Glad you found it helpful, but be careful: the torus has no boundary! It is a disc glued to S^1 v S^1, but the S^1 v S^1 isn't 'boundary' (in the usual sense of the word). If you draw the S^1 v S^1 on the surface of the torus, you will hopefully see what I mean.
@@jonathanevans27 Yes, I understand what you mean - the torus in R^3 has no boundary! Poor choice of words. What I initially meant was that if you remove the 2-cell from your CW decomposition of a torus, you are left with a 0-cell and two 1-cells, which is homeomorphic to S1 V S1.
at 11:16, and before I didn't understand how that is disjoint union, X is two point and D^1 is closed interval, they have two boundary points as common right? how that is disjoint union
@Sayan Ghosh: it's the finest topology for which these inclusions are continuous (i.e. If you add in any more open sets then the intersection with X_i will fail to be continuous). Most people would call that strong. There was some discussion of the historical reasons for the terminology here: math.stackexchange.com/questions/3234052/confusion-about-topology-on-cw-complex-weak-or-final
It's Z: you can cover it with two open sets and then apply Van Kampen. Sit your circle in the xy plane. Take one of the sets (U) to be a neighborhood of a vertical line passing through the circle. U is contractible. Its complement V is homotopy equivalent to a torus and the intersection is homotopy equivalent to a circle. Using Van Kampen, this means you get a quotient of Z^2 which kills off one of the factors, giving Z.
@@Fatima-fn6jj You're right, but this is a common abuse of language in the context of Van Kampen's theorem. I mean: take a neighbourhood of the complement so that they do overlap. To see the relationship with a torus, notice that the whole situation is rotationally symmetric around the vertical axis. Take a vertical plane and slice R^3 using this plane. The circle hits the plane at two points, the vertical line sits in the plane and separates these points. Now U is obtained by rotating a neighbourhood of this vertical line around the line itself (giving a solid cylinder) and V is obtained by taking a half-plane minus a point and rotating it around. A half plane minus a point is homotopy equivalent to a circle, and the rotation turns it into a torus.
At 1:20 that is a definition of a sphere, not a disk, right? For disk it should be |x|
I cannot put into words how awesome this is. Is the uploader God? Cause he is doing the Lord's work.
Just awesome.
Its feels me like complete use of the 20:13 minutes of my life.
Thanks a lot Sir.
You are very welcome!
Thank you very much ! I understood in 20 minutes what I couldn't understand for three months...
Glad you found it useful!
Same for me
thanks a lot! you are good at explaining things
Thanks a ton! really was struggling to understand this. This really helped!
You are great teacher! Thanks a lot!
Really clear exposition.
Very good video !
Thanks a lot!
I finally understand why the boundary of a torus is S1 V S1. Thank you!
Glad you found it helpful, but be careful: the torus has no boundary! It is a disc glued to S^1 v S^1, but the S^1 v S^1 isn't 'boundary' (in the usual sense of the word). If you draw the S^1 v S^1 on the surface of the torus, you will hopefully see what I mean.
@@jonathanevans27 Yes, I understand what you mean - the torus in R^3 has no boundary! Poor choice of words. What I initially meant was that if you remove the 2-cell from your CW decomposition of a torus, you are left with a 0-cell and two 1-cells, which is homeomorphic to S1 V S1.
@@demetrasiountris8260 Exactly! I thought that that was what you meant.
Thank you !!!
Great video but the sound level is too low.
at 11:16, and before I didn't understand how that is disjoint union, X is two point and D^1 is closed interval, they have two boundary points as common right? how that is disjoint union
THANKS!
Your disc should be less than or equal to
Woops, thanks! Well spotted.
Great and clear exposition
Thank you
Is the definition in the beginning of Dk correct? Isnt it less than 1?
Yes, you're right.
@@jonathanevans27 great video btw
is it called the weak topology because its the weakest topology on X such that the inclusion maps from each X_i to X is continuous?
@Sayan Ghosh: it's the finest topology for which these inclusions are continuous (i.e. If you add in any more open sets then the intersection with X_i will fail to be continuous). Most people would call that strong. There was some discussion of the historical reasons for the terminology here: math.stackexchange.com/questions/3234052/confusion-about-topology-on-cw-complex-weak-or-final
@@jonathanevans27 Ahhh yes its the finest. Sorry for my incorrect use of words
妙啊！torus那个地方太妙了！
Why you define disk with circle equation? 0:55 So is it a disk or circle?
Indeed, it's a typo (or the videographic equivalent thereof).
@@jonathanevans27 regardless, this video is great! Thank you for your time and effort 😊
Dear ,
What is the fundamental group of R3-S1 ?
It's Z: you can cover it with two open sets and then apply Van Kampen. Sit your circle in the xy plane. Take one of the sets (U) to be a neighborhood of a vertical line passing through the circle. U is contractible. Its complement V is homotopy equivalent to a torus and the intersection is homotopy equivalent to a circle. Using Van Kampen, this means you get a quotient of Z^2 which kills off one of the factors, giving Z.
@@jonathanevans27 Thank you so much for your reply god bless you
But if V is the complement of U then the intersection is empty . And how can V be homotopy eq to a torus it seems its not that easy .
@@Fatima-fn6jj You're right, but this is a common abuse of language in the context of Van Kampen's theorem. I mean: take a neighbourhood of the complement so that they do overlap. To see the relationship with a torus, notice that the whole situation is rotationally symmetric around the vertical axis. Take a vertical plane and slice R^3 using this plane. The circle hits the plane at two points, the vertical line sits in the plane and separates these points. Now U is obtained by rotating a neighbourhood of this vertical line around the line itself (giving a solid cylinder) and V is obtained by taking a half-plane minus a point and rotating it around. A half plane minus a point is homotopy equivalent to a circle, and the rotation turns it into a torus.
@@jonathanevans27 You are amazing , thank you sir