AlgTop2: Homeomorphism and the group structure on a circle

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This is the first video of the second lecture in this beginner's course on Algebraic Topology. We give the basic definition of homeomorphism between two topological spaces, and explain why the line and circle are not homeomorphic.

Then we introduce the group structure on a circle, or in fact a general conic, in a novel way, following Lemmermeyer and as explained by S. Shirali.

This gives a gentle intro to the definition of a group. It also uses Pascal's theorem in an interesting way, so we give some background on projective geometry.

This lecture is part of a beginner's course in Algebraic Topology given by N J Wildberger at UNSW.

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Another way to see this: continuous functions take points close to each other (in the domain; in this case, the circle) onto points close to each other (in the range; here, it is the half-open line.) Now look at a small interval containing 0 (=1) on the circle. The part of the interval "above" 0 gets mapped closer to 0; but the part "below" the interval gets mapped closer to the other end of the line, near 1. So a small interval on a circle will get mapped to a "large" interval [0, 1).

nbk
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Excellent video!
Hope Prof Wildberger can make more of these videos on undergraduate mathematics!

Mathtuition
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Take such a hexagon on a circle with opposite sides parallel, mark an arbitrary vertex 'O', and mark the rest of the vertices, clockwise or counterclockwise, as B*C, A, B, C, A*B, then back to O. You'll see that the segment from A to B*C is parallel to the segment from A*B to C, so they will have the same product, and associativity holds.

albertsteiner
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Hi ctressle Yes this is definitely a taste of algebraic geometry: an analogous group operation on a cubic curve is one of the central topics in that subject.

unswelearning
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sir, it feels good that u feel that Indian high school teaching is of very high standard(coz i m an Indian!!)....but the competition to study in a good college is way tougher than in australia....only students in an around 99.5 percentile get to study in good is crazy...isnt it?btw i m a stud mad about modern physics and from that ....in topology....take a bow, sir!!!

koustubh
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So much useful information that was completely understandable.

TheVopepigota
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@tinlyx Since in a circle, the points 0 and 1 were the same, you are unable to "go back" from the circle to the line in a continuous way. That is why the inverse fails to be continuous.

nbk
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If you denote the identity by 0, then it would be more natural to call the operation an addition. Also, in group theory it is quite common to call a commutative group operation an addition. (see article "Elementary Group Theory" on wikipedia)

howeworth
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Does the group presented here match the standard multiplicative group on the unit circle in the complex plane?

robfrost
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in the very beginning when we started group structure on a circle..we kinda did geometrical thing to obtain (a*b)*c=a*(b*c) and then afterwards we said that we need pascals thm for that associativity. i am getting confused between these..was the first one not enough or that was just for circle and pascals was for any conic(is this the difference?)...please explain the difference

RizwanAli-ibmg
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This stuff is awesome, glad I found it! The group structures on the conics, is that also a taste of algebraic geometry?

ctressle
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Around 26:00 he's talking about inverses, but what would the inverse of the point exactly across from the origin be? Is it it's own inverse?

theleastcreative
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The circle group is isomorphic to R/Z. I wish the professor had arrived at this result and talked some more about the unitary group.

nbk
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The series makes topology as familiar as the air around the watcher.

hamidrezarezazadeh
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Couldn't the non-homeomorphism from the closed walk to S1 result in the non-uniqueness of solutions to the inverse function? If we map f:A->S, then for any argument of A it has a corresponding point in S but not vice versa; f inverse is 2pi periodic.

peterclark
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why is A*A a tangent and not "all the lines that go through A"? I was a bit surprised because A' is not A, however close they might be, and we do know A.

SalsaTiger
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why define all these as mulitpcation, can we define all these as addition?

ybc
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Does anyone solve the problems? Please show me. Thanks.

lukealucard
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Anyone can help me to solve the problems? I'm starting to study algebraic topology =)

LuisRamirez-gcds
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Does anyone think that he sounds a little bit like Alex Trebek?

julie