Homology: Faces, Boundaries & Degenerate Cubes in Topology (Sec. 7.2) by J. Simon - Prof. , Study notes of Topology

Download Homology: Faces, Boundaries & Degenerate Cubes in Topology (Sec. 7.2) by J. Simon - Prof. and more Study notes Topology in PDF only on Docsity! 22M:201 Fall 04 J. Simon Notes on Homology (Sec. 7.2) November 1, 2004 0.1. Don’t worry yet about “degenerate” cubes. I suggest you study this section in the order: first = class notes, second = this handout, third = read the text and do the homework problems. 0.2. Just to make sure you understand how “faces” are defined. . . Suppose T is a singular 3-cube in X, that is T : I3 → X. Then T has six “faces”. Each face of T is a singular 2-cube, that is a map from I2 → X. The idea is easier than the notation we eventually end up using. There are 3 directions in R3. In each direction, T has a “far” face and a “near” face: the text calls these the “back” and “front” faces respectively. We have to specify six functions from I2 into X. So for each of the six functions, we have to decide where to send the points (a, b) ∈ I2. Keep saying it over and over again: “Each face is a function. A given face is the function that sends points (a, b) to . . . ”: Direction front face back face x−direction T (0, a, b) T (1, a, b) y−direction T (a, 0, b) T (a, 1, b) z−direction T (a, b, 0) T (a, b, 1) 0.3. Define the “boundary”. The boundary of an n−cube T is a linear combination of (n− 1)−cubes. The boundary of the boundary of T is then a linear combination of (n− 2) cubes. We want that boundary to be 0. The “boundary” is actually a homomorphism from the group of n−cubes to the group of (n− 1)− cubes, denoted ∂n : Qn(X) → Qn−1(X). The claim is that the homomorphism ∂n−1 ◦ ∂n : Qn(X) → Qn−2(X) is the zero homomorphism. Consider the case where T is a 2-cube. Then ∂2(T ) is a linear combination of 1− cubes, and the boundary of that is a linear combination of 0−cubes, essentially a linear combination of points of X. Each point in this combination appears twice, and we need to make sure that it appears once with a (+) and once with a (−). That is accomplished by the identities AiAj(T ) = Aj−1Ai(T ) etc. displayed in text (7.2.1). Here is an example: T is a 3-cube. What is (A1A3T )(s)?? (Note: don’t think of this as functional composition that is associative; A1 of 1 (A3T )(s) would be A1 of a point in X, which doesn’t make sense. Also A3T (s) would not make sense, since the domain of A3T is I 2, not I1.) A3T is a 2−cube. A1(A3T ) is the front face of that 2−cube in the first coordinate direction. So A1(A3T )(s) = (A3T )(0, s). Now, what does (A3T ) do to a pair (t, s)? It takes the pair to a point on the front face of I3 in the third direction, and then applies T . So (A3T )(0, s) = T (0, s, 0). Similarly, A1B3T (s) = T (0, s, 1). HOMEWORK: Prove the fourth identity in (7.2.1), that is, prove that the functions BiAj(T ) and Aj−1Bi(T ) are identical. (Hint: Both of these are (n− 2)−cubes. So start with a point (a1, . . . , an−2) and see where it is sent under each of the two maps.) 0.3.1. NOW we define the boundary homomomorphism. The function ∂n : Cn(X) → Cn−1(X) is given by its action on each basis element, that is its action on each n−cube: ∂n(T ) = n∑ i=1 (−1)i((AiT )− (BiT )) . HOMEWORK: Prove that (boundary)(boundary)=0. Since this will be true if an only if it is true for generators of Qn(X), it is necessary and sufficient for you to prove that for each singular n−cube T , ∂n−1(∂n(T )) = 0 ∈ Cn−2(X) . (Hint: The proof follows directly from the identities (7.2.1); you just need to do the careful bookkeeping.) 0.4. Provisional definition of cycles, bounds, and homology groups. We are going to define the cycles, Zn(X) to be those combinations of n−cubes that have 0 boundary. That is, Zn(X) = ker ∂n . These n−cycles are trying to record the existence of n−dimensional “holes” in X. But if a cycle is actually the boundary of something, then there’s no “hole” to record. With this in mind, we define a c©2004, J. Simon, all rights reserved page 2