Multiplicativity of the Euler Characteristic, a Question Which Has Remained Open for Almost 35 Years [4]

Abstract

Covering maps and fibrations of spaces fulfilling certain technical conditions are known to satisfy a multiplicative formula relating the Euler characteristic of the domain to that of the codomain. An open question posed by Albrecht Dold in 1980 asks in general when this is true and can be stated as: for which classes of maps is it true that if χ(X) denotes the Euler characteristic of a space X, and if f : X → Y has the property that χ(f −1(y)) = k for all y ∈ Y and for some integer k, the multiplicative formula χ(X) = k · χ(Y ) holds? A corroborative answer is given herein for simplicial maps of finite simplicial complexes, while counterexamples are constructed for cellular maps of finite CW complexes, continuous maps of closed topological manifolds, and even smooth maps of smooth manifolds.

I want to express my deepest appreciation to everyone who has helped me with this thesis and supported me throughout my years at Tulane. Thank you, Slawomir Kwasik, for your patience, support, and wonderful advice on matters of acedemia, love, and life. It has been a pleasure working with you.

Thank you to everyone in the mathematics department at Tulane for all of your help and for making our department a second home to me. Special thanks to my committee members Albert Vitter, Maurice Dupre, Michael Mislove, and Rafal Komendarczyk. Also to Andreas Michaelides, Ellis Fenske, and Franz Hoﬀman, not only for their help on this project, but for their friendship during our time in New Orleans.

Lastly, thank you to Casey Tatangelo, my parents Douglas and Dana Johnson, and my family and friends for your love and support.

ii List of Figures

2.1 Homotopy Lifting Property Diagram [1]...... 7

4.1 Graph of g : I → I ...... 17

6.1 Graph of f : R → R ...... 29 6.2 Graph of α, β, and f ...... 30

iii Contents

Acknowledgement ii

List of Figures iii

1 Introduction1

2 Established Conditions for Multiplicativity5

3 Simplicial Maps 11

4 Cellular Maps 16

5 A Map of Topological Manifolds 24

6 Smooth Maps 26

References 35

iv 1

Chapter 1

Introduction

In 1758 Euler published a simple yet remarkable formula which states that for a convex polyhedron:

V − E + F = 2, (1.1) where V , E, and F are the number of polyhedral vertices, edges, and faces, respectively.

It was very surprising at the time that this very basic formula should apply to all convex polyhedra and that any triangulation may be used to find V, E, and F. Today, many topology students learn this formula as a special case of the Euler characteristic. The modern definition, given below extends the notion of the polyhedral formula to a much larger class of spaces, topological spaces of any dimension that are of bounded finite type (meaning a space whose homology groups are all finitely generated, with only finitely many being nonzero).

Definition 1.1. The Euler characteristic of a space of bounded finite type, X, is ∞ P i th χ(X) = (−1) Rank(Hi(X)), where Hi(X) is the i homology group of X. i=1 It is evident from this definition that the Euler characteristic depends only on the 2 homology of X. Spaces that are homeomorphic or homotopy equivalent have isomorphic homology groups, so homeomorphic spaces and homotopy equivalent spaces have the same Euler characteristic. Thus it is a topological and homotopical invariant [2].

Being able to encode topological data in a single integer via the Euler characteristic has proven very useful. There have been applications of Euler characteristic, not only in pure mathematics, such as in the proof of the Lefschetz ﬁxed point theorem, or in classifying surfaces, but also in applied math and science. For example, as stated in [3], the Euler characteristic has been a valuable tool for target ennumeration in sensor networks, and an “important parameter in the physics of porous materials.”

Fortunately, the Euler characteristic has many properties that make its computation for complicated spaces easier by using the Euler characteristics of simpler related spaces. For example, if the topological space X is the union of subspaces A and B, then χ(X) = χ(A) + χ(B) − χ(A ∩ B), provided that X, A, and B are of bounded ﬁnite type. Additionally, χ(A × B) = χ(A) × χ(B)[2]. If instead, X is the connected sum of closed manifolds A and B of dimension n, then χ(X) = χ(A) + χ(B) − χ(Sn).

In 1980, Albrecht Dold posed the following question about the multiplicativity of the Euler characteristic, a question which has remained open for almost 35 years [4].

Question 1.1. Motivated by the Vietoris mapping theorem and analogous results, we ask whether the following is true: “if f : X → Y is a continuous map between compact metric spaces such that χ(f −1(y)) = 1 for all y ∈ Y then χ(X) = χ(Y ).”

The question is of interest also for simpler spaces X,Y , say compact CW-spaces or manifolds. On the other hand, the question loses much of its interest if the continuous map f is further restricted (for example simplicial or ﬁbration). Answers for restricted 3 classes of maps will not be counted towards a solution!

A little more general, one can ask whether

χ(X) = k · χ(Y ) (1.2) if χ(f −1(y)) has the same value k ∈ Z for all y ∈ Y .

Here is a brief description of our results.

In chapter 2 we recall well known conditions which imply a positive answer to Dold’s question. These include coverings (ﬁnite) and ﬁbrations.

In chapter 3 we provide the following new result for simplicial maps of ﬁnite simplicial complexes:

Theorem 1.2. Let f : X → Y be a simplicial map of ﬁnite simplicial complexes such that χ(f −1(y)) = k ∀y ∈ Y . Then

χ(X) = k · χ(Y ). (1.3)

This gives a positive answer to Dold’s question for a class of maps which, despite being restricted, allows more variation among preimages and is closely related to cellular and smooth classes of maps. The argument depends heavily on the special properties of simplicial maps.

Chapter 4 gives a negative answer to Dold’s question even for cellular maps of CW- complexes. A counterexample to the proposed multiplicative formula is constructed 4 for each integer k, where k is the Euler characteristic of the preimage of any point under the speciﬁed cellular map. The results are then summarized by the following theorem:

Theorem 1.3. For any k ∈ Z, there is a cellular map f : X → Y of ﬁnite CW complexes X,Y such that χ(f −1(y)) = k ∀y ∈ Y and χ(X) 6= k · χ(Y ).

This is in sharp contrast with Theorem 1.2.

In Chapter 5 we present an example of a map of closed topological manifolds under which the preimage of each point has identical Euler characteristic equal to 3, but does not satisfy equation (1.2), thus giving a counterexample to the formula in Dold’s question for maps between members of a very nice class of spaces.

Lastly, we reﬁne one of our counterexamples for CW-complexes to obtain a smooth map between smooth manifolds, hence obtaining the following theorem, which is un- expected in view of the result for simplicial maps:

Theorem 1.4. There is a smooth map f : X → Y of smooth manifolds X,Y such that χ(f −1(y)) = k ∀y ∈ Y and χ(X) 6= k · χ(Y ). 5

Chapter 2

Established Conditions for Multiplicativity

An alternate formulation of the Euler characteristic deﬁnition is given for CW complexes by the subsequent theorem due to Euler and Poincare [1]:

Theorem 2.1. For a ﬁnite CW complex, X

∞ X i χ(X) = (−1) ci (2.1) i=1

where ci is the number of i-cells in X.

This characterization is often more convenient, since, at times, counting numbers of cells can be considerably easier than computing homology. Theorem 2.3 is a nice result about the multiplicativity of the Euler characteristic for covering maps that most ﬁrst year algebraic topology students are familiar with and whose proof (as outlined in [1]) makes use of this new Euler characteristic formula. First, recall the deﬁnition of a covering map:

Deﬁnition 2.2. Let X and Y be topological spaces. A map p : X → Y is a 6

covering map if for each point y ∈ Y there is an open neighborhood U of y in Y such that p−1(U) is a union of disjoint open sets in X, each of which is mapped homeomorphically onto U by p. If there are k such disjoint open sets in X, then X is called a k-sheeted covering of Y .

So a k-sheeted covering of a ﬁnite CW complex Y , can be thought of locally as a disjoint union of k homeomorphic copies of a neighborhood of any point in Y .

Theorem 2.3. If X is a k-sheeted covering of a ﬁnite CW complex, Y , then χ(X) = k · χ(Y )

Proof. It follows from the Lifting Lemma [5] that for each i-cell in Y , the characteristic

i map from the unit disc into Y , φα : D → Y , lifts to a map whose image is contained in a speciﬁed homeomorphic copy of that cell, contained in X. Since there are k copies to choose from, the characteristic map from the i-dimensional unit disc, Di, to Y lifts exactly k ways into X. Thus X inherits a CW complex structure with k i-cells for each single i-cell of Y (here the weak topology on X comes from the weak topology on Y , the continuity of characteristic and covering maps, and the

fact that each neighborhood of any chosen basepoint is evenly covered). Let ciY and ciX be the number of i-cells in Y and X, respectively. By the CW deﬁnition for Euler characteristic, we then have:

∞ X i χ(X) = (−1) ciX i=1 ∞ X i = (−1) (k · ciY ) i=1 ∞ ! X i = k (−1) ciY ) i=1 = k · χ(Y ) 7

A similar result about the multiplicativity of the Euler characteristic is known for ﬁbrations.

Deﬁnition 2.4. A map p : E → B is a ﬁbration if for any homotopy , ht, from any

space X to B, and any lifting of h0 there always exists a lifting of ht completing the commutative diagram in Figure 2.1 as indicated. The preimage of any point in B is then called a ﬁber.

he0 X × {0} E h et p

X × I B ht

Figure 2.1: Homotopy Lifting Property Diagram [1]

Theorem 2.5 ([6]). Let p : E → B be a fibration that is orientable over a field, k, with fiber, F , and with path connected base space, B. Assume χ(F ) and χ(B) are both defined (over the field). Then the Euler characteristic of E is defined and:

χ(E) = χ(F )χ(B). (2.2)

Thus the Euler characteristic of E is determined by the Euler characteristic of B and of each fiber, F . Note that since all fibers are homotopy equivalent by definition, they all have the same Euler characteristic. If we think about the geometry of this situation, the theorem seems plausible, but we dont have such a clear proof in this case. A more algebraic argument is needed, because there is not an immediate way to show how all the fibers piece together geometrically to form E and how this will affect the computation of the Euler characteristic. The proof of Theorem 2.5 presented here, which appears (more concisely) in [6], relies on a spectral sequence. 8

Deﬁnition 2.6. A spectral sequence consists of:

(i) a sequence of bigraded modules Er (each of which is an indexed collection of

modules Es,t for every pair of integers s, t)

r (ii) a collection for every E of homomorphisms from Es,t to Es−r,t+r−1 such that the the composition of any consecutive such homomorphisms is the 0 map (these homomorphisms, for each r, are collectively called a diﬀerential of bidegree (−r, r − 1))

(iii) some integer k such that if r ≥ k, there is an isomorphism from the bigraded

r+1 module deﬁned by Hs,t(E) = [ker(d : Es,t → Es−r,t+r−1)]/d(Es+r,t−r+1) to E .

Proof of Theorem 2.5. To every fibration satisfying the theorem’s conditions, one may assign a first-quadrant spectral sequence, which converges to a bigraded module, E∞, associated to a filtration on the homology of E (with coefficients in k), and whose E2 term is isomorphic to the homology of B with coefficients in the homology of F (see Theorem 9.2.17 of [6]). The module Ht(F, k) over k is torsion free, and it follows from the Universal Coefficients Theorem and KünnethTheorem for homology

2 2 that Hs(B,Ht(F, k)) ≈ Hs(B, k) ⊗ Ht(F, k). So Es,t ≈ Hs(B, k) ⊗ Ht(F, k), and E is finitely generated. For the remainder of the proof homology coefficients will be understood to be in k, and shall be omitted from notation. For our purposes, the significance of the spectral sequence being a first-quadrant spectral sequence is that

r ∞ for any given s, t if r is large enough, Es,t ≈ Es,t. Now, since we are working with a spectral sequence, we will need the following deﬁnition:

Definition 2.7. For a finitely generated bigraded module Er, define the Euler characteristic of Er as:

r X s+t r χ(E ) = (−1) dim(Es,t) s,t 9

Note that by this deﬁnition,

2 X s+t 2 χ(E ) = (−1) dim(Es,t) s,t

X s+t = (−1) dim(Hs(B, k) ⊗ Ht(F, k)) s,t

X s+t = (−1) dim(Hs(B, k)) · dim(Ht(F, k)) s,t

= χ(B)χ(F )

Also, since E2 is ﬁnitely generated, Er is for all r, as well. Taking into account the ﬁrst-quadrant properties of the spectral sequence, we then have Er = E∞, and sub- sequently χ(Er) = χ(E∞)) for r large enough.

Using the deﬁnition of H(Er), the First Isomorphism Theorem for Algebra, and properties of dimension, χ(Er) = χ(H(Er)) for all r (see theorem 4.3.14 of [6] for an outline of the proof). Now, since Er+1 = H(Er), we have χ(E2) = χ(H(E2) = χ(E3) = ··· = χ(Er). By the above, χ(E∞) = χ(E2) = χ(B)χ(F ).

∞ P ∞ ∞ By definition, dim(Hn(E )) = dim(Es,t). E is associated to a filtration of the s+t=n ∞ th homology of E, so for each s, Es,t is isomorphic to the quotient of the the s succes- sive pair of terms in the filtration. Since the dimension of a quotient of modules over a field is the difference of their dimensions, there is pairwise cancellation of consecutive

P ∞ terms in the sum of dimensions of such quotients corresponding to dim(Es,t), s+t=n P ∞ excluding the last term, dim(Hn(E)). Hence dim(Hn(E)) = dim(Es,t). s+t=n

Combining this with our deﬁnition of the Euler characteristic of a ﬁnitely gener-

P n P n ated bigraded module yields χ(E) = (−1) rank(Hn(E)) = (−1) dim(Hn(E)) = n n P n P ∞ P s+t ∞ ∞ (−1) dim(Es,t) = (−1) dim(Es,t) = χ(E ) n s+t=n s,t 10

Therefore, χ(E) = χ(B)χ(F ) 11

Chapter 3

Simplicial Maps

The category of simplicial complexes and simplicial maps provides a good starting point to explore the possible answers to Dold’s question. Although its maps are a restricted class of maps, the ability to control how preimages of points in the codomain intersect in the domain of a simplicial map allows more freedom in the “shape” of the preimages themselves. Unlike fibrations and k-sheeted coverings, preimages of points under simplicial maps need not be homotopy equivalent and may have differing homology groups. For reference, we now include the definition of a simplicial map.

Deﬁnition 3.1. A map f : X → Y of simplicial complexes is a simplicial map if f takes all vertices of X to vertices of Y and is linear on each simplex with respect to its vertices[1].

Also, there is another convenient reformulation of the Euler characteristic deﬁnition for simplicial complexes [7]:

Deﬁnition 3.2. If X is a ﬁnite simplicial complex of dimension n, the Euler characteristic of X, χ(X), is: n X i χ(X) = (−1) ci (3.1) i=0 where ci is the number of i-simplices in X. 12

Using this new deﬁnition in conjunction with the deﬁning properties of simplicial maps leads to the result below:

Theorem 1.2. Let f : X → Y be a simplicial map of ﬁnite simplicial complexes such that χ(f −1(y)) = k ∀y ∈ Y . Then

χ(X) = k · χ(Y ). (1.3)

Proof. We proceed by induction on n = dim(Y )

Base case: n = 0

If n = dim(Y ) = 0, then Y is a disjoint union of vertices. Let c0 be the number of

Fc0 such vertices. Then Y = i=1 vi and χ(Y ) = c0. Every ﬁnite simplicial complex is a ﬁnite disjoint union of path connected subcomplexes. Simplicial maps are continuous, hence preserve path-connectedness. Thus each path-connected component of X maps

−1 to exactly one vi ∈ Y , from which it follows that f (vi) is a ﬁnite disjoint union of path connected components of X for every i. Also, X is the disjoint union of the preimages of the vi’s: c0 G −1 X = f (vi) i=1

−1 −1 By assumption, χ(f (y)) = k ∀y ∈ Y , so χ(f (vi)) = k ∀i such that 1 ≤ i ≤ c0. Therefore, using the fact that the Euler characteristic of a disjoint union is the sum of the Euler characteristics, we have:

c0 c0 c0 G −1 X −1 X χ(X) = χ( f (vi)) = χ(f (vi)) = k = k · c0 = k · χ(Y ) i=1 i=1 i=1 13

Induction step: Assume true for dim(Y ) = n − 1.

n−1 Let cn be the number of n-simplices in Y , and let Y be the (n − 1)-skeleton of Y . Then by deﬁnition:

n−1 n χ(Y ) = χ(Y ) + (−1) cn

n−1 For each i ≥ 0, let bi be the number of i-simplices in X that f maps to Y \ Y . Then: m −1 n−1 X i χ(X) = χ(f (Y )) + (−1) bi i=n where m = dim(X).

Consequently, the induction assumption that χ(f −1(Y n−1)) = k · χ(Y n−1) implies:

m n−1 X i χ(X) = k · χ(Y ) + (−1) bi (3.2) i=n

Note that f, being simplicial, maps any i-simplex in X to a single simplex of Y , and

n−1 since any i-simplex among the bi does not map into any simplices in Y , it must map speciﬁcally to a single n-simplex.

cn n cn n−1 S n Let {σj }j=1 be the set of n-simplices of Y , let Y \ Y = σj , and let bij be the j=1 n number of i-simplices of X that f maps to σj .

c Pn Then bi = bij. Combining this with equation (3.2) gives: j=1

m cn cn m n−1 X i X n−1 X X i χ(X) = k · χ(Y ) + (−1) bij = k · χ(Y ) + (−1) bij (3.3) i=n j=1 j=1 i=n 14

Claim 3.1. Let f : X → Y be a simplicial map, and let y ∈ int(σn), where σn is any n-simplex of Y . Then f −1(y) is a cell complex with one (i − n) cell for each i-simplex in X which maps to σn and whose image is not contained in a sub-simplex of σn.

Proof of Claim. This follows by induction on n. First, if n = 0, then f −1(y) is obvi- ously a cell complex with one i-cell for each i-simplex which maps to σ0.

Assume the claim is true for each σn−1.

Let y ∈ int(σn). Then, using barycentric coordinates, one obtains that f −1(y) ∩ σn, for an n-dimensional simplex σn ∈ X, is homeomorphic to a product of simplexes

n n i−n {σk } in σ with the total dimension i − n and hence homeomorphic to D . More-

i−n n over, points on the boundary of D lie on the boundary of some σk , and thus on the boundary of σn, so the cells are properly attached (i.e. ∂(Di−n) is attached to the (i − n)-skeleton). Consequently, one gets a cell complex, as claimed.

n −1 So by Claim 3.1, for any yj ∈ int(σj ), f (yj) is a cell complex with one (i − n)-cell

n for each i-simplex mapping to σj . Thus:

m −1 X i−n χ(f (yj)) = (−1) bij i=n

m X i−n ⇒ k = (−1) bij i=n

m n X i ⇒ (−1) k = (−1) bij i=n This produces: cn m cn X X i X n n (−1) bij = (−1) k = k(−1) cn j=1 i=n j=1 15

Equation (3.3) now becomes:

n−1 n n−1 n χ(X) = k · χ(Y ) + k(−1) cn = k(χ(Y ) + (−1) cn) = k · χ(Y )

So χ(X) = k · χ(Y ) as desired. 16

Chapter 4

Cellular Maps

Having shown there is a multiplicative result for appropriate simplicial maps, we now turn our attention to cellular maps of finite CW-complexes, which were specifically of interest to Dold in regards to question 1.1. A cellular map f : X → Y of CW- complexes X and Y is required only to be continuous and to have the image of the n- skeleton of X, f(Xn), contained in the n-skeleton of Y . The simple conditions of this definition allow for a great amount of flexibility. Note that simplicial maps between simplicial complexes are cellular for the natural CW structures on the simplicial complexes. One thing a cellular map can do that a simplicial map cannot do is send an interior point to of an i-cell (simplex) of X to Y m, for some m < n, without sending all of the specified i-cell (simplex) to Y m. The examples constructed in the remainder of chapter 4 exploit this fact to prove:

Theorem 1.3. For any k ∈ Z, there is a cellular map f : X → Y of ﬁnite CW complexes X,Y such that χ(f −1(y)) = k ∀y ∈ Y and χ(X) 6= k · χ(Y ).

Proof:

We ﬁrst consider the case k = 3 and construct an interesting continuous map, g : I → I, where I is the unit interval, [0, 1]. 17

∞ Let the sequence {xi}i=1 be deﬁned as follows:

i x = i i + 1

∞ Let the sequence {yi}i=1 be deﬁned as follows:

  0 i = 0   − i+1 yi = 1 − e 2 i odd   − i−2  1 − e 2 i even

One can then deﬁne a continuous, piecewise linear function g : I → I whose graph appears to “connect the dots” formed by the points (xi, yi) with line segments. The graph of g can be seen below, and g shall be deﬁned as follows:

 yi+1−yi  yi + (x − xi) x ∈ [xi, xi+1] g(x) = xi+1−xi (4.1)  1 x = 1

Figure 4.1: Graph of g : I → I 18

It is clear from the graph of g that the preimage of any x ∈ (0, 1) is comprised of 3 points (observe that the bottom point of each “valley” in the graph is equal to the top point of the preceeding “peak”), that the preimage of x0 = 0 is comprised of two

1 points, and the preimage of x1 = 1 is a single point in [0, 1]. Let f : I → S be defined as f(x) = e2iπg(x). Then f is a map from the unit interval to S1 such that f(0) = f(1), so f −1(y) consists of three points ∀y ∈ S1. Hence the Euler characteristic of the preimage of any point in S1 under this map f is 3; χ(f −1(y)) = 3 for all y ∈ S1. Give I the CW-decomposition in which both endpoints are identified with two 0-cells, to which a single 1-cell is attached. View S1 ∈ C1 as having a CW-decomposition with one 0-cell identified with z = 1 to which a single 1-cell is attached. Then f(0) = f(1) = 1 implies f maps the 0-skeleton of I to the 0-skeleton of S1. It is clear that f also maps the 1-skeleton of I to the 1-skeleton of S1 and is consequently a cellular map. Moreover, f is a cellular map I → S1 for which χ(f −1(s)) = 3 for all s ∈ S1, and χ(I) = 1 6= χ(f −1(s)) · χ(S1) = 3 · 0 = 0. Thus equation (1.2) does not necessarily hold for cellular (hence also for continuous) maps.

For k > 3

Let g(x) be as above. To deﬁne a map f : I → S1, under which the preimage of each point is exactly k distinct points, one can allow f to send [0, 1/2] to S1 in the same manner as I was mapped to S1 in the previous example, and then wrap the

1 remainder of I,[ 2 , 0], around the circle k − 3 times. Then f is deﬁned accordingly as:

  e2πi·g(2x) x ∈ [0, 1 ] f(x) = 2 4(k−3)πi(x− 1 ) 1  2  e x ∈ [ 2 , 1]

1 Note that f( 2 ) is well deﬁned, and that f is therefore continuous. Furthermore, if I and S1 are given the aforementioned CW-decompositions, then f(0) = f(1) = 1 19

implies f maps the 0-skeleton of 1 to the 0-skeleton of S1. It is again clear that f also maps the 1-skeleton of I to the 1-skeleton of S1 and is consequently a cellular map.

Using the CW-complex formula for computation of the Euler characteristic, it is easy to see that χ(I) = 1 and χ(S1) = 0. Thus, ∀k > 3 we have constructed a cellular map f : I → S1 with f −1(s) made up of exactly k points (so χ(f −1(s)) = k) ∀s ∈ S1, and χ(I) = 1 6= χ(f −1(s)) · χ(S1) = k · 0 = 0.

For k ≤ 2

We cannot construct counterexamples using continuous maps from I to S1 under which the preimage of any point is exactly 2 or fewer distinct points. For k ≤ 0 this is because a preimage cannot consist of a negative number of points, and the preimage cannot be empty (i.e. contain zero points) for all points in Y . For k = 1, this is due to the fact that a continuous, bijective map from a compact space, X, to Hausdorﬀ space, Y , is necessarily a homeomorphism, hence induces isomorphisms on homology groups, resulting in the equality of Euler characteristics of X and Y . For k = 2, diﬃculty arises due to the ensuing lemma.

Instead, let X be the wedge sum of n copies of S1, for n = 4 − k. Let CXe be

the “cone with antenna” over X; CXe = X × I/ ∼, where (x0, t) ∼ (x1, t) ∀x0, x1 ∈ X

1 and ∀t ∈ I such that 2 ≤ t ≤ 1. We shall use the map g constructed above to create a new map sending a copy of X plus three distinct points to each point of S1. Deﬁne f : CXe → S1 as:   e4πit t ∈ [0, 1 ] f(x, t) = 2  2πi·g(2t−1) 1  e t ∈ [ 2 , 1]

4πi( 1 ) 2πi 0 2πi·g(0) 2πi·g(2( 1 )−1) Note that e 2 = e = e = e = e 2 , so f is continuous by the 20

gluing lemma. Additionally, f is cellular since it maps all 0-cells of CXe to the zero cell of S1, if CXe is given the CW-decomposition consisting of three 0 cells, (n + 2) 1-cells, and n 2-cells, and S1 is given the aforementioned decomposition. It is obvious that the 1-skeleton, and 2-skeleton of CXe are then mapped to the 1-skeleton and 2-skeleton of S1 (which both equal S1).

Also, for any point s ∈ S1, f −1(s) consists of a “slice” through CXe (which is a wedge sum of n copies of S1), and the disjoint union of three distinct points (since

2πi·g(2t−1) 1 e wraps [ 2 , 1] around the circle the same way I is wrapped around the circle in the case that k = 3). Therefore, ∀s ∈ S1,

_ G _ χ(f −1(s)) = χ(( S1) {3 points}) = χ( S1) + 3 = −(n − 1) + 3 = 4 − n = k n n

It follows from the cellular decomposition of CXe that χ(CXe ) = 1. As a result:

χ(CXe ) = 1 6= 0 = k · 0 = χ(f −1(s)) · χ(S1)

Lemma 4.1. There is no continuous map f : I → I such that the preimage under f of 0 is a single point, of 1 is a single point, and of all other values is comprised of two points.

Suppose such a map exists. Then the map, f : [0, 1] → [0, 1], cannot be strictly increasing on all of I = [0, 1] or strictly decreasing on all of I, else the preimage of each interior point would consist only of one point. Hence, f must change from increasing to decreasing or from decreasing to increasing at least once, so has a local max or 21

min, say at (x0, f(x0)).

Now, suppose the function has another local max or min after (x0, f(x0)) (with no others in between), call the point where this occurrs (x1, f(x1)). Without loss of

generality, let f be ﬁrst increasing (making (x0, f(x0)) a local max and (x1, f(x1)) a

local min). Let y ∈ (0, 1) be any number such that f(x0) > y > max(f(0), f(x1)).

Then since f is continuous on [0, x0] and [x0, x1], there is at least one x value in each

interval, call them xa and xb, so that f(xa) = f(xb) = y, by the intermediate value theorem. Thus the inverse image of y already must contain two points.

Also, if f(x0) < 1, there must exist some value xM ∈ I such that f(xM ) = 1 (or

−1 else f (1) = ∅). This value, xM , must occurr in the interval (x1, 1] due to the fact

that f increases to f(x0) (which is less than 1) on [0, x0] and decreases on [x0, x1] (so

f(x) < 1 ∀x ∈ [0, x1]). Then f(x1) < y < f(xM ) = 1, and by the intermediate value

theorem, again, we have a value xc in [x1, xM ] with f(xc) = y. So the preimage of y contains at least 3 points. This is a contradiction.

If f(x0) = 1, then we consider the following three cases:

i) f(0) < f(1)

Then there is some value y with max(f(0), f(x1) < y

[x1, 1] (note: since f(x1) is a local minimum, it cannot be an absolute maximum

on [x1, 1]). Then by the intermediate value theorem and the following inequalities,

f(0) < y < 1 = f(x0), f(x1) < y < 1 = f(0), and f(x1) < y

f on [x1, 1], there exists one value in each of the intervals (0, x0), (x0, x1), and (x1, 1), call them x, x0, and x00 such that f(x) = f(x0) = f(x00) = y. Thus the preimage of y contains three distinct points. Again, we have a contradiction.

−1 ii)f(0) = f(1) First note that since f (1) is a single point and f(x0) = 1, then 22 f(0) 6= 1 6= f(1). Similarly f(0) 6= 0 6= f(1). Let y = f(0) = f(1). Also, y < f(x0) = 1, and f must attain the value of 0 at some other value, call it x2.

Then we have f(x2) = 0 < y < f(x0) = 1, and by the intermediate value theorem, there is some other number, call it x0 for which f(x0) = y = f(0) = f(1). So the preimage of y contains three points, which is still a contradiction. iii) f(0) > f(1)

Again, f(x1) cannot be the absolute maximum value of f(x) on [x1, 1] because it is a local minimum. So the absolute maximum value of f(x) on [x1, 1] must either be f(1) or occur at a value xM ∈ (x1, 1).

If the absolute maximum value of f(x) on [x1, 1] is f(1), then f(x) ≤ f(1) ∀x ∈ [x1, 1].

−1 Let y be any value such that f(1) < y < f(0). Then f (y) ∩ [x1, 1] = ∅. Also f

−1 increases from f(0) to 1 on [0, x0], so f (y) ∩ [0, x0] = ∅. Since f is strictly decreasing on [x0, x1], there can only be one value in [x0, x1] for which f(x) = y. Since

I = [0, x0] ∪ [x0, x1] ∪ [x1, 1] the preimage of y contains at most one value in I. This results in a contradiction.

If the absolute maximum value of f(x) on [x1, 1] is f(xM ), then let y be any value such that max{f(x1), f(1)} < y < f(xM ). Using the inequalities f(x1) < y < f(x0) = 1, f(x1) < y < f(xM ), and f(xm) < y < f(1) and the intermediate value theorem on the intervals [x0, x1], [x1, xM ], and [xM , 1], respectively, there must be at least three distinct points in the preimage of y. We have a contradiction, once more.

The local max or min is also an absolute max or min on [0, 1] since f is strictly increasing to its left and strictly decreasing to its right (or vice versa). So it must be either 0 or 1, else any values below (above) an absolute min (max) would have 23

empty preimage. Withouot loss of generality, let the extrema be an absolute max, hence equal to 1. Then for all values in (0, 1) to have two points in their preimage, the graph must start at 0, increase to 1, then decrease all the way down to 0 again. This means that there are two points in the preimage of 0. Contradiction!

Chapter 5

A Map of Topological Manifolds

In light of the results of chapter 4, it is clear that a multiplicative formula for conti- inuous maps f : X → Y of topological manifolds X,Y with f −1(y) = k for all y ∈ Y does not hold for all examples (take I and S1 to be manifolds with the usual atlas on each, then the cellular map f : I → S1 deﬁned in the previous chapter is also a continuous map of manifolds). However, we include the example below to show that there are counterexamples to this type of multiplicativity even when both spaces are without boundary.

2 Let X be the closed manifold X = (S2 ×S2)#CP and let Y = S4. We now construct φ : X → Y as follows:

4 4 Let f : I → I be deﬁned as in the previous example, and let f4 : I → I be

the product of maps f × IdI × IdI × IdI . f4 is a product of continuous maps, hence a continuous map of topological manifolds. Note that since f is 3-to-1 on the interior

4 of I, f4 is 3-to-1 on the interior of I . Since f maps 0 to 0 and 1 to 1, as does IdI ,

4 4 −1 1 1 3 ∂I is contained in the preimage of ∂I . Recall that f (0) = {0, 2 }, so { 2 } × I is in the preimage of ∂I4, as well. 25

4 4 4 4 Let g : I → S be the continuous map contracting ∂I to a single point, y0 ∈ S . Also, let h : X \ int(I4) → S4 be the constant map sending all of h : X \ int(I4) to

4 y0. Then by the gluing lemma [8], the map φ : X → S such that φ|X\int(I4) = h and φ|I = g ◦ f4 is a continuous map of closed manifolds. The preimage of any point

4 y ∈ S such that y 6= y0 is then a union of 3 points, hence χ(y) = 3.

−1 4 3 Let Z = φ (y0) = [X \ int(I )] ∪ int(I ). To compute the Euler characteristic of Z we use a Mayer-Vietoris Sequence, where we view Z as the union of A ≈

2 [(S2 × S2) \ int(I4) ∪ int(I3)] \ D4 and B ≈ CP \ D4.

The non trivial homology groups of A (also computed by using a Mayer-Vietoris

4 2 2 3 sequence and noting that A ∪ D is homeomorphic to S ∨ S ∨ S ) are H0(A) =

Z,H2(A) = Z ⊕ Z, and H3(A) = Z ⊕ Z. It is also worth noting that B is homeomorphic to S2.

Using this information in the Mayer-Vietoris sequence for X shows that the only non-

trivial homology groups of X are H0(X) = Z,H2(X) = Z ⊕ Z ⊕ Z, and H3(X) = Z. So χ(X) = 1 + 3 − 1 = 3. Therefore φ−1(y) = 3 for all y ∈ S4. Now, using the properties of Euler characterstic stated in the introduction: χ(X) = χ(S2 × S2) +

2 2 χ(CP ) − χ(S4) = χ(S2) · χ(S2) + χ(CP ) − χ(S4) = 4 + 3 − 2 = 5. It is now clear that 5 = χ(X) 6= k · χ(Y ) = 3 · 2 = 6. 26

Chapter 6

Smooth Maps

Lastly, we come to the smooth category. Since any smooth map between smooth manifolds of ﬁnite diemnsion can be approximated by a simplicial map (as a conse- quence of all smooth manifolds having a trianglation and the simplicial approximation theorem [9]) , and there is additional structure provided by the smooth charts, one would think that a product formula would hold for smooth maps between smooth manifolds having the same Euler characteristic for the preimage of each point. On the contrary, we are able to use bump functions to reﬁne a previous construction and show that:

Theorem 1.4. There is a smooth map f : X → Y of smooth manifolds X,Y such that χ(f −1(y)) = k ∀y ∈ Y and χ(X) 6= k · χ(Y ).

Proof. Consider the unit interval I ⊂ R (with the standard smooth atlas) as a smooth manifold with boundary. It is then evident that to prove this theorem one need only to concoct a “smooth version” of the function connecting the dots deﬁned in the preceeding section.

Claim 6.1. There is a smooth function f : I → I, such that: 27

1. f −1(1) = 1

2. f −1(0) consists of two points

3. f −1(x) consists of three points ∀x ∈ (0, 1)

Construction of f:

∞ Let {xi}i=1 be deﬁned as follows:

i x = i i + 1

∞ Let {yi}i=1 be deﬁned as follows:

  0 i = 0   − i+1 yi = 1 − e 2 i odd   − i−2  1 − e 2 i even

Note that lim xi = 1, and lim yi = 1, as 1 is the limit of both its odd and even i→∞ i→∞ sub-sequences.

Now, let h : R → R be the smooth bump function:

 − 1  e x x > 0 h(x) =  0 x ≤ 0 28

and g : R → R be the smooth cutoﬀ function (as in [8]) deﬁned as:

h(x) g(x) = h(x) + h(1 − x) for all x ∈ R. Then g has the following properties:

i. g is smooth on [0, 1]

ii. g(0) = 0 and g(1) = 1

iii. g0(x) ≥ 0 and g(n)(0) = g(n)(1) = 0 for every positive integer, n

Note that if 0 and 1 are replaced with xi and xi+1, respectively, in properties i), ii), and iii) above, then the same properties hold for g( x−xi ) for all i (since we are xi+1−xi just composing g with a linear function), except property ii) would be changed to

x−xi x−xi g( )(xi) = 0 and g( )(xi+1) = 1. xi+1−xi xi+1−xi

x−xi Let fi : → be the function fi(x) = yi + (yi+1 − yi)g( ) for all x ∈ R R xi+1−xi R

x−xi Observe that fi(x) is a composition of a linear function with g( ) and con- xi+1−xi sequently has the following properties:

i. fi is smooth on R, and particularly on [xi, xi+1]

ii. fi(xi) = yi and fi(xi+1) = yi+1

(n) (n) yi+1−yi (n) x−xi iii. For n ≥ 1, f (x) exists and f (x) = n g i i (xi+1−xi) xi+1−xi

Finally, let f : R → R be deﬁned as

  0 x ≤ 0   f(x) = fi(x) x ∈ [xi, xi+1]    1 x ≥ 1 29 for all x ∈ X.

Figure 6.1: Graph of f : R → R

Properties i) and ii) of the fi together imply that f is continuous on R, except possibly at x = 1. In the same fashion, property (iii) of g and property (iii) of fi together im-

(n) (n) ply that fi (xi) = fi−1(xi) = 0 for each i. This new fact, combined with property (i)

(n) of each fi, shows that f (x) exists and is continuous on R, except possibly at x = 1 .

We now show by induction that f (n)(x) exists at and is continuous at x = 1 for all n ≥ 0. n = 0:

Clearly lim f(x) = 1. x→1+ 30

To compute the limit as f approaches 1 from the left, we shall employ the squeeze

theorem. Let α : R → R and β : R → R be deﬁned for all x ∈ R as follows:

− 3x−2 α(x) = 1 − e 4−4x

− 1 β(x) = e 4−4x + 1

Straightforward computation shows that lim α(x) = 1 = lim β(x). x→1− x→1−

3 It is clear from Figure 6.2 that near x = 1, speciﬁcally on the interval [ 4 , 1), the following inequality holds for all x.

α(x) ≤ f(x) ≤ β(x)

Figure 6.2: Graph of α, β, and f

So by the Squeeze Theorem: lim f(x) = 1 (6.1) x→1−

Thus, by the deﬁnition of f, equation (6.1), and equation (6.2), f is continuous because lim f(x) = 1 = f(1). x→1 31

Assume f (n) exists and is continuous at x = 1 for a general n ∈ N.

(n) (n) (n) (n) By deﬁnition f (n+1)(1) = lim f (x)−f (1) = lim f (x)−0 = lim f (x) . Here the x→1 x−1 x→1 x−1 x→1 x−1 second equality follows from our assumption that f (n) is continuous at x = 1. From the deﬁnition of f, one can see that the limit as x goes to 1 from the right of f (n)(x) must be 0, and so must f (n)(1). And since f (n) is continuous

! f (n)(x) f (n)(x) lim = lim lim x→1− i→∞ − x − 1 x→xi x − 1 ! f (n) (x) = lim lim i−1 i→∞ − x→xi x − 1

! ! (n) (n) fi−1(x) fi−1(x) To compute this limit, we’ll show that lim lim = 0, hence lim lim = i→∞ − x−1 i→∞ − x−1 x→xi x→xi 0.

In the computations below, the ﬁrst inequality is due to the fact that the distance from x to 1 must be less then than or equal to the length of the interval [xi−1, xi] containing x. 32

! (n) x−xi−1 ! f (n) (x) y − y g ( ) i−1 i i−1 xi−xi−1 lim lim = lim lim n i→∞ x→x− x − 1 i→∞ x→x− (x − x ) x − 1 i i i i−1 (n) x−xi−1 ! y − y g ( ) i i−1 xi−xi−1 ≤ lim lim n i→∞ x→x− (x − x ) x − x i i i−1 i i−1 ! |y − y | x − x = lim lim i i−1 g(n)( i−1 ) i→∞ − n+1 x→xi (xi − xi−1) xi − xi−1 ! |y − y | x − x = lim i i−1 lim g(n)( i−1 ) i→∞ n+1 − (xi − xi−1) x→xi xi − xi−1 " !# |y − y | x − x = lim i i−1 · lim lim g(n)( i−1 ) i→∞ n+1 i→∞ − (xi − xi−1) x→xi xi − xi−1 " !# |y − y | x − x = lim i i−1 · lim lim g(n)( i−1 ) i→∞ n+1 i→∞ − (xi − xi−1) x→xi xi − xi−1 " # " !# |y − y | x − x = lim i i−1 · lim lim g(n)( i−1 ) i→∞ i i−1 n+1 i→∞ − x − x ( i+1 − i ) x→xi i i−1 " # " !# |y − y | x − x = lim i i−1 · lim lim g(n)( i−1 ) i→∞ 1 n+1 i→∞ − x − x ( i(i+1) ) x→xi i i−1 " !# h n+1i (n) x − xi−1 = lim |yi − yi−1|(i(i + 1)) · lim lim g ( ) i→∞ i→∞ − x→xi xi − xi−1 " !# h 2(n+1)i (n) x − xi−1 ≤ lim |yi − yi−1|i · lim lim g ( ) i→∞ i→∞ − x→xi xi − xi−1

If i is even, say i = 2l for some l ∈ N, then:

2(n+1) − 2l−2 − 2l−1+1 2n+2 |yi − yi−1|(2l) = |(1 − e 2 ) − (1 − e 2 )|(2l)

1 − e 2n+2 = (2l) el (2l)2n+2(1 − e) = el

The limit of this even subsequence, obtained by repeated application of l’Hospital’s 33

(2x)2n+2(1−e) rule to the limit lim x is: x→∞ e

(2l)2n+2(1 − e) lim = 0 l→∞ el

2(n+1) A similar argument shows that the odd subsequence of |yi − yi−1|i converges to

2(n+1) 0, too. The limit of the sequence |yi − yi−1|i is therefore necessarily 0. It can ﬁnally be seen that:

(n) ! " !# fi−1(x) h 2(n+1)i (n) x − xi−1 lim lim ≤ lim |yi − yi−1|i · lim lim g ( ) = 0 · 0 = 0 i→∞ x→x− x − 1 i→∞ i→∞ x→x− x − x i i i i−1

(n) This implies that lim f (x) = 0, and from the deﬁnition of f it is easily seen that x→1− x−1 (n) lim f (x) = lim 0 = lim 0 = 0. Therefore, f (n+1)(1) = 0. x→1+ x−1 x→1+ x−1 x→1+

Additionally, f (n+1)(x) is continuous at x = 1 because f (n+1)(1) = 0, lim f (n+1)(x) = x→1+ lim 0 = 0, and x→1+

! (n+1) (n+1) lim f (x) = lim lim fi−1 (x) x→1− i→∞ − x→xi ! y − y x − x = lim lim i i−1 g(n+1)( i−1 ) i→∞ − n+1 x→xi (xi − xi−1) xi − xi−1 " !# |y − y | x − x = lim i i−1 · lim lim g(n+1)( i−1 ) i→∞ n+1 i→∞ − (xi − xi−1) x→xi xi − xi−1

= 0 · 0

= 0

This shows that f is smooth on R, so is certainly smooth when restricted to I, as well. As in the previous example (for k = 3), one can see from the graph of f that the preimage of any point in (0, 1) will be comprised of three points, the preimage of x = 0 is two discrete points, and the preimage of x = 1 is a single point. Thus the 34 claim is justiﬁed.

Now let π : I → S1 be the smooth map deﬁned as π(x) = e2iπf(x). Recall that this map identiﬁes the endpoints of I. Then the composition, F = π ◦ f, is a smooth map from I to S1 under which the preimage of any point in S1 is made up of 3 discrete points. It follows that χ(I) = 1 6= χ(F −1(y))(S1) = 3 × 0 = 0.

Together, theorems 1.2 and 1.3 produce the ﬁnal corollary:

Corollary 6.1. If f : X → Y is a smooth map of smooth manifolds X,Y where χ(y) = k for all y ∈ Y and χ(X) 6= k · χ(Y ), then for all triangulations of X and Y , f is not simplicial. 35

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Biography

The author was born in Augusta, Maine in 1985 and graduated from Northeastern University with a BS in Mathematics in 2008. The author started the PhD program at the Tulane University mathematics department in August 2009, eventually completing the program in August 2015.