TITLE: Graph homomorphisms and line graph
QUESTION [8 upvotes]: If $G,H$ are simple, undirected graphs, we say that they are in a hom(omorphism)-relation if there is a graph homomorphism from $G$ to $H$ or from $H$ to $G$.
For any graph $G$ let $L(G)$ denote its line graph.
If $G,H$ each have at least one edge, and they are in a hom-relation, does the same hold for $L(G), L(H)$?

REPLY [7 votes]: Things are even worse than you imagine.
Tony's answer shows that the homomorphism order of line graphs can be partitioned into intervals whose endpoints are the complete graphs. I made use of this fact in my thesis. Then the line graphs in the interval $[K_d, K_{d+1})$ are the line graphs of graphs with maximum degree $d$ (ignoring exceptions caused by $K_3$). For $d = 1$, this interval contains only $K_1$, for $d = 2$ it contains only $K_2$ and the odd cycles of length at least 5 (up to homomorphic equivalence). After that however things get more complicated. In fact, it is shown here that every other such interval is universal, meaning that ANY countable partial order can be embedded order-preservingly into it. Finally, the graphs in that paper whose line graphs are used to prove universality of the interval $[K_d, K_{d+1})$ are all homomorphically equivalent to $K_d$.
All together this means that not only are there two homomorphically equivalent graphs whose line graphs have no homomorphisms between them in any direction, as Gordon's answer shows. But for any $d \ge 3$, there are an infinite number of graphs all homomorphically equivalent to $K_d$, but whose line graphs form a partial order that any countable partial order can be embedded in.