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In light of my tweet thread about the category of finite sets and commutative monoids (https://t.co/jnY0wZZbxq), I thought I might try to say what the analogue is for braided monoidal things (although much of this is still somewhat hypothetical).


It's also just kind of a cool combinatorial structure! I've been talking to @CreeepyJoe about this lately, as well as @grassmannian.

The first thing you have to know is that, in a braided monoidal category you can still have commutative monoids. Since a braided monoidal category C has a "twist" map for every object β(x):x⊗x→x⊗x, if x is a monoid you can ask for the following diagram to commute:


Remember that being symmetric monoidal just means that if you take the twist map above and do it twice, you get the identity map, but braided monoidal doesn't mean that. But it's okay! You can still define commutative monoids here.

But so anyway, we can talk about commutative monoids in braided monoidal categories.
OK, I may be guilty of a DoS attack attempt on mathematicians' brains here, so lest anyone waste too much precious brain time decoding this deliberately cryptic statement, let me do it for you. •1/15


First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15

Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15

Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15

… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
It is trying when mathematicians declare condescendingly that there is no point doing things because their models tell them so. Well maybe some of the assumptions don't hold up. How did that work out for the no additional risk from large events and no point in border controls...


During wave 1 cases fell very fast, faster than I think most people were expecting. Particularly in Scotland. Rt was probably ~0.5 until we started easing off.

This was despite a constant leak of cases coming out of hospitals and LTC facilities as we were rationing PPE and are policies were nowhere near ideal. There was insistence from infection control that droplet protections were sufficient. We have all learned a lot since then.

Not to mention we have learned to avoid the shit show of actively importing cases into care homes. We've learned not to repeat that. Other sectors have learned too.

We've learned a lot and there's no reason we can't control this new variant. But we will not manage if we don't try and act with clarity of purpose.