Timeline for $K$-theory backwards
Current License: CC BY-SA 3.0
11 events
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| Dec 14, 2017 at 18:59 | history | edited | John Berman | CC BY-SA 3.0 |
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| Dec 14, 2017 at 17:36 | comment | added | John Berman | Let us continue this discussion in chat. | |
| Dec 14, 2017 at 17:32 | comment | added | John Berman | I don't see why you can apply the adjoint functor theorem. We would need objectwise group complete symmetric monoidal $\infty$-categories to be closed under arbitrary small limits and filtered colimits, and also to be accessible. | |
| Dec 14, 2017 at 17:23 | comment | added | Tim Campion | Maybe you're looking for the notion of a dinatural transformation. In any event, let's say that $(C,\otimes)$ is objectwise group complete if every object has an $\otimes$-inverse. The objectwise group complete symmetric monoidal $\infty$-categories are closed among all symmetric monoidal $\infty$-categories under products and pullbacks and filtered colimits, and so by the adjoint functor theorem there is a left adjoint to the inclusion functor. This left adjoint is what I'm referring to above. | |
| Dec 14, 2017 at 17:07 | comment | added | John Berman | Because of this issue with mixed functoriality, you might have trouble making sense of the free "group-completion" construction. A similar question was asked a couple years ago mathoverflow.net/questions/214767/… | |
| Dec 14, 2017 at 17:02 | history | edited | John Berman | CC BY-SA 3.0 |
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| Dec 14, 2017 at 17:00 | comment | added | John Berman | I was assuming the inverse is a covariant functor. How would you write down axioms for a grouplike symmetric monoidal $\infty$-category with a contravariant inverse? We would like to have an axiom like $-X\otimes X\cong 1$, but the functoriality is wrong. | |
| Dec 14, 2017 at 16:53 | comment | added | Tim Campion | In fact, more generally, if $(C,\otimes)$ has duals for objects, taking the dual is a functor $C^{op} \to C$. In particular, if $(C,\otimes)$ has inverses for objects, then taking the inverse is a functor $C^{op} \to C$. Are you requiring there to be a functor $C \to C$ rather than $C^{op} \to C$? | |
| Dec 14, 2017 at 16:38 | comment | added | Tim Campion | I'm confused -- let $Pic_R$ denote the full subcategory of $Mod_R$ on the $\otimes$-invertible objects. Then the $\otimes$-inverses in $Pic_R$ are functorial, since $L^{-1} = Hom(L,R)$, and $Hom(-,R): (Pic_R)^{op} \to Pic_R$ is a functor. But $Pic_R$ has non-invertible morphisms. | |
| Dec 14, 2017 at 1:40 | history | edited | John Berman | CC BY-SA 3.0 |
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| Dec 14, 2017 at 0:20 | history | answered | John Berman | CC BY-SA 3.0 |