#colimits

[aufenthaltsraumspizza]

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

[aufenthaltsraumspizza]

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.

[aufenthaltsraumspizza]

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

[aufenthaltsraumspizza]

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

[eka-mark]

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.

[isomorphismes]

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!

[eatsleepmath]

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.