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# Anindya Bhattacharyya

Anindya Bhattacharyya
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• Coincidentally I came across some very useful notes from John on "groupoidification" the other day which spells out this profunctors = matrices thing in more detail. Unfortunately they are sitting on the math.ucr.edu server, which is still kaput, bu…
• think there's a typo here: There's a tensor product $$V \otimes W$$ of vector spaces $$V$$ and $$W$$, which has dimension equal to the dimension of $$\mathcal{V}$$ and the dimension of $$\mathcal{W}$$. should read "the dimension of $$\mathcal{… • @Michael – I think we can simplify your answer slightly: [ (X \times Y) \times Z = \lbrace \langle\langle x, y\rangle, z\rangle : x \in X, y \in Y, z\in Z \rbrace ] [ X \times (Y \times Z) = \lbrace \langle x, \langle y, z\rangle\rangle : x \in X,… • @John, thanks for that. It strikes me that this looks very much like the comparison functor between monoids (on a set) and T-algebras, in that we start with a set + associative binary operation + unit and end up with a set + an n-ary operation for e… • @Michael – 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, eg \(A\otimes (B\otimes C)$$ and $$(A\otimes B)\otimes C$$ for all objects $$… • The final bit of the puzzle is proving the second triangle equation from the first one and the pentagon. I'll just sketch a proof here. The trick here is to tensor the second triangle equation by \(A$$ on the left. This looks like it's making it mo…
• OK with a few pointers from that paper I've got a proof that $$\lambda_I = \rho_I : I\otimes I\to I$$. First we note that $$\rho$$ is natural, so for any $$\phi : I\otimes I\to I$$ we have $$\phi\circ\rho_{I\otimes I} = \rho_I\circ(\phi\otimes 1_I)… • Thanks for those pointers @John – fyi the nLab page says the fact we can derive \(\lambda_I = \rho_I$$ from the pentagon and triangle was first proved by Kelly in this 1964 paper: On MacLane's conditions for coherence of natural associativities,…
• oh of course, it's a cube! I missed that simplification.
• ok I think I've got a solution to Puzzle 280 Suppose $$f : I \to I$$ and $$g : I \to I$$ are morphisms in a monoidal category going from the unit object to itself. Show that $$fg = gf$$. We've seen earlier how to prove [(1\otimes f)\circ(k…
• re Puzzle 281 Going from left to right we have three main blocks: $$\qquad\Phi \otimes 1_b : a \otimes b \to (c \otimes d) \otimes b$$ $$\qquad1_c \otimes \Psi : c \otimes (d \otimes b) \to c \otimes (e \otimes f)$$ $$\qquad\Theta \otimes 1_f : … • I have a question: how is the "strictification" related to "coherence" (in the sense of the "coherence theorem" in Mac Lane's CWM for instance) – is it a stronger version of coherence? or a modern reworking of coherence? or are the two concepts quit… • re Puzzle 280 I think @Reuben is right that we can solve this using the Eckmann-Hilton argument mentioned earlier in this course. Here's as far as I've got. First note that Puzzle 278 is basically an "exchange law": [(f\otimes g)\circ(h\otimes k) … • there's a more revealing diagram for 278 if you use the boxes-and-wires notation @John sketches out at the start of the lecture • @Igor – here's a one-line answer to Puzzle 278: \((f\otimes g)\circ(f'\otimes g') = \otimes(f, g)\circ\otimes(f', g') = \otimes((f, g)\circ(f', g')) = \otimes(ff', gg') = ff'\otimes gg'$$
• @Igor note that $$\otimes$$ is a functor – which tells us it preserves composition...
• yes that's true. but tbh I don't really have much of a feel for what a relation is, beyond the formal definition. I mean the relations I'm familiar with are peculiar ones like ordering and equivalence. What's a relation in general tho? umm... err...…
• One of the things I've found slightly tricky about profunctors is avoiding the temptation to imagine $$\Phi : \mathcal{X} \nrightarrow \mathcal{Y}$$ as some sort of map taking things in $$\mathcal{X}$$ to things in $$\mathcal{Y}$$. This picture kick…
• hm well the smallest such $$a$$ is $$\text{max}(2b - 3b', 0)$$. and I suppose this represents the minimum revenue we need to recycle if we want to get at least $$b$$ out while putting at most $$b'$$ in. eg we have \$5 to put into the system, can we …
• David wrote: So, the big feasability relation is always $$\mathrm{True}$$ (for valid input values). This makes sense if one views the feasability relation as saying what possible stationary states the system can have. This is what I ended up g…
• OK, recapitulating the composition of conjoints rule one last time using opposite profunctors... We've shown that $$\check{F}(y, z) = \widehat{F^\text{op}}(z, y)$$. This is equivalent to $$(\check{F})^\text{op} = \widehat{F^\text{op}}$$. From whic…
• And on a similar note: Given a profunctor $$\Phi : \mathcal{X} \nrightarrow \mathcal{Y}$$, define $$\Phi^\text{op} : \mathcal{Y}^\text{op} \nrightarrow \mathcal{X}^\text{op}$$ by $$\Phi^\text{op}(y, x) = \Phi(x, y)$$. It's easy to prove $$\Phi^\tex… • Addendum to the above: Given a \(\mathcal{V}$$-functor $$F : \mathcal{Z} \to \mathcal{Y}$$, define $$F^\text{op} : \mathcal{Z}^\text{op} \to \mathcal{Y}^\text{op}$$ by $$F^\text{op}(z) = F(z)$$. It's easy to prove $$F^\text{op}$$ is indeed a $$\mat… • Let's work out what \(\check{F} \circ \check{G}$$ is, where $$F : \mathcal{Z} \to \mathcal{Y}, G : \mathcal{Y} \to \mathcal{X}$$. [(\check{F} \circ \check{G})(x, z) = \bigvee\big(\check{G}(x, y) \otimes \check{F}(y, z)\big) = \bigvee\big(\mathcal{X…
• @John wrote: Here's a thought: you could try to define the associator and its inverse as companions of enriched functors that are clearly inverses of each other, and then prove $$\widehat{FG} = \widehat{F}\circ\widehat{G}$$... This might require…
• Pedantic addendum to @Matthew's solution to Puzzle 224 – we need to prove the associator really is a profunctor. Suppose $$\langle\langle a, b\rangle, c\rangle \leq \langle\langle x, y\rangle, z\rangle$$ and $$\langle x', \langle y', z'\rangle\rang… • This is great proof but I can't help noticing that the argument doesn't work in Top because it's not cartesian closed. Nevertheless the result holds. Now I'm wondering why! • Thought I'd update/fix my previous attempt at proving the first snake inequality given the conventions we've settled on for cap and cup. I've also restricted it to the case where \(\mathcal{V} = \textbf{Bool}$$, although the proof works in the gener…
• @Keith – surely we already have an isomorphism of $$\mathcal{V}$$-categories $$X \times Y \cong Y \times X$$ given by $$(x, y) \mapsto (y, x)$$. We can check that this is in fact a $$\mathcal{V}$$-functor: \[ (Y \times X)((y, x), (y', x')) = Y(y, y…
• thing is directed graphs have loops in them, but I'm a bit unsure what it would mean to plug the output of a blob directly into one of its inputs. for instance suppose you had $$\Phi : X \nrightarrow Y$$ and $$\Psi : Y \nrightarrow Z$$ – you could t…