I start watching your series here. I learnt the relationship of cohomology and homology. I will watch the whole series because it helps a lot and saves me some time to read books later. Good work!
I have an additional question, how is your boardwork so clean? Do you have a rough idea of how much you are going to fit in one board? I just wing it and it ends up being very messy.
I sketch notes on my iPad ahead of the lecture, giving me a rough idea of how I'll arrange the content on the boards. But I also do plenty of winging it.
I am from India and have advanced topology along with Algebraic Topology in my MSc last year . Your videos are amazing and helping me a lot to overcome the difficulties in Algebraic Topology .
Great Video! But what does it mean for the group hom(Z,Z) to be isomorphic to Z? Aren’t they groups of maps to groups of functions? Sorry if this is a dumb question, as I have a weak abstract mathematical background
hom(Z,Z) is the group of homomorphisms from Z to Z. As explained in the video each homomorphism f in Hom can be characterized by an integrer via f(1) =n. Therefore there is a bijection n f between Z and Hom(Z,Z).
Not a dumb question; they’re isomorphic but certainly not ‘equal’. One consists of a set of integers, the other a set of maps from the integers to the integers. To add to the above comment: they’re not just in bijection (non-isomorphic groups can certainly be in bijection); the natural map that bijects them is a homomorphism since f_(n + m) = f_n + f_m since they agree on 1 (they both send it to n + m).
Sorry professor late this time not gonna watch this video right now currently enjoying my vaccation travelling vietnam. My favourite hobbie after math is travelling. To this date i have travelled 2 countries internationally thailand and vietnam. Being in vietnam i feel like a billionare. Also thanks for the lecture in advance.
30:48 Yes, you can check it if you like… but you can instead just remember that contravariant represented functors preserve coproducts 😉 (a fact I learnt only very recently!).
I start watching your series here. I learnt the relationship of cohomology and homology. I will watch the whole series because it helps a lot and saves me some time to read books later. Good work!
These series just keeps getting better and helping me understand topological data analysis and geometric deep learning. Have a wonderful Easter Sir.
Aahh! Finally a lecture on cohomology! Really excited to learn this from you! Thank you professor ♥️
This series has been helping me ton with my PhD oral exam preparations! I just hope a couple more get released before Tuesday... :D
Unfortunately we won't have the next lecture until later next week. All the best on your oral exam and PhD. You'll rock it! 👊🏼
How many more videos will there be in this series? I hope we get to see more of chapter 3 (and 4), these lectures are gold..
Around 3 more. We'll spend the next couple lectures on the cup product.
I have an additional question, how is your boardwork so clean? Do you have a rough idea of how much you are going to fit in one board? I just wing it and it ends up being very messy.
I sketch notes on my iPad ahead of the lecture, giving me a rough idea of how I'll arrange the content on the boards. But I also do plenty of winging it.
I am from India and have advanced topology along with Algebraic Topology in my MSc last year . Your videos are amazing and helping me a lot to overcome the difficulties in Algebraic Topology .
absolutely love these lectures!
Finished it! It was an amazing lecture! ♥️
These videos are amazing. Thank you very much !!
No, they aren't.
Great Video! But what does it mean for the group hom(Z,Z) to be isomorphic to Z? Aren’t they groups of maps to groups of functions? Sorry if this is a dumb question, as I have a weak abstract mathematical background
hom(Z,Z) is the group of homomorphisms from Z to Z. As explained in the video each homomorphism f in Hom can be characterized by an integrer via f(1) =n. Therefore there is a bijection n f between Z and Hom(Z,Z).
Not a dumb question; they’re isomorphic but certainly not ‘equal’. One consists of a set of integers, the other a set of maps from the integers to the integers. To add to the above comment: they’re not just in bijection (non-isomorphic groups can certainly be in bijection); the natural map that bijects them is a homomorphism since f_(n + m) = f_n + f_m since they agree on 1 (they both send it to n + m).
Much appreciated thing. Thank you Professor! 🙂
Sorry professor late this time not gonna watch this video right now currently enjoying my vaccation travelling vietnam. My favourite hobbie after math is travelling. To this date i have travelled 2 countries internationally thailand and vietnam. Being in vietnam i feel like a billionare. Also thanks for the lecture in advance.
Enjoy your travels!
Thanks for this!
Anthony 🙇
Waiting for The Sun, waiting for next lecture😊
30:48 Yes, you can check it if you like… but you can instead just remember that contravariant represented functors preserve coproducts 😉 (a fact I learnt only very recently!).
really, its unexpected.