Algebraic Topology 1: Homotopy Equivalence 

CORRECTION: The very last lemma (time 1:04:00) stated that if you have a quotient map X -> X/A, then X and X/A are homotopy equivalent. I forgot to add that this is only true if the subcomplex A is contractible (homotopy equivalent to the point). Without the requirement that A is contractible, the lemma may not hold. For example, the circle S^1 is a subcomplex of the disk D^2, namely its boundary. Therefore we can consider the quotient map D^2 -> D^2/S^1. The image of this map is itself just the sphere S^2 (think about gluing all the points along the boundary of a disk together to give you a single point - the point becomes a pole of the sphere). But D^2 is *NOT* homotopy equivalent to S^2. (Why not? One is contractible and the other is not.) Why doesn't the lemma work here? Because we quotiented out by the subcomplex S^1 which is not contractible violating the requirement.

Protec this man at all costs

Many many thanks to you professor Andrew, you are truly a gifted professor

This is awesome, thanks so much! Looking forward to future videos. Keep up the great work.

Amazing lecturer

Hi, just discovered this channel and loving it so far! I'm self-studying algebraic topology with Hatcher's book, and solutions for the exercises are notoriously hard to find. Could you go through some of the exercises for each chapter ? On a personal note, I'm stuck on some of the first few exercises of chapter 1.1 - namely exercises 1, 4 and 6. Also, some category theory might be beneficial to students. Once you get the hang of it, I find that it helps put things together in a way that reduces a lot of the mental load for AT.

Brilliant lecture. Really useful with a large clear pad to sketch these concepts. The 'north' and 'south' pole of the sphere finally touching as the string shrinks and disappears. Seemed to 'click' conceptually, very good introduction.

THANKS A LOT FOR EXCELLENT LECTURE!!!

excellent lecture

Thanks for the intresting lecture ........

Thank you very much

We love proof by example

How exactly is R2 contractible? Or D2\S1? The problem for me is not that it's unbounded, but that it doesn't have a boundary. If f0 is mapping all R2 to (0,0), while f1 is the identity on R2, don't {fi((0,0)):i@[0,1]} have to be closed?

Thank you very much for this! I have a question. I am interested in the application to learn physics, do you think the course and the book of Hatcher would be a good approach? Thanks!!

46:18 Conjecture: if X and Y are homotopy equivalent, then there is a continuous image of Y (resp. X) that is a deformation retraction of X (resp. Y). This is prolly true... is the converse true?

1:06:51

0:34

i think you made a mistake in the definition of homotopy ( at 14:00)

32:05

Retraction (convergence, syntropy) is dual to inclusion (divergence, entropy). "Always two there are" -- Yoda. Attraction is dual to repulsion -- forces are dual!

What’s the name of the lecturer?