4 ms·
These posts always seem so damn interesting, but I get lost very early on. In this case it was at "Product is functorial in both arguments so, in particular, w
by omginternets 6y ago
These posts always seem so damn interesting, but I get lost very early on. In this case it was at "Product is functorial in both arguments so, in particular, we can define a functor".
Any suggestions for introductory material that might help me appreciate Milewski's writings?
- derekmcloughlin 6y agoHis own introduction to category theory for programmers is good: https://bartoszmilewski.com/2014/10/28/category-theory-for-programmers-the-preface/ https://bartoszmilewski.com/2014/10/28/category-theory-for-p...
- omginternets 6y agoNot bad! Looks like I've found my next 20%-time project! :)
- kerkeslager 6y agoThis feels like the category theory version of the Feynman Lectures.
- henearkr 6y agoSteve Awodey's "Category Theory" is an excellent book to learn the basics, and there are no requirements to know beforehand topology or higher abstract algebra in order to follow it.
- omginternets 6y agoAaaaaaaand purchased! Thanks!
- deleted 6y ago[deleted]
- henearkr 6y agoI've read it once, so feel free to contact me if you need any help!
- Twisol 6y agoTo be super clear, the quoted sentence continues, including the typeset formula below it. (It's an academic convention that I'm not a particular fan of.) > Product is functorial in both arguments so, in particular, we can define a functor L_a(c) = c * a. Bartosz is saying that we can fix one side of the product, giving us a one-argument specialization that is a functor. In this case, we're fixing `a` (subscript binds tighter than parameters, in some sense) and letting `c` vary. That's what it means to be functorial in both arguments: fix either one and you get a functor. The same phenomenon shows up in other fields where you have n-ary functions. Bilinear maps (in linear algebra) are linear in each argument. You could come up with something similar involving continuity in each argument for a binary function in real analysis (although I think that would actually be a weaker property than claiming the binary function is continuous...)
- omginternets 6y agoYup, I had gleaned that much from context. My issue is taht I don't know what a functor is! Sure, I could look it up, but my problem is that I spend all my time looking things up. At some point, this gets in the way of proper understanding. It's a strong signal that I'm missing some foundational knowledge. (Your reply is most appreciated, though!)
- Twisol 6y agoFair! Sorry for belaboring an unnecessary point ^_^;
- omginternets 6y agoNot at all!