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100 days of productivity (337/primes)
I continued with the exercises on (co)limits, but then got stuck at an obvious isomorphism for most of the day. This problem was eventually solved with the help of a random internet stranger. (It may be more accurate to say that a symptom was solved, as the real problem is my stupidity.)
Artists of the day: Run the Jewels
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100 days of productivity (311/primes)
I continued reading through the chapter on (co)limits. I turns out that I’m surprisingly bad with how limits and colimits interact.
My cold is mostly gone, so that my tiredness and feeling-like-shit-ness are back to the usual level; I therefore hope that I will actually get something done tomorrow.
100 days of productivity (331/primes)
I did some more exercises on (co)limits.
Artist of the day: Aesop Rock
Song of the day: None Shall Pass by Aesop Rock
100 days of productivity (317/primes)
I started doing the exercises on (co)limits and started preparing the algebra review course.
Albums of the day: The Inevitable Rise and Liberation of NiggyTardust! by Saul Williams, The Impossible Kid by Aesop Rock, The Life of Pablo by Kanye West
It has long been appreciated that the theory of colimits captures essentially everything about constructions by gluing - you use coproducts to add new bits and coequalizers to attach them - but I think they can also do the same for course-graining. Coproducts collect degrees of freedom together and coequalizers “blur them together”, so to speak. Nobody seems to be working on making this precise. I will probably think about it a lot more.
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This person not only indexed all of the videos, but also wrote down summary notes of each of the episodes! Big thanks to Brent Yorgey!
The colimit of subgroups of \(\mathbb{Z}\)
Compute the colimit (in the category of abelian groups) of the diagram \[\mathbb{Z} \overset{2!}{\to} \mathbb{Z} \overset{3!}{\to} \mathbb{Z} \dots\] where the map \(\mathbb{Z} \overset{n}{\to} \mathbb{Z}\) factors into \(\mathbb{Z} \to n \mathbb{Z} \hookrightarrow \mathbb{Z}\). The answer's \(\mathbb{Q}\), it takes a little work, in general you just need your collection of transition maps to intersect every nontrivial subgroup of \(\mathbb{Z}\), because then you can specify where each \(\frac{1}{n}\) in \(\mathbb{Q}\) gets sent to to define the unique mediating morphism.
