Algebraic Topology 14: Exact Sequences & Homology of Spheres
Playlist: • Algebraic Topology
We introduce exact sequences and a particular long exact sequence on the (reduced) homology groups for a subspace A of the space X and its quotient X/A. Then we use this to calculate the (singular) homology of the spheres S^n. We also discuss the homology of the suspension SX of a space X and give a topological proof of Brouwer's fixed point theorem.
Presented by Anthony Bosman, PhD.
Learn more about math at Andrews University: www.andrews.edu/cas/math/
In this course we are following Hatcher, Algebraic Topology: pi.math.cornell.edu/~hatcher/...
Пікірлер: 9
0:00 Recap on homology of spheres 04:15 Motivating reduced homology 05:45 Defining reduced homology Note: You can also define reduced homology with the augmentation map from the 0-chain group to Z, which takes all chains to the sum of the coefficients of the simplices. Then the homology of this new chain complex is the reduced homology. 07:10 Reduced homology for contractible spaces 09:00 Introducing exact sequences 13:00 Properties of exact sequences 18:50 Examples of exact sequences 21:20 Motivation for exact sequences by stating a theorem: the singular homology groups of a subspace A in a space X and the quotient by that subspace X/A form a long exact sequence 26:00 Using this theorem to prove that the singular homology of spheres is what we expect it to be 35:00 Generalization of this method of calculating singular homology of spheres to more general spaces and topological cones and suspensions over them 43:05 Brouwer’s fixed-point theorem in n dimensions as a corollary of the singular homology of spheres
Absolutely amazing
Thanks for sharing!
As always, thanks for the great lecture!! One thing that bothered me both with the original and this proof of Brouwer's fixed-point theorem is that there there wasn't any discussion about whether D^n is convex or not, but it was always drawn as such. In the case D^n isn't convex, couldn't there be multiple r(x) points on the boundary? (I guess that it's fine because D^n is homotopy equivalent to a convex space, but I'd expect a bit of discussion about that...?)
@MathatAndrews
5 ай бұрын
Recall D^n is defined to be the n-dimensional disk: the set of points of radius less than or equal to one from the origin in R^n. This is certainly convex!
@DDranks
5 ай бұрын
Oh, I was thinking that the proof was more generally about spaces that are homeomorphic to D^n since this is a topology class, but I shouldn't assume things too much :)
I suppose this is not the last lecture for the serie? Will all lectures be uploaded and how many lectures in total?
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