I swear these lectures are god gifted

I've noticed that the universal cover for the wedge of two copies of S_1 is literally just the cayley graph for the free group Is this a coincidence? Because the way you drew R as a helix (with "vertices" corresponding to the base point in S_1) is technically just the cayley graph of , or Z.

Great lecture. Its very clear. He/ him is less common for referring to objects than she/ her, because relationships (ie in graph theory) are feminine by default. But, that has no affect on the topic. I just thought it was interesting.

Probably the nost accesible way to learn algebraic topology.

Amazing

An absolute gem of lecture series❤

Tried to do an Escher staircase style version of winding number helix going from one chirality to another. The attempt at representing an R3 object in R2 only demonstrates itself at the interstes of an even sided diagram on paper. May be fun for artwork when unable to absorb algebraic topology swiftly.

Thank you for an amazing course! I wonder if you have any lecture notes?

@MathatAndrews

4 ай бұрын

Thanks! Unfortunately I don't have lecture notes - though in this series I am following Hatcher's text closely, which is linked to in the video description.

I dont understand the example at around minute 29, the figure on the right, with two points and 4 paths: Why can we call two paths a (respectively b) when they are not the same?

@dlitvinov28

2 ай бұрын

It's not two actual 'a' paths, but rather two paths that both cover the 'a' path in the original space X.

@greenland8376

2 ай бұрын

@@dlitvinov28 i see thanks!

Your illustration of the one with four loops and 3 vertices has totally lost me. The vertex y0 has two arrows labeled b, one ingoing and one outgoing, on the left. How can one of them be b and the other be b inverse if they are not just one arrow? The inverse of an arrow goes back along the arrow, right? That's part of the definition of a group, right? Am I overthinking this? Does everything work if I just mentally remove one of the arrows? No, I don't think so, if b squared is a loop; what am I missing here? Wait, wait, I think I've got it now. The inverse of b _is_ b, you just made the second arrow to make it look like a circle, showing that by0 = x0 and bx0 = y0. It still seems like this would complicate contracting some loops, though. b b inverse ought to be contractable, right? Now I'm stuck on the fact that a does something totally different at x0 than y0. Is that just a side effect of the way you've drawn it? Like, either way, a squared does get you back to y0 or x0. Does this have something to do with the fact that we're looking at this system as a double cover of the simpler space where both a and b lead to the same point?