A **graph homomorphism** $f$ is a function $f : V(X) \to V(Y)$ such that if $uv \in E(X)$, then $f(u)f(v) \in E(Y)$. If such an $f$ exists, write $X \to Y$. $X$ and $Y$ are **hom-equivalent** if $X \to Y$ and $Y \to X$—write $\def\fromto{\leftrightarrow} X \fromto Y$. $X$ and $Y$ are **hom-incomparable** if $X \not\to Y$ and $Y \not\to X$.

The **direct product** $X \times Y$ of two graphs $X$ and $Y$ is defined by $V(X \times Y) = V(X) \times V(Y)$ and $(x,y) \sim (x',y')$ iff $x \sim x'$ and $y \sim y'$. The poset of hom-equivalence classes of graphs (under the ordering $\to$) forms a lattice with disjoint union $+$ as the join and $\times$ as the meet.

The hom-equivalence classes of (finite) graphs each have unique-up-to-isomorphism representatives called **cores**. Among other characterizations, $X$ is a core iff every endomorphism is an automorphism.

If $X$ and $Y$ are cores and $X \to Y$, then $X + Y \fromto Y$ and $X \times Y \fromto X$, so neither can be a core. However, if $X$ and $Y$ are connected hom-incomparable cores, it is easy to show $X + Y$ is also a core. More generally, if $X$ and $Y$ are (possibly disconnected) cores, $X+Y$ will be a core if each component of $X$ is hom-incomparable with each component of $Y$—equivalently, $X+Y$ is a core if $X$ and $Y$ are hom-incomparable cores that are "$+$-coprime".

If $X$ and $Y$ are cores, when is $X \times Y$ also a core? Is the condition that $X$ and $Y$ be hom-incomparable and $\times$-coprime sufficient?

As a related but more open-ended question, we have unique prime factorization on connected bipartite graphs in the graphs-with-loops (according to the *Handbook of Product Graphs* by Hammack, Imrich, and Klavžar, section 8.5) so we also have it for the connected cores; can this be extended to all cores, and do we require loops?

The motivation is to see how nicely the lattice structure of the hom-equivalence classes plays with cores. I have been unable to find references to this question anywhere.

I originally posted this question on math.SE, but brought it here upon the advice of a professor.