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.ed...
In this course we are following Hatcher, Algebraic Topology: pi.math.cornel...

[@-minushyphen1two379]

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

[@herodeonhopeful]

Thanks for sharing!

[@コンティオプル]

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]

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]

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 :)

[@ompatel9017]

Absolutely amazing

[@juanmipesan313]

My novel!

[@xinhaofan396]

I suppose this is not the last lecture for the serie? Will all lectures be uploaded and how many lectures in total?

[@张金龙-t9u]

nice🥰🥰🥰🥰