Since closed geodesics are not nullhomotopic, the intersections of these geodesics with will be arcs which, together with an arc in would form a geodesic digon. Such digons are forbidden in hyperbolic geometry. ∎
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