Algebraic Topology 4: Brouwer Fixed Point Theorem & Borsuk-Ulam

Подписаться 17 тыс.

Просмотров 7 тыс.

50% 1

Playlist: • Algebraic Topology
We use the fundamental group to prove the Brouwer Fixed Point Theorem which states that any continuous map from the disk to itself leaves at least one point fixed. We also prove Borsuk-Ulam which tells us that there are two points on the opposite sides of the earth with the exact same temperature and barometric pressure.
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...

Опубликовано:

5 окт 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 25

@md.mehedihasanrasel9684 2 месяца назад

Esteemed sir, you are one of the finest teacher i have ever seen. Please provide the new lectures of this series. Complete the book of hatcher please.

@sinclairabraxas3555 9 месяцев назад

those are very good! having a blast here binge watching this series

@johnkaylor8670 6 месяцев назад

Yes. This series is wonderfully written and entertaining.

@berlinisvictorious Месяц назад

15:44 how do we know r * h_t is well-defined? What property it follows from?

@YouandYoussYouandYouss 5 месяцев назад

In your proof of Brouwer's thm, you forgot to talk about the case where the image of a point of the circle (border) by f is a point of the circle.

@עמיתלרמן Месяц назад

I might be missing something but f(x)=x doesn't matter if the image is or is not on the border. So this still implies a retraction so I don't see a problem with f(x) being on the border.

@עמיתלרמן Месяц назад

I will not forget this quote: "The good news is, while there still are constructivist mathematicians, everyone recognizes them as a big kookie"

@richardchapman1592 5 месяцев назад

Don't understand how you can use a unit sphere with it's central point and a straight line from a point on the surface to an anipodal point as an ideal to homotopic other paths across through S2. Is their any correspondence with describing bent spacetime using relationship to orthogonal axes?

@-minushyphen1two379 Год назад

33:25 alternative proof of Borsuk-Ulam

@muralidharansomasundaram1509 7 месяцев назад

Brilliant proof.

@vekyll 4 месяца назад

It seems I have a very bad intuition regarding continuous functions. :/ For example, I really can't convince myself that D1 without origin is homeomorphic to S1. Still there are arbitrarily close points in D1 without origin that map to antipodal (far away) points on S1.

@MrMetrizable 3 месяца назад

Do you mean D2 with origin removed? S1 is not homeomorphic to D2 minus origin, you can disconnect S1 by removing two points, not so for D2. They are homotopic by deformation retract though

@vekyll 3 месяца назад

@@MrMetrizable yeah, sorry, I meant D2. Still, I don't know what "homotopic by deformation retract" means. :/

@kovatembel 2 месяца назад

I’m not very knowledgeable about topology, but I guess your concept of continuity is a slightly inaccurate. Your statement on arbitrarily close points in D2-{0} mapping to antinodal points is true, but I guess that what matters for continuity is that given a point x in D2-{0}, there is some neighborhood that gets mapped to an arbitrary small neighbourhood of the image f(x). Intuitively, for a point (x=epsilon,y=0), your map looks continuous on an open circle of radius epsilon or smaller. (The map around 18:00 shows that these spaces are homotopic, not homeomorphic, as that would require a unique inverse)

@sallylauper8222 6 месяцев назад

Formally it contains all it's Limit points...

@MarioPucci_mamio Год назад

Proofs via the lifting technique are always a bit weird I mean couldn't one just formulate the Theorem directly with respect to the lifted space, give it a meaning of its own right, and then obtain the Theorem about the original space as a corollary via projection?

@MathatAndrews Год назад

I'm not exactly sure what you have in mind, but I believe the order here is important to give the contradiction. Recall the contraction is produced by showing that the equivalence class of the loop (in S^1) is both the trivial [w_0] and non-trivial. We show that it is not trivial by lifting it, showing that it must be equivalent to [w_m] for odd m and hence cannot be [w_0].

@davidhand9721 4 месяца назад

I think your definition of f in the antipodal points theorem is wrong; it should be S2 -> R, not R2, right? The example you gave was temperature, which is decidedly in R. Shouldn't the g in your proof also be R2 -> S1? S2 doesn't even fit in R2.

@abebuckingham8198 4 месяца назад

He's proving it for S1 -> R not S2 -> R2. That's what allows him to use the intermediate value theorem. The higher dimensional proofs take a little more work.

@richardchapman1592 6 месяцев назад

What happens if the winding number helix is observed from an S1 in a constant force field.

@abebuckingham8198 4 месяца назад

The winding number is a topological property and independent of the embedding. Since all elements of the fundamental group are loops the work done along them is always zero in a conservative vector field but that's more of a physics thing and requires a metric.

@richardchapman1592 4 месяца назад

@@abebuckingham8198 not really grasped much of this yet but surely a metric is implied when there is any homo....(?) with R1.

@blizzard_inc Год назад

i'd like to point out that you might have accidentally misrepresented constructive math and constructive mathematicians. in constructed math, you don't say "the law of excluded middle is false", or "proof by contradiction not a real proof". rather, you just don't *use* the law of excluded middle. As a matter of fact, in constructive math you can prove that the law of excluded middle is not *not* true, which is done by contradiction. there is only a specific kind of proof by contradiction which is not allowed, and that is a proof where you suppose "not A", arrive at a contradiction, and then conclude "therefore, A". this kind of proof relies on the law of excluded middle, or equivalently, double negation elimination. however, another kind of proof by contradiction *does* still hold, where you suppose "A is true", arrive at a contradiction, and then conclude "not A". as a matter of fact, this is how falsehood is defined; "A implies contradiction" is equivalent to "not A". Also, calling constructive mathematicians kooky is uncalled for. you wouldn't call a mathematician who would rather not use the Axiom of Choice that either. you might want to watch this video on the topic, as it points out wonderfully what constructive math is and isn't about: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-21qPOReu4FI.htmlsi=kG_StLZwHi3zhr8q

@MathatAndrews Год назад

I was being lighthearted - your point is well taken.