Algebraic Topology 5: Homeomorphic Spaces have Isomorphic Fundamental Groups

[@carl6167]

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.

[@nahomdejene903]

I love this channel so much❤😊. Keep doing all pure mathematics courses👍

[@OedipusLee]

Thank you for this amazing course.

[@-minushyphen1two379]

at 36:40, what about space-filling curves? Wouldn’t they be continuous surjections from the closed unit interval to the sphere?

[@enzoregna3521]

No

[@enpeacemusic192]

The inverse wouldn't be continuous

[@richardchapman1592]

Guessing that in R3 space can be filled continuously with membranes from R2. That would involve a discontinuity at infinity when mapping to S1.

[@carl6167]

You are right! There is a technicality he seemingly forgotten. If you choose your 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, with each of their image being a connected path in the respective disk. You can "straigten out" each such path. That way you show that this space filling curve is homotopic to a map that certainly misses at least one point. You can then apply his argument.

[@JennieXie-rg2xd]

amazing!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@richardchapman1592]

Is learning a meaningful function topologically and in what sense can it be be considered continuous?

[@abebuckingham8198]

A function must have a domain and codomain to be defined. I don't see any obvious domain or codomain for learning.

[@richardchapman1592]

@@abebuckingham8198 could look at that the words used in various topics of learning to define several domains. The scopes of those words would have to be defined by other words and not sure if hierarchy could be handled with winding numbers in Z.

[@abebuckingham8198]

@@richardchapman1592 Math is a social discipline and it's up to you to explain your ideas to others in a way that's convincing. If you think there's a domain then articulate it. Vague notions of what might be possible aren't enough.

[@richardchapman1592]

Apologies for being inadequate. I can only attend to thought matters sporadically due to dissemblement by a range of other matters associated with what is loosely termed, reality. Am trying to plough through the Andrews lectures though and maybe will be, more cogent at a later date. I do however appreciate your feedback at my confusions of concept. Thanks.

[@richardchapman1592]

@@abebuckingham8198 My original query concerned the extent of which the use of words to describe concepts may be considered as elements of topological groups of those concepts. Whether the notions of homology and rings apply rather depends upon if there are inverse functions that take similar concepts from one to the other and back. One responder to the lectures has already briefly alluded to this.

[@SphereofTime]

0:07

[@MarioPucci_mamio]

As for the last proof, wouldn't one actually need to prove that there are no such phi and psi while you only proved the two shown are not what you'd be looking for?

[@hbghlyj]

consider a point and R.

[@xanderlewis]

You're right, but the two shown are sufficiently general (because S^2 has a lot of symmetry) so as to be equivalent to having done the proof for 'any two maps'. It's like starting a proof by saying 'without loss of generality, let's assume we have...'. The only choice we have in such maps is which point of S^2 to pick out with psi.

[@markuseisgruber]

Biggggg.

[@SphereofTime]

14:07