Timeline for Is there an infinite chain of endofunctors of finite sets?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2023 at 19:10 | history | became hot network question | |||
| Sep 25, 2023 at 14:03 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 1 character in body
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| Sep 25, 2023 at 12:31 | vote | accept | Sebastian Meyer | ||
| Sep 25, 2023 at 12:28 | comment | added | David Wärn | Indeed there is no natural transformation $\mathcal P \to \mathrm{id}$, but there is a natural transformation $\mathcal P \mathcal P \to \mathcal P$, which sends everything to the empty set. | |
| Sep 25, 2023 at 12:25 | answer | added | Tom Goodwillie | timeline score: 18 | |
| Sep 25, 2023 at 12:21 | comment | added | Federico Cantero | To prove that, let $\alpha: F\to 1$. Then the composition $\eta\circ\alpha$ yields a choice of an element $x_A$ in each set $A$, and this choice is natural with respect to any morphism. Therefore, this element must be fixed by all automorphisms of the set $A$, but this is only possible if the set has at most one element. | |
| Sep 25, 2023 at 11:53 | comment | added | R. van Dobben de Bruyn | One candidate could be $F = \mathcal P$ (the covariant power set) and $F_i = F^i$. The natural transformation $\eta \colon 1 \to F$ taking $a \in A$ to the singleton $\{a\} \in \mathcal P(A)$ induces multiple natural transformations $F_i \to F_{i+1}$ (namely $F^j \eta F^{i-j}$ for $j \in \{0,\ldots,i\}$). It seems to me that there should be no natural transformations in the other direction, but I don't know how to prove this (nor whether this is true, for that matter). | |
| Sep 25, 2023 at 11:11 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, fixed formatting issue
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| Sep 25, 2023 at 11:09 | history | asked | Sebastian Meyer | CC BY-SA 4.0 |