Timeline for Can the Category of Schemes be Concretized?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2013 at 10:41 | comment | added | Adam Epstein | Indeed. I flubbed the obvious point of how $\Phi$ is all about nonabelian groups. | |
| Sep 10, 2013 at 3:15 | comment | added | Todd Trimble | (Undeleted now; I hope I've fixed my earlier try.) | |
| Sep 10, 2013 at 2:33 | comment | added | Todd Trimble | @AdamEpstein: Unfortunately, this idea doesn't quite work since $U$ factors through the continuous inclusion $i: \textbf{Ab} \to \textbf{Grp}$, and continuous functors of the form $\textbf{Ab} \to \textbf{Set}$ (including for example $\Phi \circ i$) are representable, hence are right adjoints. I'm hoping the basic idea can be modified (and tried and deleted something, which those with 10k rep can see). | |
| Sep 8, 2013 at 21:30 | history | edited | Tom Leinster | CC BY-SA 3.0 |
Updated Latex for MO 2.0
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| Sep 6, 2013 at 10:49 | comment | added | Adam Epstein | Just a quick idea for a candidate. Start with a continuous functor $\Phi:\mathfrak{Group}\rightarrow\mathfrak{Set}$ which has no left adjoint (for example, the product of $Hom_{\mathfrak{Group}}(\Gamma_\alpha,\cdot)$ where $\Gamma_\alpha$ is a simple group of cardinality $\aleph_\alpha$) and compose with the group of units functor $U:\mathfrak{Ring}\rightarrow\mathfrak{Group}$ which has left adjoint given by the group ring functor $Z:\mathfrak{Group}\rightarrow\mathfrak{Ring}$. | |
| Sep 6, 2013 at 9:16 | comment | added | Martin Brandenburg | What is an example of a continuous functor $\mathsf{Ring} \to \mathsf{Set}$ which doesn't have a left adjoint? | |
| Oct 23, 2009 at 7:41 | history | answered | Tom Leinster | CC BY-SA 2.5 |