Timeline for Presentation of the covariant power set functor
Current License: CC BY-SA 3.0
17 events
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| Nov 1, 2017 at 11:32 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Nov 1, 2017 at 11:25 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Nov 1, 2017 at 11:18 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Nov 1, 2017 at 10:55 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Nov 1, 2017 at 10:50 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 30, 2017 at 14:36 | comment | added | მამუკა ჯიბლაძე | (And also that $\hom(1,P^+)=\varnothing$, but this is clear since $P^+(0)=\varnothing$) | |
| Oct 30, 2017 at 13:41 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 30, 2017 at 13:26 | comment | added | მამუკა ჯიბლაძე | In other words, there are no non-identitiy morphisms $P^+\to P^+$, where $P^+$ is the subfunctor of nonempty subsets (since $P=1\sqcup P^+$). | |
| Oct 30, 2017 at 13:23 | comment | added | HeinrichD | Alright, and there are exactly two morphisms $P \to P$. | |
| Oct 30, 2017 at 13:22 | vote | accept | HeinrichD | ||
| Oct 30, 2017 at 11:23 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 30, 2017 at 10:37 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 30, 2017 at 10:31 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 30, 2017 at 10:28 | comment | added | მამუკა ჯიბლაძე | Good point. As said, we just need the free semilattice on $\hom(1,-)$. So this just amounts to ensuring commutativity and $xx=x$, so you must be right here. | |
| Oct 30, 2017 at 10:26 | comment | added | HeinrichD | Ok, this is good. Can we reduce the set of surjections here? Probably we only need the transpositions (as generators for the permutations) and the maps $\{1,\dotsc,n+1\} \to \{1,\dotsc,n\}$, $k \mapsto k$ (for $k \leq n$) resp. $n+1 \mapsto n$? | |
| Oct 30, 2017 at 10:17 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Oct 30, 2017 at 10:10 | history | answered | მამუკა ჯიბლაძე | CC BY-SA 3.0 |