A function $f:P\to Q$ from a poset $(P,\le_P)$ to a poset $(Q,\le_Q)$ is an *order-embedding* if, for all $p,p'\in P$, $p\le_P p'$ if and only if $f(p)\le_Q f(p')$.

We partially order the Cartesian product $P\times Q$ of posets $(P,\le_P)$ and $(Q,\le_Q)$ as follows: for $(p,q),(p',q')\in P\times Q$, $(p,q)\le_{P\times Q}(p',q')$ if $p\le_P p'$ and $q\le_Q q'$.  Similarly, we can define the product of more than two posets.

A *chain* is a poset $(C,\le)$ that is totally ordered: that is, for all $c,c'\in C$, either $c\le c'$ or $c'\le c$.  For $n\in\mathbb N$, we denote the $n$-element chain by $\bf n$; its elements are $\{0,1,\dots,n-1\}$.

The *dimension* of a poset $P$ is the smallest number of chains into the product of which we can order-embed $P$.

Consider the poset $S_3:=({\bf 2}\times{\bf 2}\times{\bf 2})\setminus\{(0,0,0),(1,1,1)\}$.  It is called the *standard poset of dimension 3*.

The poset $S_3\times{\bf 2}$ is shown in the figure below. The colors of the lines are merely for ease of reading.

[![enter image description here][1]][1]

If you put lines between the "corresponding" elements of the two parts of the diagram below, you will have $S_3\times{\bf 2}\times{\bf 2}$.

[![enter image description here][2]][2]

Find an order-embedding of $S_3\times{\bf 2}\times{\bf 2}$ into a product of 4 chains.

According to the middle of page 80 of Lin ("The dimension of the Cartesian product of posets," *Discrete Mathematics* **88** (1991), 79-92), there should be one.

  [1]: https://i.sstatic.net/fdhYw.jpg
  [2]: https://i.sstatic.net/ZgZyZ.jpg