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This question is related to a question recently asked by Joel David Hamkins.

Let $(P,\leq)$ be a poset. We call it surjectively rigid if the only order-preserving surjective map $f:P\to P$ is the identity $\text{id}_P$. Given an infinite set $P$, is there an ordering relation $\leq$ such that $(P,\leq)$ is surjectively rigid?

(A positive answer would imply a positive answer for Joel's question: we can take the topology of upper sets.)

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  • $\begingroup$ If you agree to $(P,<)$ instead of $(P,\leq)$ then take a rigid undirected graph and view it as a poset. $\endgroup$ – Adam Przeździecki Aug 14 '15 at 11:35
  • $\begingroup$ Is there an example of a countable surjectively rigid graph? $\endgroup$ – Ehud Meir Aug 14 '15 at 11:39
  • $\begingroup$ @Ehud Meir -- yes, there are rigid graphs of any cardinality. It is recursively linked from this question, a more direct link is mathoverflow.net/a/6300/1946 This is about directed graphs but the same is true for infinite undirected ones. $\endgroup$ – Adam Przeździecki Aug 14 '15 at 13:16
  • $\begingroup$ If you work with strict orders it is trivial; you can just take well-orders. $\endgroup$ – Eric Wofsey Aug 14 '15 at 13:17
  • $\begingroup$ But there is a big difference between the case $\leq$ and $<$. If we just consider $<$ then any ordinal will do, but ordinals are not surjectively rigid. $\endgroup$ – Ehud Meir Aug 14 '15 at 14:28
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I claim that if $X$ is a linear ordering and $f:X\rightarrow X$ is an order preserving surjective mapping which is not the identity mapping, then there is an order preserving injective mapping $g:X\rightarrow X$ which is not the identity mapping.

If $f$ is surjective but not the identity function, the there is some function $g:X\rightarrow X$ such that $f\circ g$ is the identity mapping. The mapping $g$ is clearly injective and not the identity mapping. I now claim that $g$ is order preserving. Suppose that $x<y$. Then $f(g(x))=x<y=f(g(y))$. Therefore, since $f$ is order preserving, we have $g(x)<g(y)$ since if $g(x)\geq g(y)$, then $f(g(x))\geq f(g(y))$ which is a contradiction. Therefore, $g$ is our desired injective order preserving mapping that is not the identity mapping.

On the other hand, by this answer, if $2^{<\kappa}=\kappa$, then there is a linear order $X$ of cardinality $2^{\kappa}$ with no order preserving injections $f:X\rightarrow X$. Furthermore, under GCH, for every uncountable cardinal $\kappa$ there is a total order $X$ with $|X|=\kappa$ and where there is no order preserving injective function from $X$ to $X$. As a consequence, if $2^{<\kappa}=\kappa$, then there is a linear order $X$ of cardinality $2^{\kappa}$ with no order preserving surjections $f:X\rightarrow X$, and under GCH for each uncountable cardinal $\kappa$ there is a total order $X$ with $|X|=\kappa$ and where there is no order preserving surjective mapping $f:X\rightarrow X$.

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  • $\begingroup$ Joseph, could you clarify exactly what you mean in the first sentence? (The identity map is an order-preserving injective mapping, so what are claiming about $g$?) $\endgroup$ – Joel David Hamkins Aug 14 '15 at 18:43
  • $\begingroup$ I guess you are saying that there is an order-preserving injective mapping $g:X\to X$ for which $f\circ g$ is the identity? $\endgroup$ – Joel David Hamkins Aug 14 '15 at 19:01
  • $\begingroup$ Joel David Hamkins. I forgot to mention that the mappings $f$ and $g$ are not the identity mapping. I edited the answer to cover this case. $\endgroup$ – Joseph Van Name Aug 14 '15 at 21:16
  • $\begingroup$ Very nice, thanks @JosephVanName! That may be useful for proving that there are no infinite surjectively rigid posets. $\endgroup$ – Dominic van der Zypen Aug 17 '15 at 6:28

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