Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets

G. Grätzer

Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets

Rings and Algebras 2025-10-02 v2

Authors: G. Grätzer

Abstract

Let $A$ be a finite ordered set. Define the ordered set $A^A$ as the set of all maps from $A$ to $A$ , ordered pointwise. Let ${}^{A} A$ be the dual of $A^A$ . We prove results in the spirit of Parts~I--III, but now using both $A^A$ and ${}^{A}A$ . For example, if $\Bigl({}^{{}^{ {}^{ {}^{A}A}A}A}A\Bigr)^{A^{A^{A}}}$ is isomorphic to $\Bigl({}^{ {}^{ {}^{ {}^{B}B}B}B}B\Bigr)^{B^{B^{B}}}$ for finite ordered sets $A$ and $B$ , then $A$ is isomorphic to $B$ .

Keywords

mesh seam generation UV unwrapping reinforcement learning Direct Preference Optimization point cloud encoder UV distortion texture mapping

Cite

@article{arxiv.2509.20726,
  title  = {Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets},
  author = {G. Grätzer},
  journal= {arXiv preprint arXiv:2509.20726},
  year   = {2025}
}

Comments

Based on Part I, incorrect

Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets

Abstract

Keywords

Cite

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