New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Introduction to Algebraic Topology

by: Virgil Wyman

Introduction to Algebraic Topology 22M 201

Marketplace > University of Iowa > Mathematics (M) > 22M 201 > Introduction to Algebraic Topology
Virgil Wyman
GPA 3.97

Jonathan Simon

Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Jonathan Simon
Class Notes
25 ?




Popular in Course

Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Virgil Wyman on Friday October 23, 2015. The Class Notes belongs to 22M 201 at University of Iowa taught by Jonathan Simon in Fall. Since its upload, it has received 32 views. For similar materials see /class/227987/22m-201-university-of-iowa in Mathematics (M) at University of Iowa.

Popular in Mathematics (M)


Reviews for Introduction to Algebraic Topology


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/23/15
22M201 Fall 08 J Simon Homology Notes 1 1 FIRST DEFINITIONS De nition 1 Standard n cells Inis the cube 01 gtlt gtlt 0 1 C R We want 0 to be a single point Let us denote this as 0 6 just to emphasize that 0 is a point De nition 2 Singular n cubes in a space X A continuous function T I a X is called a singular n cube in X Remark Informally we can think of a singular U cube as a point in X there are as many maps T 0 a X as there are point in X Remark Formally we can think of a singular Z cube in X as a path in X We have studied paths a lot already and so you are already an eppert on singular Z cubes De nition 3 Group of singular n cubes in X Let Qn denote the abstract free abelian group with basis the set of all singular n cubes in X The group QnX consists of nite linear combinations of functions two different functions do not interact at this level but coef cients can cancel For example suppose X is R1 and ST are singular 2 cubes in X given by recipes Tuo ui and Sui 7uii Then in Q2X the subgroup generated by S and T has rank 2 The fact that there is some arithmetic relation between the recipes de ning S and T does not produce a relation between S and T as elements of Q2R unless the two recipes worked out to exactly the same function which here they don7t On the other hand relations like 2T 7 3T 7T are valid in Q2 De nition 4 Degenerate and nondegenerate cubes A singular cube T I a X is called degenerate if the function is independent of one of the coordinate ualues J Simon all rights reserved page 1 Remark A singular U cube cannot be degenerate A singular Z cube is degenerate if and only if it is a constant function Remark Example In Q3R the 3 cube Tui w w2u cosu 7 w3 is degenerate because it is independent ofy The singular 3 cube Tui w u i w is NON degenerate even though it is afunction from a 3 dimensional cell to a Z dimensional space De nition 5 Singular n chains On Since Qn is generated by all singular cubes we can write Qn as a direct sum of the free abelian group generated by degenerate cubes and the free abelian group generated by nondegenerate cubes which we call 0X Alternatively we can de ne 0X as the quotient QnX BAX 39 Technically an element of 0 is an equivalence class or an ordered pair whose rst coordinate is 0 Two elements of 0 are equal if their difference is a linear combination of degenerate n cubes OnX Remark Different homology theories There are other ways to develop homology theories for epample one can work with triangulated spaces where the cells are actual segments triangles etc quotsimplicial homology one can work with continuous maps from simplices into X quotsingular homology usually refers to this and there are generalizations to spaces that are more general topologically see ech or Vietoris homology theories Each approach has some technical prices to pay for the ways in which it is nice The price we pay for singular cubical homology is that we have to fuss with these degenerate cubes 11 The boundary of a singular cube If T I 7 X is a singular cube we want to de ne the boundary of T as an algebraic object speci cally an element of Qn1X that somehow captures our intuitive idea of boundary Geometrically the boundary of a cube I is the set of its faces If we restrict T to one of the faces of I we can de ne a singular n 7 1 cube and by combining these a certain way we will get a linear combination of 2n singular n 7 1 cubes representing the boundary of T De nition 6 The it front and back faces of a singular n cube Suppose T I 7 X is a singular n cube Assume n 2 1 we will J Simon all rights reserved page 2 consider the case of 0 cubes later Fix an index i 1 S i S n De ne the n 7 1 cube AfT In 1 a X by AT818n71 T81 8i7108i 8 Note T0 51 sn1 and T51 sn10 are the extreme cases i 1n Similarly we de ne BT818n71 T81 8i7118i 8 We now de ne the boundary of a singular cube T I a X as M ZEDKAfT i ByT i1 In this way we de ne the boundary of a singular n cube to be a certain linear combination of singular n 7 1 cubes Since the n cubes are a basis for the free abelian group Q the function 3 extends to a homomorphism 3n QnX a Qn1X Lemma 01 HOMEWORK IfT is a degenerate n cube in X then EXT 6 Dn1X Remark Note the above lemma does not claim that each face of a degenerate cube is degenerate some faces are degenerate and the non degenerate ones cancel Once we know the above lemma that 3nan D a Dikl we see that 3 induces a homomorphism 3n CnX a n1X Theorem 1 The composition of homomorphisms 3 3n On1X1 n71X is the zero map Put brie y 3n 0 an 0 Even more cryptically but a nice way to remember this key parts of any homology theory 36 0 Proof HOMEWORK To prove this theorem you need to understand faces of faces In the next section we show how to derive the face identities77 that you can use to prove the theorem You dont have to prove the face identities D J Simon all rights reserved page 3 12 Face identities We can View the operations of taking faces as homomorphisms AKB QnX a Qn1X What happens when we take a face of a face If T is an n cube7 then AfT is an n 7 1 cube7 and AgilAyT is an n 7 2 cube For example7 if T is a singular 5 cube7 then B3AgT51752753 T5101752753 We see this as followsA T is the 4 cube t17t27t37t4 Tt107t27t37t4 The B face of of AST is the 3 cube 515253 a AgT511752753 T5101752753 If we use the dimension indicating superscripts as B4 and A5 above7 we can see that BiAj is not the same as figBi and establish the following identities Proposition 11 Eqns 7217 Sec Vll21 of Massey AiAj AHA BiBj Brig A149 EMA 31A AME 13 Cycles Bounds and Homology Groups lf 2 E On such that anzn 07 we call 2 a n cycle The group of n cycles is Zn ker n The image group 6n10n1 Q On is denoted Bn7 the group of n bounds Since an 0 6711 07 we have Bn Q Zn Thus the quotient group zltXgt makes sense this is the ndimensional singular cubical homology group of X J Simon all rights reserved page 4 Remark As with the fundamental group it will be easy to show that homology groups are topological inuariants But it will be di cult to show that any of them are nontriuial beyond dimension 0 Be patient Meanwhile let s establish a relation between 7T1X and H1X The full theorem the Hurewicz theorem says that for a path connected space H1 X E abelianized 7T1X In the eccercise below you are asked to establish a part of this relationship Proposition 12 1 Suppose f 01 a X is a loop based at x0 6 X Then uiewing f as a singular Z cube f is a cycle 2 Suppose the loop f 01 a X is homotopically triuial rel endpoints in X That is f 1 E 7T1Xz0 Then the Z cycle f is homologically triuial ie f 0 E H1X Proof HOMEWORK You should do this exercise by working directly from the careful de nitions of boundary cycle etc To show that an n chain is a cycle show that its boundary 0 E Owl but recall that all degenerate chains are 0 in Owl To show that an n cycle is homologically trivial you need to show that it lies in B but again to be77 an element of B the chain just needs to differ from an actual boundary by 0 or by something degenerate End of Handout J Simon all rights reserved page 5 22M201 Fall 06 J Simon De ning Cell Complex and Why We would like to have a clean de nition of cell complecc to go along with our intuitive idea of what it means to cut a space into cells that intersect each other in a controlled way Making this intuitive idea precise is more complicatedsubtle than one might rst expect This often happens in mathematics Think about the intuitive idea of a continuous function as a graph that can be drawn without lifting your pen from the paper compared to the E 7 6 de nition We want to describe a class of spaces that is 0 general enough to include many familiarimportant spaces 0 narrow enough to allow us to prove theorems of the general form IfX A U B where AB and A B are nice and we understand something about each of A B and A B and we understand how A B sits within each of A B then we can deduce something about for X People take two general approaches to de ning some kind of cell complex At one extreme we restrict the cells to be very special in particular triangles or other simplexes simplices for the language puristsbut why dont we say complices 7 see the de nition of triangulation of a surface in the Massey text At the other extreme we allow the individual cells to be topological cells attached to one another in an orderly way see the de nition of CW complex in the Hatcher text note Massey also gives a de nition of CW complex but it is stated in a spread out way that may be hard to decipher In our course I want us to try to use the idea of cell complex in an informal way but have the machinery of CW complex as developed in Hatcher as something we can fall back on ifas needed One of the main theorems of our course is the theorem saying that Euler characteristic of a cell complex is a topological invariant all rights reserved page 1 Theorem Suppose X is a space realized as a nite dimensional cell complecc of dimension N with a nite number of cells in each dimension For each i let BAX denote the rank of the ith homology group Then the alternating sum of the numbers of cells in each dimension equals the alternating sum of the 61 That is if n denotes the number of cells of dimension j N 00 ZH nj ZenoX 70 i0 I used different letters7 i7j to index the above sums to emphasize that the respective numbers nk and Bk generally are not the same 7 though they are related in particular7 BAX S nk Thus7 in fact7 the sum of Betti numbers is not really in nite it stops at N or sooner For example7 we can represent the 2 sphere 52 as the surface of a cube7 with 710 87711 127712 6 This gives no 7 n1 n2 2 On the other hand7 we will have later in the semester H0SZ 2 Z H1SZ 0 H2SZ 2 Z and zero thereafter7 so 5017 5107 521 and zero thereafter For the solid cube 37 7 we have the same numbers as for the surface of the cube of cells in dimensions 07 17 and 27 and acquire one new cell in dimension 3 Meanwhile the solid cube has 60 1 and all other Betti numbers 0 All this will come much later I am hoping that by introducing some words and notation now7 they will seem normal to you later7 so you7ll be able to focus on the ideas and not get tied up inby the syntax all rights reserved page 2


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Janice Dongeun University of Washington

"I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.