Algebraic Topology 20: Introduction to Cohomology

  Рет қаралды 4,174

Math at Andrews University

Math at Andrews University

Күн бұрын

Playlist: • Algebraic Topology
We give a brief recap of homology and then show how dualizing the chain complex by Hom(--,Z) gives a cochain complex with coboundary maps that we use to calculate cohomology. We show that for finitely generated chain groups, we can calculate the cohomology in terms of the homology groups. Then we dualize with other coefficient groups G and discuss the universal coefficient theorem for cohomology.
Presented by Anthony Bosman, PhD.
Learn more about math at Andrews University: www.andrews.ed...
In this course we are following Hatcher, Algebraic Topology: pi.math.cornel...

Пікірлер: 25
@ikechukwumichael1383
@ikechukwumichael1383 5 ай бұрын
These series just keeps getting better and helping me understand topological data analysis and geometric deep learning. Have a wonderful Easter Sir.
@swaruppaul4417
@swaruppaul4417 5 ай бұрын
Aahh! Finally a lecture on cohomology! Really excited to learn this from you! Thank you professor ♥️
@Oreo_od50
@Oreo_od50 5 ай бұрын
This series has been helping me ton with my PhD oral exam preparations! I just hope a couple more get released before Tuesday... :D
@MathatAndrews
@MathatAndrews 5 ай бұрын
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! 👊🏼
@AzizBouland
@AzizBouland 5 ай бұрын
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..
@MathatAndrews
@MathatAndrews 5 ай бұрын
Around 3 more. We'll spend the next couple lectures on the cup product.
@UncoveredTruths
@UncoveredTruths 5 ай бұрын
absolutely love these lectures!
@swaruppaul4417
@swaruppaul4417 5 ай бұрын
Finished it! It was an amazing lecture! ♥️
@sayanghosh6544
@sayanghosh6544 5 ай бұрын
Much appreciated thing. Thank you Professor! 🙂
@gabesorci1638
@gabesorci1638 5 ай бұрын
These videos are amazing. Thank you very much !!
@sallylauper8222
@sallylauper8222 5 ай бұрын
No, they aren't.
@ompatel9017
@ompatel9017 5 ай бұрын
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.
@MathatAndrews
@MathatAndrews 5 ай бұрын
Enjoy your travels!
@xanderlewis
@xanderlewis 2 ай бұрын
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!).
@aviralsood8141
@aviralsood8141 5 ай бұрын
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.
@MathatAndrews
@MathatAndrews 5 ай бұрын
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.
@aviralsood8141
@aviralsood8141 5 ай бұрын
Thanks for this!
@IshouldGetQualified
@IshouldGetQualified 5 ай бұрын
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
@thomasbastos3869
@thomasbastos3869 5 ай бұрын
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).
@xanderlewis
@xanderlewis 2 ай бұрын
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).
@broccoloodle
@broccoloodle 5 ай бұрын
Anthony 🙇
@algebraist_24
@algebraist_24 5 ай бұрын
Waiting for The Sun, waiting for next lecture😊
@Desidarius_Erasmus99
@Desidarius_Erasmus99 5 ай бұрын
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 .
@depressedguy9467
@depressedguy9467 5 ай бұрын
really, its unexpected.
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