中文

自由格函子弱保持满射拉回

环与代数 2021-03-18 v1

摘要

p(x,y,z)p(x,y,z)q(x,y,z)q(x,y,z)为项。若存在公共“祖先”项s(z1,z2,z3,z4)s(z_{1},z_{2},z_{3},z_{4}),通过识别某些变量分别特化为ppqq\begin{align*} p(x,y,z) & \approx s(x,y,z,z)\\ q(x,y,z) & \approx s(x,x,y,z), \end{align*}则方程p(x,x,z)q(x,z,z) p(x,x,z)\approx q(x,z,z) 可由s(x,y,z,z)s(x,y,z,z)s(x,x,y,z)s(x,x,y,z)的语法合一平凡得到。在本注记中我们证明,对格项,更一般地对格序代数的项,其逆亦真。给定项p,qp,q及方程\begin{equation} p(u_{1},\ldots,u_{m})\approx q(v_{1},\ldots,v_{n})\label{eq:p_eq_q} \end{equation}其中{u1,,um}={v1,,vn}\{u_{1},\ldots,u_{m}\}=\{v_{1},\ldots,v_{n}\},总存在“祖先项”s(z1,,zr)s(z_{1},\ldots,z_{r})使得p(x1,,xm)p(x_{1},\ldots,x_{m})q(y1,,yn)q(y_{1},\ldots,y_{n})作为ss的代换实例,其合一给出原方程。在范畴论术语中,上述命题限于格时有更简洁的表述:自由格函子弱保持满射的拉回。

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引用

@article{arxiv.2103.09566,
  title  = {Free-lattice functors weakly preserve epi-pullbacks},
  author = {H. Peter Gumm and Ralph Freese},
  journal= {arXiv preprint arXiv:2103.09566},
  year   = {2021}
}