The <span class="mathjax-formula">$ {C}^{1}$</span> closing lemma for generic <span class="mathjax-formula">$ {C}^{1}$</span> endomorphisms

www.elsevier.com/locate/anihpc

The C ¹ closing lemma for generic C ¹ endomorphisms

Alvaro Rovella

, Martín Sambarino

CMAT – Fac. de Ciencias, Univ. de la República, Iguá 4225, CP:11400 Montevideo, Uruguay Received 13 April 2010; received in revised form 2 September 2010; accepted 16 September 2010

Available online 18 September 2010

Abstract

Given a compactm-dimensional manifoldMand 1r∞, consider the spaceC^r(M)of self mappings ofM. We prove here that for every mapf in a residual subset ofC¹(M), theC¹closing lemma holds. In particular, it follows that the set of periodic points is dense in the nonwandering set of a genericC¹map. The proof is based on a geometric result asserting that for genericC^r maps the future orbit of every point inMvisits the critical set at mostmtimes.

©2010 Elsevier Masson SAS. All rights reserved.

MSC:37Cxx; 58K05

Keywords:Closing lemma; Critical points; Transversality

1. Introduction

LetMandN be real compactC^∞manifolds without boundary. For every 1r∞,C^r(M, N )will stand for the set of classC^r maps endowed with the usualC^r topology. WhenM=N, the spaceC^r(M, M)will be denoted byC^r(M).

The problem known asthe Closing Lemmastates that: given a nonwandering pointxof aC¹mapf, is it possible to find aC¹perturbationgoff such thatxisg-periodic? This was solved by C. Pugh for diffeomorphisms in 1967 [4].

Since then, many generalizations were obtained in different contexts besides of diffeomorphisms (see [5]). The case ofC¹endomorphisms was not included in the above contexts.

Nevertheless, forC¹endomorphisms of a compact manifold and by a clever extension of Pugh’s technique, L. Wen obtained aC¹closing lemma valid for those having a finite number of critical points, see [7] and [8]. However, the set ofC¹maps for which this hypothesis holds is not dense. Here we will prove that:

Theorem 1.There exists a residual set RinC¹(M)such that givenf ∈R,U a C¹ neighborhood off andx a nonwandering point off, there exists a mapg∈U such thatxisg-periodic. Moreover, there exists a residual subset DofRsuch that for everyf ∈Dthe set of periodic points is dense in the nonwandering set.

✩ Partially supported by ANII, proyecto FCE2007 No. 577.

* Corresponding author.

E-mail addresses:leva@cmat.edu.uy (A. Rovella), samba@cmat.edu.uy (M. Sambarino).

0294-1449/$ – see front matter ©2010 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2010.09.003

This result answers positively the questions posted by M. Shub in the final section of [6]. The first main ingredient in the proof of Theorem 1 is Wen’s result on closing with finitely many singularities. The second one is a geometric result regarding the study of the singularities of a map. The study of the singularities of smooth mappings has been largely considered in the literature. It is well known, for example, that the singular set of a genericC^∞map admits a stratification in submanifolds. The singular or critical set of a mapf∈C¹(M)is defined as the set of points where the differential off is not invertible, and is denoted byS_f. Our interest here is to determine how frequently the future orbit of a point intersects the critical set off, at least restricting the study to a residual set of maps.

Theorem 2.LetMbe a compact manifold of dimensionm. Given any positive integerr1it holds that:

1. There exists a residual subsetRr ofC^r(M)such that, for everyf ∈Rr, and every z∈M, the set of integers n0such thatfⁿ(z)∈Sf contains at mostmelements.

2. There exists a residual subsetPr ofC^r(M)such that for every mapf ∈Pr it holds that

n0fⁿ(Sf)does not intersect the set of periodic points off.

2. Proof of Theorem 1 and some consequences

In this section we prove our generic C¹ closing lemma assuming that Theorem 2 holds. We first state Wen’s result. To do that we need some definitions. A treeF=(Q, f )is an infinite sequence of mutually disjoint nonempty finite setsQ₀, Q₁, . . . , Q_n, . . .whereQ₀consists of a single pointq₀together with a mapf :Q− {q₀} →Qwhere Q=_∞

n=0Q_n such thatf mapsQ_n intoQ_n−1for eachn=1,2, . . .. An infinite sequenceq₀, q₁, . . . , q_n, . . .is an infinite branch ofFiff (q_n)=q_n−1for eachn=1,2, . . .. A treeFis called complete iff is onto.

We also denote byO⁻(x)=

n0f⁻ⁿ(x)the negative orbit ofx.

Theorem 3. (See [8].) Let f ∈C¹(M) and let x be a nonwandering point of f. Assume that there exists y∈ O⁻(x)∩Ω(f )such that

1. O⁻(y)∩Ω(f )∩S_f = ∅.

2. (O⁻(y)∩Ω(f ), f )is a complete tree.

Then, given anyC¹neighbourhoodUoff inC¹(M)andUopen set containingx, there existg∈U andp∈Usuch thatpis a periodic point ofg.

This theorem corresponds to the Case 1 in the proof of Theorem A in [8]. In that case it was assumed that O⁻(y)∩S_f = ∅; the arguments used there can easily be adapted.

Lemma 1.LetR=R1∩P1the residual set inC¹(M)as in Theorem2. Letf ∈Rand letx be a nonwandering point off. Then, one of following statements holds:

1. There existsy∈O⁻(x)∩Ω(f )such thatf⁻¹(y)∩Ω(f )= ∅. 2. There existsy∈O⁻(x)∩Ω(f )such that

(a) O⁻(y)∩Ω(f )∩Sf = ∅.

(b) (O⁻(y)∩Ω(f ), f )is a complete tree.

Proof. Assume that (1) does not happen. Then, for anyy∈O⁻(x)∩Ω(f )we have thatf⁻¹(y)∩Ω(f )= ∅.Thus, any point y ∈O⁻(x)∩Ω(f ) is contained in an infinite branch Σx= {x0=x, x1, x2, . . .}withf (xn)=xn−1for n1 and wherexi∈Ω(f ).Sincef ∈Rit follows that any branchΣx containsat mostmsingular points. Among the branchesΣx⊂Ω(f )take one with a maximal cardinality of singular points in it, sayΣ= {x0=x, x1, x2, . . .}.

Then, there existsn0such thatxn∈/Sf for allnn0Lety=xn₀. It follows thatO⁻(y)∩Ω(f )∩Sf = ∅, otherwise we contradict the way we have chosenΣ. In particular,f⁻ⁿ(y)∩Ω(f )is finite for alln0. This and the fact that (1)does not hold imply that(O⁻(y)∩Ω(f ), f )is a complete tree. Thus, we have proved that(2)holds. 2

By virtue of Theorem 3 and the previous Lemma 1, to prove the first part of Theorem 1 we have to handle the situation where there exists y∈O⁻(x)∩Ω(f ) such thatf⁻¹(y)∩Ω(f )= ∅. We will prove this reasoning as in Case 2 of Theorem A in [8].

Letf ∈RwhereRis as in Lemma 1, and letU be a neighborhood of f. We will use without proof that there existssuch that for anyrsmall andx∈M, ifu, w∈B(x, r)then there existsh:M→M such thath(u)=w, supp(h)⊂B(x, r)and thath◦f ∈U. Heresupp(h)denotes the set of pointsysuch thath(y)=y. Letx be a non- wandering point (and non-periodic, otherwise there is nothing to prove) off and letUbe any open set containingx. Lety∈O⁻(x)∩Ω(f ) such thatf⁻¹(y)∩Ω(f )= ∅. Letk be such thatf^k(y)=x and consider V an open set containing y such that fⁱ(V ), i=0, . . . , kare disjoint and f^k(V )⊂U. Consider a decreasing sequence r_n→0 such thatB(y, r_n)⊂V for alln. Sincey is nonwandering, we can pickz_n∈B(y, r_n)such thatf^mn(z_n)∈B(y, r_n) for somem_n (which is larger thank) andfⁱ(z_n) /∈B(y, r_n)for 1im_n−1. Thus, we have thatf^mn(z_n)→y and, taking a subsequence if necessary, we have thatf^mn−1(z_n)→wwithf (w)=y. Sincew is wandering, there exists an open setW containingw such thatfⁱ(W )∩W= ∅for alli1. Therefore, fornlarge enough, we can take h1 supported onW such that h1(f^mn−1(z_n))=w and h1◦f ∈U. (We would like to push f^mn−1(z_n) to a preimage ofzn but this is not possible, there is no preimage of zn inW). Consider alsoh2supported onB(y, rn) such that h2(y)=zn such that h2◦f ∈U. Let h:M→M be h=h1 inW,h=h2 on B(y, rn)and otherwise the identity. It follows that g=h◦f ∈U. Notice thatzn, f (zn), . . . , f^mn−2(zn)does not intersectsW sinceW is wandering. On the other hand,f (zn), . . . , f^mn−1(zn)does not go throughB(y, rn)by the way we have chosenmn. Thereforegⁱ(zn)=fⁱ(zn)for i=0, . . . , mn−2. Now,g^mn−1(zn)=g(f^mn−2(zn))=h1(f^mn−1(zn))=w. And fi- nally,g^mn(zn)=g(w)=h(f (w))=h2(y)=zn. Therefore,znisgperiodic. Moreoverg^k(zn)=f^k(zn)∈U,and so we found agperiodic point inU. This completes the proof of the closing lemma forf∈R.

To complete the proof of Theorem 1, we must prove that there exist a residual subsetDofRsuch that for every f ∈Dthe set of periodic points is dense in the nonwandering set. This follows by standard arguments. Let{U_n}n∈N

be a countable basis ofM. For alln∈Nconsider:

An=

f∈C¹(M)

Perh(f )∩U_n= ∅

beingPerh(f )the set of hyperbolic periodic points off, and considerBn=C¹(M)−An. From the continuation of hyperbolic periodic points we know thatAn is an open set and thereforeAn∪Bnis open and dense∀n∈N. Then D1=

n∈N(An∪Bn)is a residual set onC¹(M). Let us see that iff∈D:=D1∩RthenΩ(f )=Per(f ).

Suppose, by contradiction, that exist somex∈Ω(f )−Per(f ), and letn∈Nsuch thatx∈UnandUn∩Per(f )= ∅ (in particularf /∈An). From the first part of Theorem 1, we know that for everyε >0 there existsgε∈C¹(M)such that theC¹distance betweenf andgεis less thanεandx∈Per(gε). Now we perturbgεto transformxon a hyperbolic periodic point keeping theC¹distance betweenf andgεless thanε. This implies thatgε∈Anand thereforef /∈Bn, thenf /∈An∪Bnwhich is absurd becausef ∈D.

Now we state some straightforward consequence of our main result.

Corollary 1.Letf∈C¹(M)aC¹−Ω-stable map. ThenΩ(f )=Per(f ).

Now, lets define the set E¹(M)=int

f ∈C¹(M): everyp∈Per(f )is hyperbolic

whereintmeans “interior”. In other words,E¹(M)is the set of endomorphisms such that there exists a neighborhood of it such that any periodic point of any endomorphism in this neighborhood es hyperbolic. Forf ∈E¹(M)we denote byPi(f )the set of periodic points of indexi(i.e.dimE_p^s =i).

Corollary 2.Letf∈E¹(M). Then,

1. There exists a neighborhoodV(f )such that ifUis a neighborhood of Per(f )andg∈V(f )satisfiesg=f inU, then Per(f )=Per(g).

2. There exists a neighborhoodV(f )such that ifUis a neighborhood ofP_i(f )andg∈V(f )satisfiesg=f inU, thenP_i(f )=P_i(g).

The proof of this corollary is simple and it is the same as for diffeomorphisms. Indeed, just take V ⊂E¹(M) connected and the result follows from the fact that for eachnthe mapsν:V→Z,ν(g)={p∈Per(g): π(p)=n} andν_i:V→Z,ν_i(g)={p∈P_i(g): π(p)=n}are continuous and hence constants.

3. Singularities of self mappings

The first assertion in Theorem 2 will be a consequence of the following transversality result:

Theorem 4.Given a compact manifoldM, and anyn >0, defineGnas the set of mapsf∈C^∞(M)such that the two following conditions hold:

1. For everyj n,f^−j(Sf)is a stratified submanifold.

2. The sequence of stratified submanifolds{f^−j(S_f)0jn}is in general position.

ThenGnis open and dense for everyn0.

3.1. Preliminaries on critical sets and transversality

In this subsection we collect some definitions and standard results, in preparation to the proof of Theorem 4.

There exist different definitions of stratifications of sets. The more adequate to our purposes is the following.

Definition 1. Given a smooth manifoldM, a pair(N, Σ )is a stratified submanifold ofM ifΣ= {N₁, . . . , N_k}is a (finite) partition of a subsetN ofM, where eachN_i is a (not necessarily closed) submanifold ofM such that the following conditions hold:

1. The union

ijNi is a closed set for everyjk.

2. For everyj k,x∈Nj, and any vectorv tangent toNj atx, there exists a sequence(xn, vn)inT Nj−1that converges to(x, v)inT M.

Note that the second condition implies that the dimensions of the submanifolds N_i are decreasing. Each of the submanifoldsN_i∈Σis called a stratum. It is also common to say thatΣ is a stratification ofN.

The most relevant example is the setL^s(R^m)of singular linear endomorphisms ofR^m. IfL_j denotes the set of linear maps having kernel of dimensionj, then{L₁, . . . , L_m}is a stratification ofL^s(R^m). Note that the codimension ofL_j in them²-dimensional space of all linear self maps ofR^misj².

Definition 2.Givenf ∈C¹(M)and 1j m, letS_j(f )be the set of pointsx∈Msuch that the kernel ofDf_xhas dimensionj.

Theorem 5.Given any compact manifoldMthere exists an open dense setGofC^∞(M)such that for everyf ∈G the following condition holds:

eitherSf is empty or there exists a maximumk=k(f )such thatSj(f )is nonempty for everyjkand is empty for everyj > k.

IfS_f is not empty, so that the second alternative holds, then(S_f, Σ_f), whereΣ_f = {S₁(f ), . . . , S_k(f )}, is a stratified submanifold ofM.

There existsG0open and dense inGsuch that for everyf ∈G0no critical point off is fixed.

The proof of the first statement consists in using jets to formalize the idea that for an open and dense set f ∈ C^∞(R^m)it holds thatDf is transverse to the stratified submanifoldL^s(R^m)of the previous example. This proof can be found in Chapter VI of [1]. The proof of the last assertion is trivial, it will be used in the proof of the second assertion of Theorem 2.

There is another local property that will be used in the proof of Theorem 4. Given a mapf∈Gand a pointx∈S_f, there exist neighborhoodsUoff andUofxsuch that ifg∈UandS_g∩U⊂S_f, thenS_g∩U=S_f∩U. This follows from the genericity of critical points.

Definition 3.LetMbe a compact manifold.

1. Any set{N_i: i∈I}of submanifolds ofMis transverse (also called in general position) if for any finite subsetJ of I such thatN_J :=

{j∈J}Nj is nonempty, it holds thatN_J is a submanifold whose codimension is equal to the sum of the codimensions of theNj withj∈J. This is denoted by{Ni: i∈I}.

2. A set{(Ni, Σi): i∈I}of stratified submanifolds ofMis transverse if{Ni: i∈I}for each choice of submani- foldsN_i∈Σ_i. The notation is{Σ_i: i∈I}.

3. LetM₀be a manifold; a smooth mapf :M→M₀ is transverse to a submanifoldN ofM₀ifDf_x(T_x(M))+ T_{f (x)}(N )=T_{f (x)}(M₀)for everyxsuch thatf (x)∈N.

4. A mapf ∈C^∞(M)is transverse to a stratified submanifold(N, Σ )if it is transverse to eachNi∈Σ. The notation isf Σ.

IfV is an open subset of the manifoldMwe say thatf is transverse toΣ inV iff|V Σ.

Theorem 6.LetMandM0be manifolds, where onlyMis assumed to be compact.

Iff :M→M0is aC^∞map transverse to a stratified submanifold(N, Σ )ofM0, then f⁻¹(Σ ):=

f⁻¹(N_i): N_i∈Σ

is a stratification off⁻¹(N ). The codimensions ofN_iandf⁻¹(N_i)are equal.

The set of smooth mapsf such thatfΣ is open and dense inC^∞(M, M₀).

The last assertion of this statement deserves an explanation. Recall that the submanifolds belonging toΣ are not necessarily closed; so it is not trivial a priori that the set of mapsf such thatfNis open whenN∈Σ. However, if f N_k, wherekis maximum such that the intersection ofN_kwith the image off is not empty, then the last item in the definition of a stratified submanifold implies that everyC^∞perturbation off is transverse toN_jfor everyjk, and does not intersectNj for everyj > k.

This theorem is a consequence of the Thom Transversality Theorem, which is stated here in a mild version, suffi- cient for all our purposes.

Theorem 7.LetΛbe an open set in an Euclidean space,Mand N compact manifolds and assume that a smooth mapF :Λ×M→N is transverse to a compact submanifoldV ofN. Then the set ofλ∈Λsuch thatFλV has total(Lebesgue)measure inΛ, whereF_λ:M→N is the map defined byF_λ(x)=F (λ, x).

One also has continuous dependence of f⁻¹(N )onf. Indeed, a possible way of defining this continuity is as follows:

Theorem 8.Letf :M→M₀be a smooth map transverse to a stratified submanifold(N, Σ )ofM₀. Given a sub- manifoldN₀⊂Mtransverse tof⁻¹(Σ ), there exists aC^∞neighborhoodU off such thatg⁻¹(Σ )is transverse to N₀for everyg∈U.

3.2. Proof of Part 1 of Theorem 2

Until now we treat withC^∞maps, in preparation for the proof of Theorem 4. Now we give the proof of Part 1 of Theorem 2 and for this it is necessary to treat maps that are notC^∞.

For maps inC¹(M), the variation of the critical sets is not continuous; however, the following semicontinuity is easy to establish:

Givenf ∈C¹(M)and a neighborhoodUofS_f, there exists a neighborhoodV0off such thatS_g⊂Ufor every g∈V0.

Theorem 4 implies Part 1 of Theorem 2:

Givenn >0, letUnbe the set off ∈C^r(M)such that sup

z∈M

card

0jn: f^j(z)∈S_f

m,

wheremis the dimension ofM.

Proof thatUnis open inC^r(M).

Letf ∈Un; givenz∈Mthere exist neighborhoodsD_zofz,U_z ofS_f andVz⊂C^r(M)off such thatg^j(D_z)∩ U_z= ∅for everyg∈Vzand every 0jnsuch thatf^j(z) /∈S_f. CoverMwith a finite number of suchD_z, and let U andVbe the intersections of the correspondingU_zandVz. LetV be the intersection ofVwith the neighborhood V0associated to the neighborhoodUofS_f given by the semicontinuity ofS_f stated above. Then, for everyg∈Vand z∈M, it holds thatg^j(z)∈S_gfor at mostmiterates between 0 andn. This proves the claim.

Proof thatUnis dense inC^r(M).

Given anyC^r open setU, use the denseness ofC^∞(M)inC^r(M)and Theorem 4 to get a mapg∈Gn∩U. To prove thatg∈Un, letx∈Mand{j₁, . . . j_k}be the set of indexesj nsuch thatg^j(x)∈S_g. LetN_i be the stratum of g^−ji(Sg)that containsx. By Theorem 4 the set{N1, . . . , Nk}is transverse (in general position), and as it has nonempty intersection, it comes that:

m=dimMcodim

i

N_i =

i

codim(Ni)k.

This proves the density ofUn.

To conclude the proof of Part 1 of Theorem 2, letRr be equal to the intersection of theUnwithn1.

3.3. Perturbation techniques

The first result needed for the proof of Theorem 4, allows us to make local perturbations without changing the critical set:

Lemma 2.Letf :R^m→R^mbe aC^∞map in the open and dense setGobtained in Theorem5and let0∈S_f be a critical point. There exist arbitrary small numbersaandb,0< a < b, and aC^∞nonnegative functionϕequal to1 inB_a, equal to0outsideB_band whose gradient is orthogonal to the kernel ofDf at each critical point off inB_b. (Badenotes the ball centered at0and having radiusa.)

Proof. Assume thatf (0)=0 and defineN (x)= f (x)². There exists a neighborhoodB of 0 such thatf⁻¹(0)∩ B= {0}, so one can choose an arbitrary small positive numberbsuch thatm=min{N (x): x∈∂B_b}is positive, where

∂B denotes the boundary of the setB. Choose anym∈(0, m)and leta∈(0, b)be such thatN (x) < mfor every x< a. Now leth: [0,+∞)→Rbe aC^∞nonnegative function such thath(t )=0 iftm, andh(t )=1 iftm. Finally define the functionϕas being equal to 0 outsideBband such thatϕ(x)=h(N (x))inBb.

Note that ∇ϕ(x)=h(N (x))∇N (x) in Bb, and that∇N (x) is a linear combination of the rows of the matrix associated toDfx, hence orthogonal to the kernel ofDfx. 2

Definition 4.Givenf∈C^∞(M),x∈Mand a neighborhoodUoff, say that a triple(V , U, f1)is adequate tox, f,U if the following conditions hold:

1. BothUandV are open neighborhoods ofxsuch thatV⊂U. 2. f1is a map inU that coincides withf outsideU.

3. S_f =S_f1.

For the proof of the theorem, we will still need two preliminary results where some local perturbations are per- formed. The first one says that we can perturb in order to make transverse the preimage of transverse manifolds, without changing the preimage of one of them. This will be used in the induction step.

Proposition 1.Letf ∈G0(G0as in the last assertion of Theorem5),U aC^∞neighborhood off andx∈S_f such thatf (x)∈N₁∩N₂, andf N₂, whereN₁ andN₂ are transverse submanifolds. There exists an adequate triple (V , U, f₁)forx, f,U such thatf₁⁻¹(N₂)=f⁻¹(N₂)and:

1. f₁N₁inV.

2. f₁⁻¹(N1)f₁⁻¹(N2)inV.

Before beginning with the proof note that the transversality ofN1andN2does not imply the transversality of its preimages. It is easy to prove that if, in addition,f is transverse toN1∩N2, thenf⁻¹(N1)f⁻¹(N2).

Proof. Let(U, τ₁)be a local chart atx such thatf (U)⊂V, where(V, τ₂)is a local chart atf (x). Asx is not fixed forf one can chooseUsuch thatf (U)∩U= ∅. By taking adequate local charts one can also assume that τ₂(N_i)is a subspaceW_i fori=1,2. Of course,W₁W₂. Letf˜=τ₂⁻¹f τ₁:R^m→R^mand define, for the functionϕ of Lemma 2, and for eachw∈W₂, the maps

F :R^m×W₂→R^m byF (X, w)= ˜f (X)+ϕ(X)w and for eachw∈W2,f˜_w:R^m→R^mbyf˜_w(X)=F (X, w).

Note thatDF_(0,0)generatesW₂, from which it follows thatF W₁in a neighborhood of(0,0), sinceW₁andW₂ are transverse. By the parametrized transversality theorem it comes thatf˜wW1for an open set of valuesw∈W2. Therefore, it is no loss of generality to assume from the beginning thatf˜is transverse toW1.

Note also thatDF(0,0)is a submersion, because (by hypothesis)∂xF =Df˜xgenerates a subspace complementary toW₂and∂_wF generatesW₂.

Letf˜_w(X)=F (X, w)and choosewsmall so thatf˜_w is transverse toW₁∩W₂in a ballB_a. Observe that every critical point off˜is a critical point off˜_wbecause the gradient of∇ϕis orthogonal to the kernel ofDf˜. This implies by genericity thatS_f˜=S_f˜

w. Definef₁as the map that coincides withf outsideU=τ₁⁻¹(B_b)and is equal toτ₂⁻¹f˜_wτ₁ inV =τ₁⁻¹(B_a). Therefore, there exist vectorswof arbitrary small norm such that the triple(V , U, f₁)is adequate to x, f,U.

By the choice ofw∈W₂it comes that the preimages ofN₂underf₁andf₂are equal. On the other hand,f₁is transverse toN₁andf₁⁻¹(N₁)f₁⁻¹(N₂), by the remark preceding the proof of the proposition. 2

The second ingredient in the proof of the induction step of Theorem 4 is the following proposition, intended to show thatS_f can be made transverse to a newf⁻ⁿ(S_f)without changingf^−j(S_f)forjn−1.

Proposition 2.Suppose thatf is a map inC^∞(M),N₁andN₂are transverse submanifolds ofMandf N₁∩N₂. If N₀is another submanifold ofM,Uis a neighborhood off, andx is a point inMsuch thatx∈N₀∩f⁻¹(N₁∩N₂), then there exists a triple(V , U, f₁)adequate tox,f,U, such that:

1. N₀f₁⁻¹(N₁), and 2. f₁⁻¹(N2)=f⁻¹(N2).

Proof. As in the proof of the above proposition, one can assume that f acts in R^m and that each N_i is a linear space. The mapf₁will be obtained from a perturbation of the typef₁(x)=f (x)+w₂ϕ(x), forw₂∈W₂andϕas in Lemma 2. DefineL=Df_xand letV be a subspace whereLis injective and such thatL(V )⊕W₁=R^mthis last exists becauseLis transverse toW₁∩W₂. Letπ:W₂→LV be the projection which assigns to eachw₂its component in LV relative to the decompositionLV ⊕W₁=R^m. Letπ₀=(L|V )⁻¹◦π. Note thatπ₀:W₂→V is surjective since W₂+W₁=R^m. Note thatL⁻¹(W₁+w₂)=L⁻¹(W₁)+π₀(w₂)forw₂∈W₂. AsV ⊕L⁻¹(W₁)=R^m, it follows thatL⁻¹(W₁)+v is transverse toW₀for almost everyv∈V. Asπ₀ is surjective, it comes thatL⁻¹(W₁+w₂)is transverse toW₀for almost everyw₂∈W₂.

This is equivalent to say thatL⁻_w¹

2(W1)is transverse toW0, whereLw=L+w. But this must hold also forfw₂, that is arbitrarily close toLw₂. 2

3.4. Proof of Theorem 4

It will be proved that ifnis a given nonnegative integer, then:

Sf, f⁻¹(Sf), . . . , f⁻ⁿ(Sf)

for everyf in an open and dense subset ofC^∞(M). The proof will be done by induction.

The initial step requires thatS_f is a stratified submanifold; this is an immediate consequence of Theorem 5.

Induction step:Assume that for everyf in some open and dense setVnit holds that for each 0jn,f^−j(S_f)is a stratified submanifold and

S_f, f⁻¹(S_f), . . . , f⁻ⁿ(S_f) .

We have to prove that there exists an open and dense setVn+1such that forg∈Vn+1we have 1. g^−j(Sg)is a stratified submanifold for everyj n+1.

2. {Sg, g⁻¹(Sg), . . . , g⁻⁽ⁿ⁺¹⁾(Sg)}.

Notice that the set of mapsgsatisfying each one of (1) and (2) is open. So, to complete the induction step it suffices to show that in any nonempty open setU contained inVnthere exists a mapgsatisfying (1) and (2).

Let us prove first that f₁⁻⁽ⁿ⁺¹⁾(S_f1)is a stratified submanifold for somef₁∈U. For this, it is sufficient to find f₁∈Usuch thatf₁f₁⁻ⁿ(S_f1).

Letf ∈U. For everyx∈S_f∩fⁿ⁺¹(S_f), letJ_xbe the set of indexesj∈ {1, . . . , n}such thatx∈f^−j(S_f), and for everyj∈J_x, letM_jbe the stratum off^−j(S_f)containingx. Take a neighborhoodUofxsuch thatf (U )∩U= ∅and the following two conditions hold: first,U∩f^−k(S_f)= ∅for every 1knthat does not belong toJ_x, and second, that U∩M= ∅for every stratumMof f^−j(S_f)of dimension less than that ofM_j. DefineN₂=

j∈J_xf (M_j) and letN1be the stratum off⁻ⁿ(Sf)containingf (x). It follows thatf (x)∈N1∩N2. Note that by the hypothesis of the induction, N₁N₂ andf N₂. So Proposition 1 can be applied to obtain a triple (V , U, f₁)adequate to x, f,U such thatf₁N₁inV, whilef₁⁻¹(N₂)=f⁻¹(N₂). The point here is that the perturbation off is supported inU, and as the preimage of N₂ is not changed when the perturbation is made, thenN₁can be thought as a fixed manifold that does not depend on the map. It follows thatf₁f₁⁻ⁿ(S_f)inV, which provides the perturbation that makesf₁f₁⁻ⁿ(S_f)locally. The way these local perturbations are accomplished in order to obtain a mapf₁that is transverse tof₁⁻ⁿ(S_f1)is standard and can be done by means of partitions of unity, as in the proof of the Transversality Theorem, see references [2, Theorem 3.2], [3, Section 3.2] or [1, Theorem 4.9].

Once it is known that f₁⁻⁽ⁿ⁺¹⁾(S_f1)is a stratified submanifold, this property persists in a neighborhood of f₁ (contained inU). For eachx∈M, defineJ (x)as the set of indexes 0jn+1 such thatf₁^j(x)∈S_f1. Let alsoM_j be the stratum off₁^−j(S_f1)that containsx. If 0∈/J_x, then there is nothing to prove sincef₁is locally a diffeomorphism aroundx and the induction hypothesis implies that the images of these submanifolds form a transverse set. On the other hand, ifn+1∈/J_x, the induction hypothesis applies directly. For the other cases take a neighborhoodUofxas above.

DefineN₁=f₁(M_n+1), and

N₂=

j∈J_x\{0,n+1}

f₁(M_j)

By hypothesis,N₁andN₂are transverse submanifolds andf₁is transverse to their intersection. We will show that we can perturbf₁to obtaingso thatS_g is transverse tog⁻¹(N₁∩N₂). First show that we can perturb to obtainf₂ such thatS_f2 is transverse to the preimage ofN₁by applying Proposition 2. Indeed, letN₀=M₀be the stratum of S_f1that containsx; thenN₀can be considered as a fixed submanifold because the perturbations leaveS_f1 unchanged by Lemma 2, as well asN₁andN₂, that do not change by the choice of the perturbations. We will now perturb this map in order to prove that the intersection of the corresponding{M_j: j ∈J_x}is transverse at x. To do this apply Proposition 2 again, takingN₁andN₂as above but definingN₀=

j∈Jx\{n+1}M_j. As before, by means of partition of unity we obtaing∈Usatisfying (1) and (2). This finishes the proof of Theorem 4.

3.5. Proof of Part 2 of Theorem 2

Proof. Given integersnandk, wheren0 andk1, denote byP_kⁿ(f )the set of pointsx∈Msuch thatfⁿ(x)is a periodic point off of period at mostk. Given any fixedk >0, there exists an open and dense setKSk such that, for everyf ∈KSk, every periodic point off of period at mostkis hyperbolic. In particular, for everyf∈KSk, the following properties hold:P_k⁰(f )is a finite set, it varies continuously withf, and it does not intersect the critical set off. Now fix somer1 and define:

Aⁿ_k=KSk∩

f: P_kⁿ(f )∩Sf = ∅ .

Note thatAⁿ_k is open inC^r(M); indeed, given disjoint neighborhoodsU of P_kⁿ(f )andV of Sf, it is clear that for every perturbationgoff it holds thatP_kⁿ(g)⊂UandSg⊂V.

Fix some positivek. We will prove now by induction onn that everyAⁿ_k is dense inC^r(M). This implies the theorem.

To prove the claim, letf ∈C^r(M)andU a neighborhood off. We can begin takingf ∈KSk∩G as both are residual sets. The initial step of the induction is then consequence off ∈KSk. So assume thatf ∈Aⁿ_k⁻¹. Note that asf ∈G, the preimage of any point is a finite set, so it is sufficient to prove that ifx∈S_f ∩P_kⁿ(f )then there exists a perturbationgoff such thatP_kⁿ(g)∩S_g∩U= ∅for some neighborhoodUofx. So begin with anx∈S_f∩P_kⁿ(f );

there exists a neighborhoodUof x such thatU∩P_kⁿ⁻¹(f )= ∅and such thatx is the unique point ofS_f ∩P_kⁿ(f ) inU. AsP_kⁿ⁻¹(f )is a (zero-dimensional) submanifold, one can use again the Thom Transversality Theorem to find a mapg∈Usuch that the restriction ofgtoUis transverse toP_kⁿ⁻¹(f )andf =goutside the closure ofU. This last assertion implies thatP_kⁿ⁻¹(g)=P_kⁿ⁻¹(f )by the choice ofU. It follows thatg_|U is transverse toP_kⁿ⁻¹(g)which is equivalent to say that every point ofP_kⁿ⁻¹(g)is a regular value ofg, or thatP_kⁿ(g)∩Sg∩U= ∅. This provides the proof of the induction step. 2

Acknowledgements

We are grateful to Sylvain Croviser, Pablo Guarino and Andrés Sambarino for useful and enlightening discussions.

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