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Sep 21, 2021 at 20:10 comment added Emily @Achim and Eric It is a weird name! I did have in mind another related statement as Eric said (that $\mathbb{Z}$ corepresents the group of units functor $(-)^\times\colon\mathsf{CMon}\to\mathsf{Sets}$ from commutative monoids to sets).
Sep 21, 2021 at 20:10 history edited Emily CC BY-SA 4.0
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Sep 21, 2021 at 17:50 comment added Eric Peterson @Achim I was also puzzled, but it is a good name for functor on rings corepresented by the group ring on Z. Given some of the other adjunctions floating around, perhaps that’s closer to what was intended?
Sep 21, 2021 at 5:42 comment added Achim Krause $(-)^\times$ is a pretty weird name for what seems to be the forgetful functor from abelian groups to sets...
Sep 21, 2021 at 3:38 vote accept Emily
Sep 21, 2021 at 3:31 answer added Tyler Lawson timeline score: 8
Sep 21, 2021 at 2:36 history edited Emily CC BY-SA 4.0
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Sep 21, 2021 at 2:31 comment added Emily @skd Sorry for the confusing notation! I didn't mean the mod 2 Moore spectrum
Sep 21, 2021 at 2:31 history edited Emily CC BY-SA 4.0
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Sep 21, 2021 at 2:31 comment added Emily @TimCampion I was writing "$\mathbb{S}/2$" to mean the connective spectrum associated to the $\mathbb{E}_\infty$-group $\Omega Q\mathbb{RP}^\infty$ because it satisfies an analogous universal property to the one for $\mathbb{Z}/2$: while morphisms of monoids from $\mathbb{Z}$ and $\mathbb{Z}/2$ to a monoid $A$ are the same as invertible/involutory elements of $A$, symmetric monoidal functors from $QS^0$ and $\Omega Q\mathbb{RP}^\infty$ to $\mathcal{C}$ are the same as invertible/involutory objects of $\mathcal{C}$. I've edited the question to use a less confusing notation. Thanks, Tim!
Sep 21, 2021 at 2:04 comment added Tim Campion "$\mathbb S / 2$" would standardly denote the mod 2 Moore spectrum, which is the same as $\Sigma^{\infty-1} \mathbb R \mathbb P^2$... but you've defined "$\mathbb S/2$" to be $\Omega Q (\mathbb R \mathbb P^\infty)$... first of all $Q(X)$ usually means $\Omega^\infty \Sigma^\infty X$, so what you've written is a space and not a spectrum, thought the (unstable) homotopy groups of $\Omega Q \mathbb R \mathbb P^\infty$ are the same as the (stable) homotopy groups of $\Sigma^{\infty-1} \mathbb R \mathbb P^\infty$. Still, that's a different spectrum from the mod 2 Moore spectrum. Could you clarify?
Sep 21, 2021 at 1:27 comment added skd S/2 doesn't have a unital multiplication. In general, S/p does have an A_{p-1}-structure, but not an A_p-structure.
Sep 21, 2021 at 1:16 history asked Emily CC BY-SA 4.0