Last weeks were very rough for me on my algebraic topology course. I honestly felt incredibly discouraged and sad, I wasn’t getting this theorem at all. This video helped me a lot :)
great video, perhaps you should mention in the last example, that = Z x Z where Z are the integers with pointwise addition. One can see this by taking these elements of the free groups from example (b) and commuting them such that we only get a^m * b^n for m,n in Z --> (m,n) in Z x Z.
at 17:14, I don't understand why we can push the loop in (U intersection V) to boundary? we have inclusion map and they induce the homomorphism, but how we know that homomorphism act in that way like we can push the loop to boundary?
can you give a small extra explanation about the homomorphism g? I see why the loop that generates pi1(U intersect V), c, is kinda retracted out to be b-1a-1ba, but when I'm thinking about V as the 8-space, and about the fact that the pair of segments a are congruent (same for b), it seems to me that g is taking c to ba. My point is that I can't match the visual explanation to the rigorous way of computing g. Thanks a lot in advance.
Maybe a more convincing demonstration is to draw the small circle on the surface of a torus, then expand it out and see that eventually you get something which does b^{-1}a^{-1}ba in the figure 8. For example, this video shows it quite convincingly czcams.com/video/P5RLa3g-nwQ/video.html In case you want to write out a more detailed proof, here's how to go about it. Warning: To be fully rigorous we'd need to be more careful about basepoints: the natural basepoint for b^{-1}a^{-1}ba is at the corner of the square, but for Van Kampen we need to take a basepoint in the intersection (i.e. the pink circle). Ignoring this subtlety, let's just convince ourselves that the pink loop in the video is freely homotopic (in the complement of U) to the commutator. Consider the linear deformation from the circle to the square (i.e. move every point in the circle along a straight line with constant speed until it hits the square; of course the speed will depend whereabouts you are on the circle, so I don't mean constant I'm that sense, but in the sense that each point will move without acceleration). You could write this down explicitly in coordinates (exercise!) and that would give you the desired free homotopy. Now the question is: why is that square path b^{-1}a^{-1}ba and not ba? Well the square path is clearly subdivided into four segments, each going from a corner to a corner (and remember the corners all get identified to a single point in the torus) so it's a concatenation of four loops. The order of the loops matters because pi_1 of the figure 8 is nonabelian. The direction of the loops also matters. Indeed, the only difference between the explicit loop you get from the aforementioned free homotopy and the honest concatenation b^{-1}a^{-1}ba is the parameterisation (which can be fixed by a homotopy that preserves the image of the loop, just like the proof of associativity of concatenation up to homotopy in pi_1).
... Now let's quickly come back and query about the point I made about basepoints. Of we show that the pink loop is freely homotopic to the commutator then the pink loop is actually homotopic to a conjugate of the commutator (and, depending on how you choose paths to identify your basepoints, it might in reality be a nontrivial conjugate: in this particular picture there's an obvious choice of very short paths which makes it honestly equal to the commutator, but there's no reason you need to make that choice). So to apply Van Kampen you would use g(c)=f^{-1}b^{-1}a^{-1}baf for some f. Well... that gives you the same answer, because in Van Kampen's theorem you quotient out by the normal subgroup generated by g(c)f(c)^{-1}, i.e. the subgroup generated by this guy and all its conjugates!
Jonathan Evans thanks alot for the comprehensive arguments. I think what made me realize I’m approaching it wrong was the fact that the pi1 of figure 8 is indeed non abelian (surprisingly enough pi1 of the torus “achieves” the abelian property).
This is the most clear explanation I've seen on this subject. Thanks a lot!
Seconding this. Thank you!!!
Last weeks were very rough for me on my algebraic topology course. I honestly felt incredibly discouraged and sad, I wasn’t getting this theorem at all. This video helped me a lot :)
Glad you found it useful! I hope the rest of your course goes better.
great video, perhaps you should mention in the last example, that = Z x Z where Z are the integers with pointwise addition.
One can see this by taking these elements of the free groups from example (b) and commuting them such that we only get a^m * b^n for m,n in Z --> (m,n) in Z x Z.
I thought the same!
Thank you for this video. You explained the theroem very clearly and the examples helped immensely!
Great, thank you so much!! This is very clear and understandable.
Thank you for the clear examples!
Thank you for this beautifull explanation!
Thanks for making this video! :)
This is great!
hope i'll get u as a teacher next year ;)
nice video!
at 17:14, I don't understand why we can push the loop in (U intersection V) to boundary?
we have inclusion map and they induce the homomorphism, but how we know that homomorphism act in that way
like we can push the loop to boundary?
Thank you so much, this vid is the best on this subject, but the voice is too low
can you give a small extra explanation about the homomorphism g? I see why the loop that generates pi1(U intersect V), c, is kinda retracted out to be b-1a-1ba, but when I'm thinking about V as the 8-space, and about the fact that the pair of segments a are congruent (same for b), it seems to me that g is taking c to ba. My point is that I can't match the visual explanation to the rigorous way of computing g. Thanks a lot in advance.
Maybe a more convincing demonstration is to draw the small circle on the surface of a torus, then expand it out and see that eventually you get something which does b^{-1}a^{-1}ba in the figure 8. For example, this video shows it quite convincingly czcams.com/video/P5RLa3g-nwQ/video.html
In case you want to write out a more detailed proof, here's how to go about it. Warning: To be fully rigorous we'd need to be more careful about basepoints: the natural basepoint for b^{-1}a^{-1}ba is at the corner of the square, but for Van Kampen we need to take a basepoint in the intersection (i.e. the pink circle). Ignoring this subtlety, let's just convince ourselves that the pink loop in the video is freely homotopic (in the complement of U) to the commutator. Consider the linear deformation from the circle to the square (i.e. move every point in the circle along a straight line with constant speed until it hits the square; of course the speed will depend whereabouts you are on the circle, so I don't mean constant I'm that sense, but in the sense that each point will move without acceleration). You could write this down explicitly in coordinates (exercise!) and that would give you the desired free homotopy. Now the question is: why is that square path b^{-1}a^{-1}ba and not ba? Well the square path is clearly subdivided into four segments, each going from a corner to a corner (and remember the corners all get identified to a single point in the torus) so it's a concatenation of four loops. The order of the loops matters because pi_1 of the figure 8 is nonabelian. The direction of the loops also matters. Indeed, the only difference between the explicit loop you get from the aforementioned free homotopy and the honest concatenation b^{-1}a^{-1}ba is the parameterisation (which can be fixed by a homotopy that preserves the image of the loop, just like the proof of associativity of concatenation up to homotopy in pi_1).
... Now let's quickly come back and query about the point I made about basepoints. Of we show that the pink loop is freely homotopic to the commutator then the pink loop is actually homotopic to a conjugate of the commutator (and, depending on how you choose paths to identify your basepoints, it might in reality be a nontrivial conjugate: in this particular picture there's an obvious choice of very short paths which makes it honestly equal to the commutator, but there's no reason you need to make that choice). So to apply Van Kampen you would use g(c)=f^{-1}b^{-1}a^{-1}baf for some f. Well... that gives you the same answer, because in Van Kampen's theorem you quotient out by the normal subgroup generated by g(c)f(c)^{-1}, i.e. the subgroup generated by this guy and all its conjugates!
Jonathan Evans thanks alot for the comprehensive arguments. I think what made me realize I’m approaching it wrong was the fact that the pi1 of figure 8 is indeed non abelian (surprisingly enough pi1 of the torus “achieves” the abelian property).
nice video, but tooo small audio