MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Here are a couple of questions:

  1. Is there a way to classify all homomorphisms between two finite posets?

  2. Same question as (1) but for infinite, locally finite countable connected posets.

These questions are relevant to causal set quantization. Thank you

PS: I mention a partial result concerning AUTOMORPHISMS of locally finite countable connected posets: any automorphism of such a poset (assuming some extra hypotheses) should either transform chains into themselves (not all chains, but a set of chains forming a partition of the poset), or tranform antichains into themselves (again a partition of the poset). So in the first case the automorphism is induced by automorphisms of a chain (which can be indexed in a clear way by integers), and in the second case the automorphism is induced by automorphisms of an antichain (that, too, satisfy further conditions). So a similar characterization in terms of homomorphisms is helpful.

For the meaning of homomorphism we can require that if $a \leq b$ then $\pi(a) \leq \pi(b)$ and if $\pi(a) \leq \pi(b)$ then there exists $a'$ and $b'$ such that $a' \leq b'$ and $\pi(a')=\pi(a)$ and $\pi(b') =\pi(b)$.

share|cite|improve this question
Could you clarify the sense of homomorphism you intend here? For relational structures, one may ask merely for $a\leq b\to \pi(a)\leq\pi(b)$, or alternatively for the stronger notion, for which $a\leq b\iff \pi(a)\leq \pi(b)$. Also, perhaps you might elaborate on what kind of classification you want? In question 1, for example, of course there is a complete computable classification, since we may uniformly compute all homomorphisms between any two finite structures. So it isn't clear to me exactly what you want. – Joel David Hamkins Sep 25 '11 at 2:47
What is a locally finite poset? A poset in which all the bounded set (from above and below) are finite? – Stefan Geschke Sep 25 '11 at 8:23
Yes, a locally finite posets is a posets in which all intervals (that is sets of the form $\{x|a\leq x \leq b\}$ ) are finite. Joel: Such a classification for finite posets would be fine if you elaborate on it a little bit. – user16974 Sep 25 '11 at 9:00
up vote -1 down vote accepted

I think you should study rather endomorphisms of posets than homomorphisms. Then you obtain an algebraic structure (the semigroup of endomorphisms) and in addition a lot of articles in this problem. See, e.g., P. M. Higgins, J. D. Mitchell, M. Morayne and N. Ruškuc, Rank Properties of Endomorphisms of Infinite Partially Ordered Sets, Bull. London Math. Soc. (2006) 38 (2): 177-191.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.