Suppose I have a lattice $\langle L, \preceq\rangle$. I know that the covering relation on the lattice is that $x\lessdot y$ if $x\prec y$ and there is no $z$ such that $x \prec z \prec y$.
Now suppose I define a new lattice as $L^{2}$, with vertices $(a\times b)$. The partial order relationship is $(a\times b)\preceq(a'\times b') \iff a\preceq a', b \preceq b'$, which is all well and good.
What I'm curious about is whether it is possible reconstruct the covering relation on $L^{2}$ from the covering relation on just $L$. Naively, I considered that, if $a\lessdot a', b\lessdot b'$, then $(a\times b)\lessdot (a'\times b')$, but working out some simple lattices by hand shows that this is not true.
Any help here would be greatly appreciated.
How about $a \times b \lessdot a' \times b' \iff a \lessdot a' \ OR \ b \lessdot b'$ ?
If $a \times b \lessdot a' \times b'$ then
there is no $c$ such that $a \prec c \prec a'$. Hence $a \lessdot a'$ .
or
there is no $d$ such that $b \prec d \prec b'$ . Hence $b \lessdot b'$ .
Since if both holds for a $c,d$, we have $(a \times b) \prec (c\times d) \prec (a' \times b')$.