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John Baez

John Baez
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• Ken wrote: So the field k itself is some kind of trivial vector space? I feel like there's a notational pun I'm missing. Yes, every finite-dimensional vector space over a field $$k$$ is isomorphic to $$k^n$$, the set of $$n$$-tuples of element…
• Great! I met her at ACT2018. I only discovered her blog after that. It's got the best introduction to the Yoneda Lemma - a 3-part series that starts with the important 'Yoneda philosophy' before getting into the mathematical details. She also r…
• Michael - yes! That's one of the main lessons of category theory! These days we use 'categorification' to mean the process of taking math done with equations and boosting it up to math done with isomorphisms, equivalences, or other such things. …
• Hey there, Jesus! You seem to be doing quite well figuring out what I was hinting at. I was too lazy to explain all this stuff, but you found a page on the nLab where I was explaining some of it to Mike Stay once upon a time, so it's okay that I'm…
• Michael wrote: John wrote: [ (X \times Y) \times Z = X \times (Y \times Z) ] but in fact this equation isn't true! Do you see why not? If you don't see why, maybe someone else can explain why. Yes, they are not equal becaus…
• Michael wrote: Basically, the theorem proved that $$\mathcal{C}$$ and $$\mathrm{str}(\mathcal{C})$$ are equivalent by showing there are two functors that are inverses between them. Close but not quite. These two functors are just 'weak' inve…
• I just found some nice slides on Mac Lane's coherence theorem, which says roughly that all diagrams built using tensor products, the unit object, associators and unitors in a monoidal category commute: Peter Hines, Reconsidering MacLane:the founda…
• Anindya wrote: from what I can make out from the proof of the strictification theorem it doesn't actually collapse all isomorphic objects into one object – it just identifies, e.g.. $$A\otimes (B\otimes C)$$ and $$(A\otimes B)\otimes C$$ for all…
• Michael wrote: It seems like there should be a skeletal category lurking in there since we are collapsing all isomorphic objects into one object. No, we're not. The actual proof of Mac Lane's strictification theorem proceeds by creating tons…
• Michael wrote: If you strictify a monoidal category, is it somehow related to a skeletal category? No. I'll start by refering you to comment #12 on Lecture 72. Then I'll say some more... Ken wrote: Theorem 1, MacLane's theorem…
• Daniel - it might (or might not) help to think about cups and caps in the category of finite-dimensional real vector spaces, where the cup is the obvious linear map [ \cup_V \colon V^* \otimes V \to \mathbb{R} ] and the cap is the linear map […
Comment by John Baez August 2018
• Igor wrote: Btw, $$fg = gf$$ implies that composition of morphisms in $$\mathcal{C}(I, I)$$ is commutative, it is interesting whether there are other implications for $$I$$. Not much: Theorem. For any commutative monoid $$A$$, there is a mono…
• Excellent solutions to Puzzle 280, folks! I guess it was a bit sadistic to assign this without mentioning that $$\rho_I = \lambda_I$$. Anindya wrote: This is listed in Mac Lane as one of the coherence equations, although more recent treatmen…
• Michael wrote: Yes! This is an almost perfect pictorial explanation of the interchange law [ (f \circ f') \otimes (g \circ g') = (f \otimes g) \circ (f' \otimes g'). ] As you note, it's a commutative square: tensoring and then composin…
• Igor wrote: Still, what is the shortest and the most correct answer to the puzzle 278, how to formulate it? 'Shortest' and 'most correct' are complementary quantities. Anindya wins the prize for brevity in comment #17: $$(f\otimes g)\circ… • Igor wrote: Let's say we have a category \(\mathcal{C}$$, which has all products. Is it true that we can define a functor $$\otimes: \mathcal{C} \times \mathcal{C} \to \mathcal{C}$$ by pointing each pair $$(x, y)$$ to the product of $$x$$ an…
• Michael wrote: It seems like once we start talking about categories, all laws or equations are presented as natural transformations like the associator and unitors. I don't get why we present them as natural transformations. Is there something…
• Our new journal is now open for submissions! For more details, go here:
Comment by John Baez August 2018
• Ken wrote: Theorem 1, MacLane's theorem - is this similar to how we got a "skeletal" poset out of a preorder by collapsing all the isomorphisms? NO!!! That's a mistake every newcomer makes. It's a natural guess, but fact Mac Lane's theorem i…
• Ken wrote: Def 12 on monoidal equivalence rubs me strangely, because FG->1 and GF->1 feel like they "want" to be a unit & counit, as from an adjunction, or like a compact category. Except it's a counit & pseudo-counit! What am I to…
• Great answer to Puzzle 278, Michael! Igor wrote: What I need to figure out what is $$(f' ,g') \circ (f, g)$$. Okay, you're trying to figure out how to compose two morphisms in $$\mathcal{C} \times \mathcal{C}$$. For this you can look at Lec…
• 21 November 2017: This week's - or really last week's - progress: 1) Daniel got this paper accepted by TAC, subject to very minor revisions. It's a great paper and you should all read it - I tend to assume you all read each other's papers to keep…
Comment by John Baez August 2018
• 12 November 2017: A huge amount of progress this week! 1) Most of the people on this list gave talks at the AMS special session on applied category theory. Almost all the talk slides from this session are here now: Applied Category Theory 2017…
Comment by John Baez August 2018
• 4 November 2017: We made a lot of progress this week, completing 3 big papers. I apologize for becoming pretty bad-tempered during this time... I was pretty stressed. But now I'm happy, because we have a lot to show the folks coming to our speci…
Comment by John Baez August 2018
• 29 October 2017: This week's progress: 1) I finally submitted my grant proposal to the NSF - it took a long time to fight its way through the UCR bureaucracy. If accepted, it could pay for 4 grad students from UCR to attend the applied category t…
Comment by John Baez August 2018
• 23 October 2017: Okay, now for this week's progress. The first item counts as next week's progress, but some of you may want to attend, to see what I do in my spare time: 1) I'm giving a talk next week: International Open Access Week, Wednesda…
Comment by John Baez August 2018
• 14 October 2018: A lot of people are trying to finish up papers, but I don't know anything that was finished this week, so I just have an interesting story about (yet another) workshop on applied category theory. Jelle Herold is a programmer who I…
Comment by John Baez August 2018
• October 5, 2017: 1) The really big news is that Blake has a postdoc job, using category theory to help design more stable electrical grids! He'll start around November 15th. I think he'll be located at the National Institute of Science and Techno…
Comment by John Baez August 2018
• Reuben wrote: Then $$id$$ = $$id \circ id$$ = $$(I \otimes id) \circ (id \otimes I)$$ = $$(I \circ id) \otimes (id \circ I)$$ = $$I \otimes I$$ = $$I$$ . Again, lots of these equalities are clearly wrong. I'm glad you say that they'…
• Reuben wrote approximately: Concretely (but with mistakes I need help with) note that $$(f \otimes g) \circ (f' \otimes g') = (f \circ f') \otimes (g \circ g')$$ which is to say that $$\otimes$$ is a functor as Anindya noted (but I suppose t…