Solution: We can cut the 2-simplex in two, perform the identification and then stitch it back up. The best way to see this is with the picture:

Size: px
Start display at page:

Download "Solution: We can cut the 2-simplex in two, perform the identification and then stitch it back up. The best way to see this is with the picture:"

Transcription

1 Samuel Lee Algebraic Topology Homework #6 May 11, 2016 Problem 1: ( 2.1: #1). What familiar space is the quotient -complex of a 2-simplex [v 0, v 1, v 2 ] obtained by identifying the edges [v 0, v 1 ] and [v 1, v 2 ], preserving the ordering of vertices? Solution: We can cut the 2-simplex in two, perform the identification and then stitch it back up. The best way to see this is with the picture: It s a mobius band! Problem 2: ( 2.1: #4). Compute the simplicial homology groups of the triangular parachute obtained from 2 by identifying its three vertices to a single point. Notation: For these problems, S denotes the free abelian group generated by elements of S. Solution: Denote A = [v 0, v 1, v 2 ], a = [v 0, v 1 ], b = [v 1, v 2 ], c = v 0, v 2 ] and then identify v 0 v 1 v 2 =: v. We have the chain complex C 2 (X) 2 C 1 (X) 1 C 0 (X) 0 0 where C 2 (X) = A, C 1 (X) = a, b, c and C 0 (X) = v. Then from 2 (A) = a + b c and 1 (a) = 1 (b) = 1 (c) = v v = 0

2 we can deduce that ker 2 = {0}, Im 2 = a + b c, ker 1 = a, b, c = C 1 (X), Im 1 = {0} and ker 0 = C 0 (X). The homology groups are H 2 (X) = ker 2 = {0} H 1 (X) = ker 1 Im 2 = a, b, c a + b c = Z Z H 0 (X) = ker 0 Im 1 = v = Z. Problem 3: ( 2.1: #5). Compute the simplicial homology groups of the Klein bottle using the -complex structure: Solution: Similar to the previous problem, we have the chain complex C 2 (X) 2 C 1 (X) 1 C 0 (X) 0 0 but this time with C 2 (X) = U, L, C 1 (X) = a, b, c and C 0 (X) = v. We have and so 2 (U) = a + b c 2 (L) = a b + c 1 (a) = 1 (b) = 1 (c) = v v = 0 ker 2 = {0} ker 1 = a, b, c Im 2 = 2a, a + b c Im 1 = {0} The homology groups are then given by H 2 (X) = ker 2 = {0} H 1 (X) = ker 1 Im 2 = a, b, c 2a, a + b c H 0 (X) = ker 0 Im 1 = v = Z.

3 Problem 4: ( 2.1: #7). Find a way of identifying pairs of faces of 3 to produce a -complex structure on S 3 having a single 3-simplex, and compute the simplicial homology groups of this -complex. Solution: Consider the -complex X obtained from [v 0, v 1, v 2, v 3 ] by identifying [v 1, v 2, v 3 ] [v 0, v 2, v 3 ] and [v 0, v 1, v 3 ] [v 0, v 1, v 2 ] by folding along [v 2, v 3 ] and [v 0, v 1 ], respectively. The orientations are given for the first identification via and the orientations for the second by With this gluing we have the identifications [v 0, v 3 ] [v 1, v 3 ] and [v 0, v 2 ] [v 1, v 2 ] [v 0, v 3 ] [v 0, v 2 ] and [v 1, v 3 ] [v 1, v 2 ]. [v 0, v 2, v 3 ] [v 1, v 2, v 3 ] and [v 0, v 1, v 3 ] [v 0, v 1, v 2 ] (1) [v 0, v 3 ] [v 1, v 3 ] [v 1, v 2 ] [v 0, v 2 ] [v 0 ] [v 1 ] and [v 2 ] [v 3 ] and will compute the homology groups from via the chain where 0 C 3 (X) 3 C 2 (X) 2 C 1 (X) 1 C 0 (X) 0 0 (2) C 3 (X) = [v 0, v 1, v 2, v 3 ] C 2 (X) = [v 0, v 1, v 3 ] [v 0, v 2, v 3 ] C 1 (X) = [v 0, v 3 ] [v 2, v 3 ] [v 0, v 1 ] C 0 (X) = [v 0 ] [v 2 ]. First, we should substantiate the claim that X is indeed S 3. To see this we consider the barycenter v of 3 and then decompose it into smaller 3-simplices [v, v 0, v 1, v 2 ], [v, v 0, v 2, v 3 ], [v, v 1, v 2, v 3 ] and [v, v 0, v 1, v 3 ]:

4 Performing the identification (1) glues the back simplex [v, v 1, v 2, v 3 ] to the left simplex [v, v 0, v 2, v 3 ] via [v 0, v 2, v 3 ] [v 1, v 2, v 3 ] and similarly the right simplex [v, v 0, v 1, v 3 ] to the bottom simplex [v, v 0, v 1, v 2 ] with the identification [v 0, v 1, v 3 ] [v 0, v 1, v 2 ]: Looking at the left of these objects, there are two faces [v, v 2, v 3 ] which in the original diagram were the one 2-simplex between [v, v 1, v 2, v 3 ] and [v, v 0, v 2, v 3 ]. Topologically the union of these forms a disc, which when contracted to a point, say v, leaves us with at solid ball D 3. By the same process, the right object can also be seen to be a D 3, and it is not hard to see that with all of the identifications made, these two D 3 s are glued along their boundaries. Hence the resulting space is X S 3. Now we compute the homology for X. Recalling the identifications (1) the values for the

5 boundary maps are 3 ([v 0, v 1, v 2, v 3 ]) = [v 1, v 2, v 3 ] [v 0, v 2, v 3 ] + [v 0, v 1, v 3 ] [v 0, v 1, v 2 ] = 0 2 ([v 0, v 1, v 3 ]) = [v 1, v 3 ] [v 0, v 3 ] + [v 0, v 1 ] = [v 0, v 1 ] 2 ([v 0, v 1, v 2 ]) = [v 1, v 2 ] [v 0, v 2 ] + [v 0, v 1 ] = [v 0, v 1 ] 1 ([v 0, v 3 ]) = [v 3 ] [v 0 ] = [v 2 ] [v 0 ] 1 ([v 2, v 3 ]) = [v 3 ] [v 2 ] = 0 1 ([v 0, v 1 ]) = [v 1 ] [v 0 ] = 0 0 ([v k ]) = 0 Now we can easily compute the homology groups for X: H 3 (X) = ker 3 = [v 0, v 1, v 2, v 3 ] = Z H 2 (X) = ker 2 Im 3 = {0} {0} = {0} H 1 (X) = ker 1 Im 2 = [v 0, v 1 ] [v 2, v 3 ] [v0, v 1 ] [v 2, v 3 ] = {0} H 0 (X) = ker 0 Im 1 = [v 0] [v 2 ] [v2 ] [v 0 ] = Z This agrees with the previously derived fact that H i (S n ) = Z exactly when i {0, n}. Problem 5: ( 2.1: #11). Show that if A is a retract of X then the map H n (A) H n (X) induced by the inclusion A X is injective. Proof. Let r : X A be a retract and i : A X be inclusion. Then r i = 1 and so the induced map of the composition is r i = 1 which implies that i : H n (A) H n (X) must be injective. Problem 6: ( 2.1: #15). For an exact sequence A e B f C g D h E (3) show that C = 0 iff the map A B is surjective and D E is injective. Hence for a pair of spaces (X, A), the inclusion A X induces isomorhpism on all homology groups iff G n (X, A) = 0 for all n. Proof. Denote the maps by e, f, g and h, respectively. If C = 0, then Im e = ker f = B so e is surjective. Also, for g to be a homomorphism g(0) = 0 and so the exactness of the sequence implies that ker h = Im g = {0} which in turn implies that h is injective.

6 Suppose that e is surjective and h is injective. So Im e = B and ker h = 0. Then by exactness ker f = Im e = B and Im g = ker h = 0. The first of these says that the image of f is 0 latter of these says that the ker g = C. By exactness, 0 = Im f = ker g = C. Problem 7: ( 2.1: #17). (a) Compute the homology groups H n (X, A) when X is S 2 or S 1 S 1 and A is a finite set of points in X. Proof. To compute the relative homology groups H n (X, A), we examine the long exact sequence... H 2 (A) i H 2 (X) j H 2 (X, A) H 1 (A) i H 1 (X) j H1 (X, A) H 0 (A) i H 0 (X) j H 0 (X, A) 0. (4) To figure out the H n (X, A), we want to fill in as much of the sequence as possible and then exploit its exactness. First we will do the case where X = S 1. As established in class, H i (S n ) = Z when i = n and is the trivial group otherwise, in this case H 2 (X) = Z. We also have that H 0 (Y ) for a space Y is just r k=1 Z where r is the number of path components of Y. Hence H 0(A) = A k=1 Z and H n (A) = 0 otherwise due to proposition 2.8. Using this information we can form a short exact sequence in the first half of (4): H 2 (A) i H 2 (X) j H 2 (X, A) H 1 (A) i 0 i j Z H2 (X, A) 0 i. (5) The fact that Im (Z) = ker = H 2 (X, A) implies that H 2 (X, A) = Z. To find H 1 (X, A) we look at the reduced homology groups of the latter half of (4). The reduced homology for a space Y is isomorphic to the regular homology for n 0 in which case H 0 (Y ) = H 0 (Y ) Z. So H 0 (A) = A 1 k=1 Z. This also means that H 1 (X) = H 0 (X) = 0. The end of the exact sequence is i H 1 (X) j H 1 (X, A) H 0 (A) i H 0 (X) j H 0 (X, A) 0 i 0 j H1 (X, A) Z A 1 i j 0 H0 (X, A) 0 Again, the short exact sequence between H 1 (X) and H 0 (X) implies that the two middle groups are isomorphic, namely that H 1 (X, A) = H 1 (X, A) = Z A 1. Lastly, each other H i (X, A) will be between two trivial groups in the long exact sequence. Thus the 0 = Im j = ker = H n (X, A) for n 1, 2. In summary: H n (S 2, A) = Z A 1 n = 1 Z n = 2 0 Otherwise. Now let X = T = S 1 S 1 and again we will use the sequence (4). To compute H 2 (T, A), it happens that (5) still holds and by the exact same reasoning, so H 2 (T, A) = Z. We are left with

7 the remainder of the sequence H 1 (A) i H 1 (T ) j H 1 (T, A) T 0 (A) i H 0 (T ) j H 0 (T, A) 0. By previous remarks, H 1 (A) = 0 and H 0 (A) = Z A and we also know (from Hatcher) that H 1 (T ) = Z Z. Since T is path connected, H 0 (T ) = Z. So 0 i Z Z j H 1 (T, A) Z A i j Z H0 (T, A) 0. Moving to reduced homology, on the right hand side of this sequence we have i j 0 H0 (T, A) 0 which tells us that H 0 (T, A) = 0 and therefore that H 0 (T, A) = 0. Lastly, to identify H 1 (T, A) notice that (T, A) is certainly a good pair since A is a deformation retract of open discs around the points that comprise it giving the isomorphism H 1 (T, A) = H 1 (T/A). Furthermore, the first homology group of a space is isomorphic to the abelianization of it s fundamental group, and I claim that π 1 (T/A) = A 1 Z. To see this we look at the topology of T/A; it is homotopic to (in fact a deformation retract of) S 1 S 1 CA with CA being the cone A I C, C identifies A 1 and C 0 having the obvious gluing to the set A of points on the torus. We can deformation retract one of the legs of the cone, a I to a point and the resulting space is the wedge of the torus with A 1 circles: ( S 1 S 1) A 1 S 1. By Van Kampens theorem, the fundamental group of this space is π 1 ( S 1 S A 1 1) S 1 = (Z Z) A 1 1=1 Z. H 1 (T/A) is the abelianization of this, and so is the group A +1 Z. To summarize: 0 n = 0 H n (T, A) = Z A +1 n = 1 Z n = 2 (6) Problem 8: ( 2.1: #18). Show that for the subspace Q R, the relative homology group H 1 (R, Q) is free abelian and find a basis. Proof. We can use the tail end of the reduced relative sequence H 1 (R) H 1 (R, Q) H 0 (Q) H 0 (R). We know that H 1 (R) = H 1 (R) = 0 since R deformation retracts to a point and H 0 (R) Z = H 0 (R) = Z implies that H 0 (R) = 0. Also H 0 counts path components so H 0 (Q) Z = H 0 (Q) = Z Q gives H 0 (Q) = Z Q 1 = Z Q. Hence the sequence above becomes 0 H 1 (R, Q) Z Q 0

8 and so which is free abelian. H 1 (R, Q) = H 1 (R, Q) = Z Q I claim that B = {[p, 0] p Q} {[0, q] q Q} is a basis. We can consider these as images of 2-simplices, and are all certainly contained in the kernal of since the endpoints are rational numbers hence elements of C 0 (Q). Suppose that m n 0 = a j [p j, 0] + b i [0, q i ] j=1 and we can also assume that p m <..., < p 1 < 0 < q 1 < < q n. We can rewrite the sum of the positive intervals [0, q j ] as n b i [0, q i ] = b 1 [0, q 1 ] + + b n 1 [0, q n 1 ] + b n [0, q n ] = b 1 [0, q 1 ] + + (b n + b n 1 )[0, q n 1 ] + b n [q n 1, q n ] (7). = (b n + + b 1 )[0, q 1 ] + (b n b 1 )[q 1, q 2 ]+ + (b n + b n 1 )[q n 2, q n 1 ] + b n [q n 1, q n ] and do the same for the sum of the negative intervals [p j, 0]. Since the q i s are ordered and any two of the [q l 1, q l ] intersect in at most a point, we know that 0 = b k + + b 1 = 0 for every k because they sum to zero. Starting from the last coefficient in the last line of (7), we have b n = 0, which implies that b n + b n 1 = b n 1 = 0 and repeating this process recursively gives b 1 = = b n = 0. The same process can be applied to the negative intervals and similarly a 1 = = a m = 0. Hence B is a set of linearly independent elements whose span is H 1 (R, Q). Problem 9: ( 2.1: #23). Show that the second barycentric subdivision of a -complex is a simplicial complex. Namely, show that the first barycentric subdivision produces a -complex with the property that each simplex has all its vertices distinct, then show that for a -complex with this property, barycentric subdivision produces a simplicial complex. Proof. Let X be a -complex and first consider the first barycentric subdivision X 1. Let σ : n X be a simplex [v 0,..., v n ] in X. After the subdivision occurs, we give barycenter b α each k-dimensional simplex [v i0,..., v ik ] and form the resulting n-simplices. Each of these simplices has only one of it s original vertices. Each of the barycenters b α is distinct because they lie in a unique face of σ, but also is clearly distinct from all of the v i by it s coordinate description thus it follows that the vertices of each n-simplex in the subdivision is distinct. Assuming that X 1 is a -complex with the above property that all it s vertices are distinct, let X 2 be it s barycentric subdivision and suppose that σ 1 and σ 2 are two n-simplices having the same set of vertices in X 2. Since they come from the subdivision, their vertex set consists of one

9 vertex of a simplex σ in the original -complex and barycenters of faces of σ. Since each point in X is in the image of exactly one restriction of a simplex, each point on the interior of Im σ must be in exactly one n-simplex of rhe subdivision. Hence σ 1 = σ 2 which implies that X 2 is a simplicial complex. Problem 10: ( 2.2: #4). Construct a surjective map S n S n of degree zero, for each n 1. Solution: Consider the sphere S n = {x R n+1 : x = 1} in R n+1 and let p : S n D n 1 be projection that forgets the last coordinate sending (x 1,..., x n, x n+1 ) (x 1,..., x n, 0) and let q : D n 1 Dn 1 S n 1 = S n be the quotient map of D n 1 by it s boundary. The composition p q : S n S n then has degree zero since the induced map can be written H n (S n ) p H n (D n 1 ) q H n (S n ) where H n (D n 1 ) = {0} since D n 1 is homotopic to a point. Then we have H n (S n ) p 0 q H n (S n ) and so the image of q is {0}, hence 1 0 so the degree is zero.

Part II. Algebraic Topology. Year

Part II. Algebraic Topology. Year Part II Year 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2017 Paper 3, Section II 18I The n-torus is the product of n circles: 5 T n = } S 1. {{.. S } 1. n times For all n 1 and 0

More information

Topology Hmwk 6 All problems are from Allen Hatcher Algebraic Topology (online) ch 2

Topology Hmwk 6 All problems are from Allen Hatcher Algebraic Topology (online) ch 2 Topology Hmwk 6 All problems are from Allen Hatcher Algebraic Topology (online) ch 2 Andrew Ma August 25, 214 2.1.4 Proof. Please refer to the attached picture. We have the following chain complex δ 3

More information

CELLULAR HOMOLOGY AND THE CELLULAR BOUNDARY FORMULA. Contents 1. Introduction 1

CELLULAR HOMOLOGY AND THE CELLULAR BOUNDARY FORMULA. Contents 1. Introduction 1 CELLULAR HOMOLOGY AND THE CELLULAR BOUNDARY FORMULA PAOLO DEGIORGI Abstract. This paper will first go through some core concepts and results in homology, then introduce the concepts of CW complex, subcomplex

More information

Math 6510 Homework 10

Math 6510 Homework 10 2.2 Problems 9 Problem. Compute the homology group of the following 2-complexes X: a) The quotient of S 2 obtained by identifying north and south poles to a point b) S 1 (S 1 S 1 ) c) The space obtained

More information

MATH 215B HOMEWORK 4 SOLUTIONS

MATH 215B HOMEWORK 4 SOLUTIONS MATH 215B HOMEWORK 4 SOLUTIONS 1. (8 marks) Compute the homology groups of the space X obtained from n by identifying all faces of the same dimension in the following way: [v 0,..., ˆv j,..., v n ] is

More information

7.3 Singular Homology Groups

7.3 Singular Homology Groups 184 CHAPTER 7. HOMOLOGY THEORY 7.3 Singular Homology Groups 7.3.1 Cycles, Boundaries and Homology Groups We can define the singular p-chains with coefficients in a field K. Furthermore, we can define the

More information

6 Axiomatic Homology Theory

6 Axiomatic Homology Theory MATH41071/MATH61071 Algebraic topology 6 Axiomatic Homology Theory Autumn Semester 2016 2017 The basic ideas of homology go back to Poincaré in 1895 when he defined the Betti numbers and torsion numbers

More information

Math 215a Homework #1 Solutions. π 1 (X, x 1 ) β h

Math 215a Homework #1 Solutions. π 1 (X, x 1 ) β h Math 215a Homework #1 Solutions 1. (a) Let g and h be two paths from x 0 to x 1. Then the composition sends π 1 (X, x 0 ) β g π 1 (X, x 1 ) β h π 1 (X, x 0 ) [f] [h g f g h] = [h g][f][h g] 1. So β g =

More information

MATH8808: ALGEBRAIC TOPOLOGY

MATH8808: ALGEBRAIC TOPOLOGY MATH8808: ALGEBRAIC TOPOLOGY DAWEI CHEN Contents 1. Underlying Geometric Notions 2 1.1. Homotopy 2 1.2. Cell Complexes 3 1.3. Operations on Cell Complexes 3 1.4. Criteria for Homotopy Equivalence 4 1.5.

More information

Math 440 Problem Set 2

Math 440 Problem Set 2 Math 440 Problem Set 2 Problem 4, p. 52. Let X R 3 be the union of n lines through the origin. Compute π 1 (R 3 X). Solution: R 3 X deformation retracts to S 2 with 2n points removed. Choose one of them.

More information

3. Prove or disprove: If a space X is second countable, then every open covering of X contains a countable subcollection covering X.

3. Prove or disprove: If a space X is second countable, then every open covering of X contains a countable subcollection covering X. Department of Mathematics and Statistics University of South Florida TOPOLOGY QUALIFYING EXAM January 24, 2015 Examiners: Dr. M. Elhamdadi, Dr. M. Saito Instructions: For Ph.D. level, complete at least

More information

Algebraic Topology Homework 4 Solutions

Algebraic Topology Homework 4 Solutions Algebraic Topology Homework 4 Solutions Here are a few solutions to some of the trickier problems... Recall: Let X be a topological space, A X a subspace of X. Suppose f, g : X X are maps restricting to

More information

MATH540: Algebraic Topology PROBLEM SET 3 STUDENT SOLUTIONS

MATH540: Algebraic Topology PROBLEM SET 3 STUDENT SOLUTIONS Key Problems 1. Compute π 1 of the Mobius strip. Solution (Spencer Gerhardt): MATH540: Algebraic Topology PROBLEM SET 3 STUDENT SOLUTIONS In other words, M = I I/(s, 0) (1 s, 1). Let x 0 = ( 1 2, 0). Now

More information

ALGEBRAICALLY TRIVIAL, BUT TOPOLOGICALLY NON-TRIVIAL MAP. Contents 1. Introduction 1

ALGEBRAICALLY TRIVIAL, BUT TOPOLOGICALLY NON-TRIVIAL MAP. Contents 1. Introduction 1 ALGEBRAICALLY TRIVIAL, BUT TOPOLOGICALLY NON-TRIVIAL MAP HONG GYUN KIM Abstract. I studied the construction of an algebraically trivial, but topologically non-trivial map by Hopf map p : S 3 S 2 and a

More information

Homework 4: Mayer-Vietoris Sequence and CW complexes

Homework 4: Mayer-Vietoris Sequence and CW complexes Homework 4: Mayer-Vietoris Sequence and CW complexes Due date: Friday, October 4th. 0. Goals and Prerequisites The goal of this homework assignment is to begin using the Mayer-Vietoris sequence and cellular

More information

FREUDENTHAL SUSPENSION THEOREM

FREUDENTHAL SUSPENSION THEOREM FREUDENTHAL SUSPENSION THEOREM TENGREN ZHANG Abstract. In this paper, I will prove the Freudenthal suspension theorem, and use that to explain what stable homotopy groups are. All the results stated in

More information

MATH 215B HOMEWORK 5 SOLUTIONS

MATH 215B HOMEWORK 5 SOLUTIONS MATH 25B HOMEWORK 5 SOLUTIONS. ( marks) Show that the quotient map S S S 2 collapsing the subspace S S to a point is not nullhomotopic by showing that it induces an isomorphism on H 2. On the other hand,

More information

Exercises for Algebraic Topology

Exercises for Algebraic Topology Sheet 1, September 13, 2017 Definition. Let A be an abelian group and let M be a set. The A-linearization of M is the set A[M] = {f : M A f 1 (A \ {0}) is finite}. We view A[M] as an abelian group via

More information

MATH 547 ALGEBRAIC TOPOLOGY HOMEWORK ASSIGNMENT 4

MATH 547 ALGEBRAIC TOPOLOGY HOMEWORK ASSIGNMENT 4 MATH 547 ALGEBRAIC TOPOLOGY HOMEWORK ASSIGNMENT 4 ROI DOCAMPO ÁLVAREZ Chapter 0 Exercise We think of the torus T as the quotient of X = I I by the equivalence relation generated by the conditions (, s)

More information

2.5 Excision implies Simplicial = Singular homology

2.5 Excision implies Simplicial = Singular homology 2.5 Excision implies Simplicial = Singular homology 1300Y Geometry and Topology 2.5 Excision implies Simplicial = Singular homology Recall that simplicial homology was defined in terms of a -complex decomposition

More information

Algebraic Topology Lecture Notes. Jarah Evslin and Alexander Wijns

Algebraic Topology Lecture Notes. Jarah Evslin and Alexander Wijns Algebraic Topology Lecture Notes Jarah Evslin and Alexander Wijns Abstract We classify finitely generated abelian groups and, using simplicial complex, describe various groups that can be associated to

More information

HOMOTOPY THEORY ADAM KAYE

HOMOTOPY THEORY ADAM KAYE HOMOTOPY THEORY ADAM KAYE 1. CW Approximation The CW approximation theorem says that every space is weakly equivalent to a CW complex. Theorem 1.1 (CW Approximation). There exists a functor Γ from the

More information

MATH 215B. SOLUTIONS TO HOMEWORK (6 marks) Construct a path connected space X such that π 1 (X, x 0 ) = D 4, the dihedral group with 8 elements.

MATH 215B. SOLUTIONS TO HOMEWORK (6 marks) Construct a path connected space X such that π 1 (X, x 0 ) = D 4, the dihedral group with 8 elements. MATH 215B. SOLUTIONS TO HOMEWORK 2 1. (6 marks) Construct a path connected space X such that π 1 (X, x 0 ) = D 4, the dihedral group with 8 elements. Solution A presentation of D 4 is a, b a 4 = b 2 =

More information

Math 752 Week s 1 1

Math 752 Week s 1 1 Math 752 Week 13 1 Homotopy Groups Definition 1. For n 0 and X a topological space with x 0 X, define π n (X) = {f : (I n, I n ) (X, x 0 )}/ where is the usual homotopy of maps. Then we have the following

More information

MATH730 NOTES WEEK 8

MATH730 NOTES WEEK 8 MATH730 NOTES WEEK 8 1. Van Kampen s Theorem The main idea of this section is to compute fundamental groups by decomposing a space X into smaller pieces X = U V where the fundamental groups of U, V, and

More information

SOLUTIONS TO THE FINAL EXAM

SOLUTIONS TO THE FINAL EXAM SOLUTIONS TO THE FINAL EXAM Short questions 1 point each) Give a brief definition for each of the following six concepts: 1) normal for topological spaces) 2) path connected 3) homeomorphism 4) covering

More information

Homology of a Cell Complex

Homology of a Cell Complex M:01 Fall 06 J. Simon Homology of a Cell Complex A finite cell complex X is constructed one cell at a time, working up in dimension. Each time a cell is added, we can analyze the effect on homology, by

More information

MAS435 Algebraic Topology Part A: Semester 1 Exercises

MAS435 Algebraic Topology Part A: Semester 1 Exercises MAS435 Algebraic Topology 2011-12 Part A: Semester 1 Exercises Dr E. L. G. Cheng Office: J24 E-mail: e.cheng@sheffield.ac.uk http://cheng.staff.shef.ac.uk/mas435/ You should hand in your solutions to Exercises

More information

7. Homotopy and the Fundamental Group

7. Homotopy and the Fundamental Group 7. Homotopy and the Fundamental Group The group G will be called the fundamental group of the manifold V. J. Henri Poincaré, 895 The properties of a topological space that we have developed so far have

More information

FUNDAMENTAL GROUPS AND THE VAN KAMPEN S THEOREM. Contents

FUNDAMENTAL GROUPS AND THE VAN KAMPEN S THEOREM. Contents FUNDAMENTAL GROUPS AND THE VAN KAMPEN S THEOREM SAMUEL BLOOM Abstract. In this paper, we define the fundamental group of a topological space and explore its structure, and we proceed to prove Van-Kampen

More information

id = w n = w n (m+1)/2

id = w n = w n (m+1)/2 Samuel Lee Algebraic Topology Homework #4 March 11, 2016 Problem ( 1.2: #1). Show that the free product G H of nontrivial groups G and H has trivial center, and that the only elements of G H of finite

More information

HOMEWORK FOR SPRING 2014 ALGEBRAIC TOPOLOGY

HOMEWORK FOR SPRING 2014 ALGEBRAIC TOPOLOGY HOMEWORK FOR SPRING 2014 ALGEBRAIC TOPOLOGY Last Modified April 14, 2014 Some notes on homework: (1) Homework will be due every two weeks. (2) A tentative schedule is: Jan 28, Feb 11, 25, March 11, 25,

More information

Algebraic Topology I Homework Spring 2014

Algebraic Topology I Homework Spring 2014 Algebraic Topology I Homework Spring 2014 Homework solutions will be available http://faculty.tcu.edu/gfriedman/algtop/algtop-hw-solns.pdf Due 5/1 A Do Hatcher 2.2.4 B Do Hatcher 2.2.9b (Find a cell structure)

More information

HOMOLOGY AND COHOMOLOGY. 1. Introduction

HOMOLOGY AND COHOMOLOGY. 1. Introduction HOMOLOGY AND COHOMOLOGY ELLEARD FELIX WEBSTER HEFFERN 1. Introduction We have been introduced to the idea of homology, which derives from a chain complex of singular or simplicial chain groups together

More information

Some K-theory examples

Some K-theory examples Some K-theory examples The purpose of these notes is to compute K-groups of various spaces and outline some useful methods for Ma448: K-theory and Solitons, given by Dr Sergey Cherkis in 2008-09. Throughout

More information

CW-complexes. Stephen A. Mitchell. November 1997

CW-complexes. Stephen A. Mitchell. November 1997 CW-complexes Stephen A. Mitchell November 1997 A CW-complex is first of all a Hausdorff space X equipped with a collection of characteristic maps φ n α : D n X. Here n ranges over the nonnegative integers,

More information

Homework 3 MTH 869 Algebraic Topology

Homework 3 MTH 869 Algebraic Topology Homework 3 MTH 869 Algebraic Topology Joshua Ruiter February 12, 2018 Proposition 0.1 (Exercise 1.1.10). Let (X, x 0 ) and (Y, y 0 ) be pointed, path-connected spaces. Let f : I X y 0 } and g : I x 0 }

More information

Homology theory. Lecture 29-3/7/2011. Lecture 30-3/8/2011. Lecture 31-3/9/2011 Math 757 Homology theory. March 9, 2011

Homology theory. Lecture 29-3/7/2011. Lecture 30-3/8/2011. Lecture 31-3/9/2011 Math 757 Homology theory. March 9, 2011 Math 757 Homology theory March 9, 2011 Theorem 183 Let G = π 1 (X, x 0 ) then for n 1 h : π n (X, x 0 ) H n (X ) factors through the quotient map q : π n (X, x 0 ) π n (X, x 0 ) G to π n (X, x 0 ) G the

More information

Math 530 Lecture Notes. Xi Chen

Math 530 Lecture Notes. Xi Chen Math 530 Lecture Notes Xi Chen 632 Central Academic Building, University of Alberta, Edmonton, Alberta T6G 2G1, CANADA E-mail address: xichen@math.ualberta.ca 1991 Mathematics Subject Classification. Primary

More information

DISCRETIZED CONFIGURATIONS AND PARTIAL PARTITIONS

DISCRETIZED CONFIGURATIONS AND PARTIAL PARTITIONS DISCRETIZED CONFIGURATIONS AND PARTIAL PARTITIONS AARON ABRAMS, DAVID GAY, AND VALERIE HOWER Abstract. We show that the discretized configuration space of k points in the n-simplex is homotopy equivalent

More information

SMSTC Geometry & Topology 1 Assignment 1 Matt Booth

SMSTC Geometry & Topology 1 Assignment 1 Matt Booth SMSTC Geometry & Topology 1 Assignment 1 Matt Booth Question 1 i) Let be the space with one point. Suppose X is contractible. Then by definition we have maps f : X and g : X such that gf id X and fg id.

More information

Algebraic Topology. Oscar Randal-Williams. or257/teaching/notes/at.pdf

Algebraic Topology. Oscar Randal-Williams.   or257/teaching/notes/at.pdf Algebraic Topology Oscar Randal-Williams https://www.dpmms.cam.ac.uk/ or257/teaching/notes/at.pdf 1 Introduction 1 1.1 Some recollections and conventions...................... 2 1.2 Cell complexes.................................

More information

The Hurewicz Theorem

The Hurewicz Theorem The Hurewicz Theorem April 5, 011 1 Introduction The fundamental group and homology groups both give extremely useful information, particularly about path-connected spaces. Both can be considered as functors,

More information

Math 6510 Homework 11

Math 6510 Homework 11 2.2 Problems 40 Problem. From the long exact sequence of homology groups associted to the short exact sequence of chain complexes n 0 C i (X) C i (X) C i (X; Z n ) 0, deduce immediately that there are

More information

THE FUNDAMENTAL GROUP AND CW COMPLEXES

THE FUNDAMENTAL GROUP AND CW COMPLEXES THE FUNDAMENTAL GROUP AND CW COMPLEXES JAE HYUNG SIM Abstract. This paper is a quick introduction to some basic concepts in Algebraic Topology. We start by defining homotopy and delving into the Fundamental

More information

Algebraic Topology M3P solutions 2

Algebraic Topology M3P solutions 2 Algebraic Topology M3P1 015 solutions AC Imperial College London a.corti@imperial.ac.uk 3 rd February 015 A small disclaimer This document is a bit sketchy and it leaves some to be desired in several other

More information

Exercise: Consider the poset of subsets of {0, 1, 2} ordered under inclusion: Date: July 15, 2015.

Exercise: Consider the poset of subsets of {0, 1, 2} ordered under inclusion: Date: July 15, 2015. 07-13-2015 Contents 1. Dimension 1 2. The Mayer-Vietoris Sequence 3 2.1. Suspension and Spheres 4 2.2. Direct Sums 4 2.3. Constuction of the Mayer-Vietoris Sequence 6 2.4. A Sample Calculation 7 As we

More information

THE COMPLEX OF FREE FACTORS OF A FREE GROUP Allen Hatcher* and Karen Vogtmann*

THE COMPLEX OF FREE FACTORS OF A FREE GROUP Allen Hatcher* and Karen Vogtmann* THE COMPLEX OF FREE FACTORS OF A FREE GROUP Allen Hatcher* and Karen Vogtmann* ABSTRACT. We show that the geometric realization of the partially ordered set of proper free factors in a finitely generated

More information

TOPOLOGY AND GROUPS. Contents

TOPOLOGY AND GROUPS. Contents TOPOLOGY AND GROUPS Contents Introduction............................... 2 Convention and Notation......................... 3 I. Constructing spaces 1. Cayley graphs of groups..................... 4 2.

More information

Homework in Topology, Spring 2009.

Homework in Topology, Spring 2009. Homework in Topology, Spring 2009. Björn Gustafsson April 29, 2009 1 Generalities To pass the course you should hand in correct and well-written solutions of approximately 10-15 of the problems. For higher

More information

Equivalence of the Combinatorial Definition (Lecture 11)

Equivalence of the Combinatorial Definition (Lecture 11) Equivalence of the Combinatorial Definition (Lecture 11) September 26, 2014 Our goal in this lecture is to complete the proof of our first main theorem by proving the following: Theorem 1. The map of simplicial

More information

Topology Hmwk 5 All problems are from Allen Hatcher Algebraic Topology (online) ch 1

Topology Hmwk 5 All problems are from Allen Hatcher Algebraic Topology (online) ch 1 Topology Hmwk 5 All problems are from Allen Hatcher Algebraic Topology (online) ch Andrew Ma November 22, 203.3.8 Claim: A nice space X has a unique universal abelian covering space X ab Proof. Given a

More information

THE FUNDAMENTAL GROUP AND SEIFERT-VAN KAMPEN S THEOREM

THE FUNDAMENTAL GROUP AND SEIFERT-VAN KAMPEN S THEOREM THE FUNDAMENTAL GROUP AND SEIFERT-VAN KAMPEN S THEOREM KATHERINE GALLAGHER Abstract. The fundamental group is an essential tool for studying a topological space since it provides us with information about

More information

Basic Notions in Algebraic Topology 1

Basic Notions in Algebraic Topology 1 Basic Notions in Algebraic Topology 1 Yonatan Harpaz Remark 1. In these notes when we say map we always mean continuous map. 1 The Spaces of Algebraic Topology One of the main difference in passing from

More information

Coxeter Groups and Artin Groups

Coxeter Groups and Artin Groups Chapter 1 Coxeter Groups and Artin Groups 1.1 Artin Groups Let M be a Coxeter matrix with index set S. defined by M is given by the presentation: A M := s S sts }{{ } = tst }{{ } m s,t factors m s,t The

More information

Math 637 Topology Paulo Lima-Filho. Problem List I. b. Show that a contractible space is path connected.

Math 637 Topology Paulo Lima-Filho. Problem List I. b. Show that a contractible space is path connected. Problem List I Problem 1. A space X is said to be contractible if the identiy map i X : X X is nullhomotopic. a. Show that any convex subset of R n is contractible. b. Show that a contractible space is

More information

INTRODUCTION TO ALGEBRAIC TOPOLOGY. (1) Let k < j 1 and 0 j n, where 1 n. We want to show that e j n e k n 1 = e k n e j 1

INTRODUCTION TO ALGEBRAIC TOPOLOGY. (1) Let k < j 1 and 0 j n, where 1 n. We want to show that e j n e k n 1 = e k n e j 1 INTRODUCTION TO ALGEBRAIC TOPOLOGY Exercise 7, solutions 1) Let k < j 1 0 j n, where 1 n. We want to show that e j n e k n 1 = e k n e j 1 n 1. Recall that the map e j n : n 1 n is defined by e j nt 0,...,

More information

On the Homological Dimension of Lattices

On the Homological Dimension of Lattices On the Homological Dimension of Lattices Roy Meshulam August, 008 Abstract Let L be a finite lattice and let L = L {ˆ0, ˆ1}. It is shown that if the order complex L satisfies H k L 0 then L k. Equality

More information

LECTURE 3: RELATIVE SINGULAR HOMOLOGY

LECTURE 3: RELATIVE SINGULAR HOMOLOGY LECTURE 3: RELATIVE SINGULAR HOMOLOGY In this lecture we want to cover some basic concepts from homological algebra. These prove to be very helpful in our discussion of singular homology. The following

More information

Math Homotopy Theory Hurewicz theorem

Math Homotopy Theory Hurewicz theorem Math 527 - Homotopy Theory Hurewicz theorem Martin Frankland March 25, 2013 1 Background material Proposition 1.1. For all n 1, we have π n (S n ) = Z, generated by the class of the identity map id: S

More information

NOTES ON DIFFERENTIAL FORMS. PART 5: DE RHAM COHOMOLOGY

NOTES ON DIFFERENTIAL FORMS. PART 5: DE RHAM COHOMOLOGY NOTES ON DIFFERENTIAL FORMS. PART 5: DE RHAM COHOMOLOGY 1. Closed and exact forms Let X be a n-manifold (not necessarily oriented), and let α be a k-form on X. We say that α is closed if dα = 0 and say

More information

arxiv: v1 [math.gt] 5 Jul 2012

arxiv: v1 [math.gt] 5 Jul 2012 Thick Spanier groups and the first shape group arxiv:1207.1310v1 [math.gt] 5 Jul 2012 Jeremy Brazas and Paul Fabel November 15, 2018 Abstract We develop a new route through which to explore ker Ψ X, the

More information

S n 1 i D n l S n 1 is the identity map. Associated to this sequence of maps is the sequence of group homomorphisms

S n 1 i D n l S n 1 is the identity map. Associated to this sequence of maps is the sequence of group homomorphisms ALGEBRAIC TOPOLOGY Contents 1. Informal introduction 1 1.1. What is algebraic topology? 1 1.2. Brower fixed point theorem 2 2. Review of background material 3 2.1. Algebra 3 2.2. Topological spaces 5 2.3.

More information

CALCULATION OF FUNDAMENTAL GROUPS OF SPACES

CALCULATION OF FUNDAMENTAL GROUPS OF SPACES CALCULATION OF FUNDAMENTAL GROUPS OF SPACES PETER ROBICHEAUX Abstract. We develop theory, particularly that of covering spaces and the van Kampen Theorem, in order to calculate the fundamental groups of

More information

10 Excision and applications

10 Excision and applications 22 CHAPTER 1. SINGULAR HOMOLOGY be a map of short exact sequences of chain complexes. If two of the three maps induced in homology by f, g, and h are isomorphisms, then so is the third. Here s an application.

More information

Algebraic Topology exam

Algebraic Topology exam Instituto Superior Técnico Departamento de Matemática Algebraic Topology exam June 12th 2017 1. Let X be a square with the edges cyclically identified: X = [0, 1] 2 / with (a) Compute π 1 (X). (x, 0) (1,

More information

HOMOLOGY THEORIES INGRID STARKEY

HOMOLOGY THEORIES INGRID STARKEY HOMOLOGY THEORIES INGRID STARKEY Abstract. This paper will introduce the notion of homology for topological spaces and discuss its intuitive meaning. It will also describe a general method that is used

More information

Pictures and Syzygies: An exploration of pictures, cellular models and free resolutions

Pictures and Syzygies: An exploration of pictures, cellular models and free resolutions Pictures and Syzygies: An exploration of pictures, cellular models and free resolutions Clay Shonkwiler Department of Mathematics The University of the South March 27, 2005 Terminology Let G = {X, R} be

More information

The Fundamental Group and Covering Spaces

The Fundamental Group and Covering Spaces Chapter 8 The Fundamental Group and Covering Spaces In the first seven chapters we have dealt with point-set topology. This chapter provides an introduction to algebraic topology. Algebraic topology may

More information

1 Whitehead s theorem.

1 Whitehead s theorem. 1 Whitehead s theorem. Statement: If f : X Y is a map of CW complexes inducing isomorphisms on all homotopy groups, then f is a homotopy equivalence. Moreover, if f is the inclusion of a subcomplex X in

More information

A Primer on Homological Algebra

A Primer on Homological Algebra A Primer on Homological Algebra Henry Y Chan July 12, 213 1 Modules For people who have taken the algebra sequence, you can pretty much skip the first section Before telling you what a module is, you probably

More information

From singular chains to Alexander Duality. Jesper M. Møller

From singular chains to Alexander Duality. Jesper M. Møller From singular chains to Alexander Duality Jesper M. Møller Matematisk Institut, Universitetsparken 5, DK 21 København E-mail address: moller@math.ku.dk URL: http://www.math.ku.dk/~moller Contents Chapter

More information

Compactifying string topology

Compactifying string topology Compactifying string topology Kate Poirier UC Berkeley Algebraic Topology: Applications and New Developments Stanford University, July 24, 2012 What is the algebraic topology of a manifold? What can we

More information

ALLEN HATCHER: ALGEBRAIC TOPOLOGY

ALLEN HATCHER: ALGEBRAIC TOPOLOGY ALLEN HATCHER: ALGEBRAIC TOPOLOGY MORTEN POULSEN All references are to the 2002 printed edition. Chapter 0 Ex. 0.2. Define H : (R n {0}) I R n {0} by H(x, t) = (1 t)x + t x x, x R n {0}, t I. It is easily

More information

Hungry, Hungry Homology

Hungry, Hungry Homology September 27, 2017 Motiving Problem: Algebra Problem (Preliminary Version) Given two groups A, C, does there exist a group E so that A E and E /A = C? If such an group exists, we call E an extension of

More information

GEOMETRY FINAL CLAY SHONKWILER

GEOMETRY FINAL CLAY SHONKWILER GEOMETRY FINAL CLAY SHONKWILER 1 Let X be the space obtained by adding to a 2-dimensional sphere of radius one, a line on the z-axis going from north pole to south pole. Compute the fundamental group and

More information

Homotopy and homology groups of the n-dimensional Hawaiian earring

Homotopy and homology groups of the n-dimensional Hawaiian earring F U N D A M E N T A MATHEMATICAE 165 (2000) Homotopy and homology groups of the n-dimensional Hawaiian earring by Katsuya E d a (Tokyo) and Kazuhiro K a w a m u r a (Tsukuba) Abstract. For the n-dimensional

More information

Tethers and homology stability for surfaces

Tethers and homology stability for surfaces Tethers and homology stability for surfaces ALLEN HATCHER KAREN VOGTMANN Homological stability for sequences G n G n+1 of groups is often proved by studying the spectral sequence associated to the action

More information

Course notes in algebraic topology

Course notes in algebraic topology Course notes in algebraic topology Benson Farb CLASS NOTES, Fall 2012 Comments/corrections/suggestions welcome November 11, 2012 2 Contents 1 Homology 5 1.1 Construction of simplicial homology....................

More information

Combinatorial Models for M (Lecture 10)

Combinatorial Models for M (Lecture 10) Combinatorial Models for M (Lecture 10) September 24, 2014 Let f : X Y be a map of finite nonsingular simplicial sets. In the previous lecture, we showed that the induced map f : X Y is a fibration if

More information

THICK SPANIER GROUPS AND THE FIRST SHAPE GROUP

THICK SPANIER GROUPS AND THE FIRST SHAPE GROUP ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 44, Number 5, 2014 THICK SPANIER GROUPS AND THE FIRST SHAPE GROUP JEREMY BRAZAS AND PAUL FABEL ABSTRACT. We develop a new route through which to explore ker

More information

Minimal free resolutions that are not supported by a CW-complex.

Minimal free resolutions that are not supported by a CW-complex. Minimal free resolutions that are not supported by a CW-complex. Mauricio Velasco Department of Mathematics, Cornell University, Ithaca, NY 4853, US Introduction The study of monomial resolutions and their

More information

1. Simplify the following. Solution: = {0} Hint: glossary: there is for all : such that & and

1. Simplify the following. Solution: = {0} Hint: glossary: there is for all : such that & and Topology MT434P Problems/Homework Recommended Reading: Munkres, J.R. Topology Hatcher, A. Algebraic Topology, http://www.math.cornell.edu/ hatcher/at/atpage.html For those who have a lot of outstanding

More information

(iv) Whitney s condition B. Suppose S β S α. If two sequences (a k ) S α and (b k ) S β both converge to the same x S β then lim.

(iv) Whitney s condition B. Suppose S β S α. If two sequences (a k ) S α and (b k ) S β both converge to the same x S β then lim. 0.1. Stratified spaces. References are [7], [6], [3]. Singular spaces are naturally associated to many important mathematical objects (for example in representation theory). We are essentially interested

More information

Tensor, Tor, UCF, and Kunneth

Tensor, Tor, UCF, and Kunneth Tensor, Tor, UCF, and Kunneth Mark Blumstein 1 Introduction I d like to collect the basic definitions of tensor product of modules, the Tor functor, and present some examples from homological algebra and

More information

Corrections to Introduction to Topological Manifolds (First edition) by John M. Lee December 7, 2015

Corrections to Introduction to Topological Manifolds (First edition) by John M. Lee December 7, 2015 Corrections to Introduction to Topological Manifolds (First edition) by John M. Lee December 7, 2015 Changes or additions made in the past twelve months are dated. Page 29, statement of Lemma 2.11: The

More information

ALGEBRAIC TOPOLOGY IV. Definition 1.1. Let A, B be abelian groups. The set of homomorphisms ϕ: A B is denoted by

ALGEBRAIC TOPOLOGY IV. Definition 1.1. Let A, B be abelian groups. The set of homomorphisms ϕ: A B is denoted by ALGEBRAIC TOPOLOGY IV DIRK SCHÜTZ 1. Cochain complexes and singular cohomology Definition 1.1. Let A, B be abelian groups. The set of homomorphisms ϕ: A B is denoted by Hom(A, B) = {ϕ: A B ϕ homomorphism}

More information

1 Spaces and operations Continuity and metric spaces Topological spaces Compactness... 3

1 Spaces and operations Continuity and metric spaces Topological spaces Compactness... 3 Compact course notes Topology I Fall 2011 Professor: A. Penskoi transcribed by: J. Lazovskis Independent University of Moscow December 23, 2011 Contents 1 Spaces and operations 2 1.1 Continuity and metric

More information

A PRESENTATION FOR THE MAPPING CLASS GROUP OF A NON-ORIENTABLE SURFACE FROM THE ACTION ON THE COMPLEX OF CURVES

A PRESENTATION FOR THE MAPPING CLASS GROUP OF A NON-ORIENTABLE SURFACE FROM THE ACTION ON THE COMPLEX OF CURVES Szepietowski, B. Osaka J. Math. 45 (008), 83 36 A PRESENTATION FOR THE MAPPING CLASS GROUP OF A NON-ORIENTABLE SURFACE FROM THE ACTION ON THE COMPLEX OF CURVES BŁAŻEJ SZEPIETOWSKI (Received June 30, 006,

More information

NOTES ON THE FUNDAMENTAL GROUP

NOTES ON THE FUNDAMENTAL GROUP NOTES ON THE FUNDAMENTAL GROUP AARON LANDESMAN CONTENTS 1. Introduction to the fundamental group 2 2. Preliminaries: spaces and homotopies 3 2.1. Spaces 3 2.2. Maps of spaces 3 2.3. Homotopies and Loops

More information

Hairy balls and ham sandwiches

Hairy balls and ham sandwiches Hairy balls and ham sandwiches Graduate Student Seminar, Carnegie Mellon University Thursday 14 th November 2013 Clive Newstead Abstract Point-set topology studies spaces up to homeomorphism. For many

More information

Pure Math 467/667, Winter 2013

Pure Math 467/667, Winter 2013 Compact course notes Pure Math 467/667, Winter 2013 Algebraic Topology Lecturer: A. Smith transcribed by: J. Lazovskis University of Waterloo April 24, 2013 Contents 0.0 Motivating remarks..........................................

More information

L E C T U R E N O T E S O N H O M O T O P Y T H E O R Y A N D A P P L I C AT I O N S

L E C T U R E N O T E S O N H O M O T O P Y T H E O R Y A N D A P P L I C AT I O N S L A U R E N T I U M A X I M U N I V E R S I T Y O F W I S C O N S I N - M A D I S O N L E C T U R E N O T E S O N H O M O T O P Y T H E O R Y A N D A P P L I C AT I O N S i Contents 1 Basics of Homotopy

More information

THE FUNDAMENTAL GROUP AND BROUWER S FIXED POINT THEOREM AMANDA BOWER

THE FUNDAMENTAL GROUP AND BROUWER S FIXED POINT THEOREM AMANDA BOWER THE FUNDAMENTAL GROUP AND BROUWER S FIXED POINT THEOREM AMANDA BOWER Abstract. The fundamental group is an invariant of topological spaces that measures the contractibility of loops. This project studies

More information

Bredon, Introduction to compact transformation groups, Academic Press

Bredon, Introduction to compact transformation groups, Academic Press 1 Introduction Outline Section 3: Topology of 2-orbifolds: Compact group actions Compact group actions Orbit spaces. Tubes and slices. Path-lifting, covering homotopy Locally smooth actions Smooth actions

More information

MAT 530: Topology&Geometry, I Fall 2005

MAT 530: Topology&Geometry, I Fall 2005 MAT 530: Topology&Geometry, I Fall 2005 Problem Set 11 Solution to Problem p433, #2 Suppose U,V X are open, X =U V, U, V, and U V are path-connected, x 0 U V, and i 1 π 1 U,x 0 j 1 π 1 U V,x 0 i 2 π 1

More information

3-manifolds and their groups

3-manifolds and their groups 3-manifolds and their groups Dale Rolfsen University of British Columbia Marseille, September 2010 Dale Rolfsen (2010) 3-manifolds and their groups Marseille, September 2010 1 / 31 3-manifolds and their

More information

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS In this section we will prove the Künneth theorem which in principle allows us to calculate the (co)homology of product spaces as soon

More information

Solutions to Problem Set 1

Solutions to Problem Set 1 Solutions to Problem Set 1 18.904 Spring 2011 Problem 1 Statement. Let n 1 be an integer. Let CP n denote the set of all lines in C n+1 passing through the origin. There is a natural map π : C n+1 \ {0}

More information

Geometric Topology. Harvard University Fall 2003 Math 99r Course Notes

Geometric Topology. Harvard University Fall 2003 Math 99r Course Notes Geometric Topology Harvard University Fall 2003 Math 99r Course Notes Contents 1 Introduction: Knots and Reidemeister moves........... 1 2 1-Dimensional Topology....................... 1 3 2-Dimensional

More information