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Suppose that $P$ and $Q$ are two posets and $f:P\to Q$ is a homomorphism (a.k.a., $f(x)\le f(y)$ whenever $x\le y$). Given a chain $C\subset P$, the image $f(C)$ automatically is a chain as well. However, if $C$ is maximal (i.e., it is not properly contained in some other chain) then the image $f(C)$ need not be maximal, even if $f$ is onto.

I am interested in knowing if there are any natural conditions on $P$, $Q$, and $f$ which guarantee that maximal chains in $P$ get mapped to chains which continue to be maximal in $Q$.

For the application I have in mind, $P$, and $Q$ are actually finite lattices and $f$ is a (surjective) lattice homomorphism. Though, I'm hoping I can get away with weaker assumptions.

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    $\begingroup$ In the lattice case (whether finite or infinite), given that your map is a surjective lattice homomorphism, then considering the condition that $f$ is also injective is not interesting, because then the lattices are isomorphic under $f$, whence the result is trivial. But if $f$ is not injective, then it can map every element of some non-trivial chain $C$ in $P$ to one element of $Q$, and it doesn't preserve maximal chains. That said, it seems you may need some special condition on $f$, rather than on $P$ and $Q$. $\endgroup$
    – amrsa
    Jul 14, 2016 at 15:50
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    $\begingroup$ @amrsa: By a lattice homomorphism I mean one which satisfies $f(0)=0$ and $f(1)=1$. So the image chain $f(C)$ has to contain both the maximum $1$ and the minimum $0$ of $Q$. $\endgroup$
    – Steve
    Jul 14, 2016 at 21:10
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    $\begingroup$ All right, so I would call that a bounded lattice homomorphism (in the sense that, although finite lattices are always bounded, in this case you're considering $0$ and $1$ as nullary operations). Thanks for clarifying :) $\endgroup$
    – amrsa
    Jul 15, 2016 at 8:58
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    $\begingroup$ Do you have access to a copy of George Grätzer's General Lattice Theory book? There, there is the concept of strongly maximal chain in distributive lattices with $0$, as those chains whose image by an onto homomorphism are maximal. Then there is a characterization of strongly maximal chains in terms of the concept of the chain $R$-generating the lattice. What is a set to $R$-generate a (distributive) lattice, doesn't seem easy to explain. $\endgroup$
    – amrsa
    Jul 15, 2016 at 9:21
  • $\begingroup$ Thanks! that sounds promising. I'm sure I can track down a copy. $\endgroup$
    – Steve
    Jul 16, 2016 at 0:09

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