Let $P$ be a finite poset. Assign the following diagram to it - put the maximal element of $P$ on the first level, the maximal of the rest to the second level, etc. Assume that the diagram is connected, that is, that between each pair of elements in the diagram we have a chain connecting them.


Let $L\_n=P\times P\times\cdots\times P$ ($n$-times) to be the product poset, that is, one element in $L\_n$ is larger than another if we have all inequalities coordinate-wise. Is it true that there exists and $N=N(P)$ such that for $n>N$ the following is true:


In the diagram of L\_n there is a perfect matching between all consecutive levels (from the one with smaller size to the one with larger size).


Note: due to the useful feedback of Andres Caicedo and David Speyer the question was altered.



Perfect matchings between levels in products of posets
Here is a counterexample. Let $P$ be the poset on $\{ 1,2,3,4,5,6 \}$ whose only nontrivial relations are $1<2$, $1<3$, $4<6$, $5<6$. Let $Q$ be the subposet on $\{ 1,2,3 \}$ and let $R$ be the subposet on $\{ 4,5,6 \}$. I claim that there is no matching between levels $\ell$ and $\ell+1$ for $n/3 < \ell < 2n/3$. (The bounds here may not be quite right.)


Suppose the matching injects level $\ell$ into level $\ell+1$. But $P^n$ is a disjoint union of $2^n$ posets, one of which is $R^n$. And level $\ell$ of $R^n$ is larger than level $\ell+1$ for $n/3 < \ell$, a contradiction.


Similarly, suppose that matching injects level $\ell+1$ into level $\ell$. But level $\ell+1$ of $Q^n$ is larger than level $\ell$ for $\ell < 2n/3$. Again, we have a contradiction.


Maybe if you asked for the poset to be connected?


