Ivan Shishkin, Rye (1878)

Problems/Category theoryNeeds work

Functor between integers with divisibility and groups

by Ancient Tree·
40
Difficulty scaleÉchelle de difficulté

This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

  1. 1–10First steps / middle schoolPremiers pas / collège
  2. 11–25Beginner / high schoolDébutant / lycée
  3. 26–50Intermediate / undergraduateIntermédiaire / licence
  4. 51–70Advanced / graduateAvancé / master
  5. 71–90Expert / specializedExpert / spécialisé
  6. 91–100Research levelNiveau recherche
These levels are approximate guides.Ces niveaux sont des repères approximatifs.
·
English
Needs work. This problem has been marked as needing work.

Consider C=(N,∣)\mathcal{C}=(\mathbb{N},|) the poset of positive integers ordered by divisibility, viewed as a category.

Find a non-constant functor from C\mathcal{C} to Grp\mathbf{Grp}, the category of groups.

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Solutions

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Solution by araucaria araucanaFR

Discussions0 useful votes

Noter que si n∣mn \mid m, alors le groupe des racines nn-ièmes de l’unité Un\mathbb{U}_{n} est un sous-groupe de Um\mathbb{U}_{m}.
Ceci motive la définition d’une application F:C→GrpF : \mathcal{C} \to \mathbf{Grp} telle que F(n)=UnF(n) = \mathbb{U}_{n} et F(n→m)=(i:Un→Um)F(n \to m) = (i : \mathbb{U}_{n} \to \mathbb{U}_{m}), pour tous n,m∈Nn,m \in \N tels que n∣mn \mid m, où ii est le morphisme d’inclusion. Il est immédiat de vérifier que F respecte les idendités et les compositions, donc c’est un foncteur.

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