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.

@hyperduality2838

14 күн бұрын

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

Protec this man at all costs

@jacksonstenger

3 ай бұрын

Agree

@johnkaylor8670

3 ай бұрын

Totally agree. This man starts at =the beginning= ! Meaning with the origin concepts of the subject he is addressing. He doesn't cavaliery assume that you already have His, personal conceptual understanding of the subject's basic essentials clear in your mind. He DEVELOPS them, step-wise in a wonderfully straightforward progression so that even persons not gifted in math can actually start to understand an advanced subject, such as this one. Kudos !

@najmussakib6337

Ай бұрын

Agree

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

@johnkaylor8670

3 ай бұрын

Yes. This man is an actual TEACHER ! Not a lofty "lecturer".

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

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.

@MathatAndrews

10 ай бұрын

Oh, that is a great idea! Let me see if I can find the time!

@John-js2uj

9 ай бұрын

@@MathatAndrewsI’m also loving the lectures and would greatly appreciate any solutions for the exercises! And I’m also up for framing some of the material in terms of category theory if that’s possible. Thanks for uploading your fantastic lectures.

Amazing lecturer

THANKS A LOT FOR EXCELLENT LECTURE!!!

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.

excellent lecture

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

Thank you very much

We love proof by example

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!!

0:34

1:06:51

32:05

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?

@xanderlewis

3 ай бұрын

The converse is not true: take X to be the one-point space and Y to be any non-contractible space (S^1, say). Then it’s certainly true that X is a continuous image of Y (under the only possible map), and X is certainly a deformation retraction of itself, but X and Y are not homotopy equivalent.

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?

@hywelgriffiths5747

5 күн бұрын

R2 can be mapped one-to-one onto the open disk

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

Why there are so many ads inside this video, it is like every 3-5 minutes 😢

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?