Thank you very much for your lecture series! You absolutely gave me the intuition and hands-on experience quickly in algebraic topology which I have been eagering for a long time. Please please keep doing it!!!

B^3 is better b/c we need the ball to be open, while D^3 is closed. But that doesn't matter!

17:20 For your context matrices are just thought as 2d arrays with a defined componentwise vector addition and scalar multiplication to form a vector space structure upon them. So it would make sense that the determinant map det: M_{n} -> R would not be a linear map. However, the determinant map is actually defined to be a alternating multilinear map with n arguments from the vector space R^n to R. In the original context of matrices, they are just an n-tuple of n-dimensional vectors (that's why determinants are only defined for square matrices). So adding those two "identity matrices" doesn't make any sense, because they are just 2-tuples of 2 dimensional vectors.They have no implied vector space structure on them and therefore your counter that the determinant is not a linear map is not well justified.

wait, can we say h(s) = -h(s + 1/2) and h is continuous so h takes 0 somewhere, but h = g \circ l, which means g takes 0 somewhere, then we also get a contradiction??

writing backwards always amazes me great vid

rewrite of the multi-index hat joke: there's a bench that seats two people. one mathematician comes and sits down on the bench. then a second mathematician sees his colleague and joins him on the bench. then a third mathematician walks up to join them and says "don't worry, I brought a hat"

soos stole mysterious mushrooms from waddles 😭😭🙏🏻🙏🏻

Beautifully explained!!!

40:35 i think it is linearly dependent because we can swap the sin^2 (x) for (1-cos2x)/2 and then solve, giving us C2 = 0 and C1 = -C0 (tanx). can someone help me out with this?

In all my years of physics and math I have never seen someone so simple and humble enough o interact and level with the class on these topics that students find hard to grasp

Uhm maybe the magician was schrodinger and he teleported the particles

According to several definitions of the Euler Characteristic of a Torus; it says is 0. However if you ( Vertices = 0 - Edges = 0 + Faces = 1) you get a 1?.

The initial vertex you remove on the inner star is of degree 3 is it not? I'm a little confused as to why you were able to do that, when you said previously the vertices had to be of degree 2.

mask clown

At 36:40 There is a technicality he didn't mention: that each loop is homotopic to a loop that misses at least one point. This ensures you deal with the case of space feeling curves. Here is how the proof goes: Consider a continuous curve from [0,1] that reaches every point. Choose to cover your sphere with finitely many disks and take their preimage with respect to the curve, this will by compactness of [0,1] and continuity of the path, this results in a finite number of open intervals, the image of a single such interval being a connected path contained in its respective disk. You can "straigten out" each such path (for example, moving it to a shortest path between the two endpoints). Repeat the operation for the finite number of intervals and we have shown that this space filling curve is homotopic to a map that certainly misses at least one point. You can then apply the argument in the video.

This was delightful

Biggggg.

I only clicked to see if you were a lady. Of course I was disturbed to see you were not. There's nothing you can teach anyone.

But what about Field extension? I know in Real Analysis, the polynomial X³ + x = 2 has exactly one real solution. But don't say that it has exactly one solution 😂😂😂

Excellent quality. Congratulations.

Great explanation. Thank you

Clarification: In the computation of the cellular differential d_1, we don't think of the coefficients as degrees of maps between spheres, because the 0-cells are just points, and not 0-spheres. The coefficients are just computed by "head minus tail" according to the orientation of the 1-cell. In particular, if there is only one 0-cell then d_1=0 always.

20 minutes in, and I can already tell that I gel with style of teaching. It builds up concepts step by step, stopping just before the next step, allowing the students to intuit the logical conclusion that the next concepts or overall rules form. This is something that can only be possible when someone both understands the concept well, and puts in a lot of work to make sure it can be explained simply. Thanks for this.

So, the Zeno paradox deals with physical objects. The premises of the "calculus" reasoning is that space and time can be divided "ad infinitum". In reality, when we reach planck scale, this is not certain. IMO, a "granularity" of space and time at a fundamental level will solve the Zeno paradox. Any take?

why is the map at 37:14 injective? Injectives maps need not be taken to injectives by a functor

these lectures are so underated! Truly a blessing to have come acrossed them. I understood things that I couldn't with my university lecturer.

It's straight to the point when it comes to intuition, well-done!

Awesome video! But I don't understand where the S1 in the X/B example comes from. If B were a 2-disk contained in X, would X/B then become S2vS2?

Man this is amazing. Where can I find exercise for this?

When will it continue?

An One hour and a half video is better than my Top2/Alg Top prof ever could be.

The explanation you gave in the beginning about why pi_1 isn't abelian is incorrect and may confuse students. It is clear that a and b^{-1} do not commute because they aren't even loops in the first place, so they don't even represent elements in pi_1. Instead you would have to look at a pair of loops, say ab^{-1} and bc^{-1}, and discuss why they don't commute in pi_1.

I start watching your series here. I learnt the relationship of cohomology and homology. I will watch the whole series because it helps a lot and saves me some time to read books later. Good work!

Ones a Japanese always a Japanese 😅 very powerful

Isn't that W example near the end hamiltonian? If you take the bottom 2 vertices as your u and v then it 6 which is greater then 5, I've found quite a few hamiltonian circuits as well.

Great man Thanks

Great vdo

I don't get it

The board!!!! please do not put it that way, it is way difficult to understand. It is way better if the board is behind you

This feels like combining the relaxing parts of asmr and the stress that I need to know this before my exam. 10/10

Good job❤

Has anybody tried projecting winding numbers on to a unit circle orthogonal to π[S¹]? You would need a function that took zero to π/2 such that it converges to 1 using an infinite series which may be quantised by values of wn according to the normalised series chosen?

Intuitively the paths of one dimension on a 2dim surface look sensible for π[S¹] when stretched into 3dim. Is a 2dim path a vector and it possible to generalise the meaning of path? Into something of ndimensions on an m>n brane in an s>mdimensional space?

Thank you very^n much!!!!

How does the figure 8 knot is 5-colorable exactly? Why is 5-colorable if you can only paint it with four colors?

How come the answer is 16 Labeles and not 12? They aren't the same shape... Do we just count them the same because of the amount of vertexe's they have is both 4?

A little slow on the uptake of this due partially to social factors (a mixed blessing) but so far I see the 'paths' as precursors of vectors and have never seen a treatment of >1dim vectors. Also we have the use of Z and R closed by using mods. Cannot see how irrationals with their infinite series definitions can be included in the metrics of mod fields.

Can see this pictorially using a 1dim path on a 2dim surface in 3dim. In larger dimensions not sure how an extrapolation is made using an analogy of an n dimensions path on a pdim brane in an sdim space.

0:09 What Kurt Godel said while completing his incompleteness theory.

Idk what other people think but topology in my eyes is the most beautiful form of magic of them all