Extend definition for $n$ number of graphs in Homomorphic product of graphs I was going through a recently introduced graph product known as Homomorphic product of Graphs, link is provided for the details.
The Homomorphic product of graphs G and H,  has the vertex set $V (G)×V (H)$ and two vertices $(a_1,a_2)$, $(b_1,b_2)$ are adjacent in Homomorphic product if

*

*$a_1=b_1$ in $G$ or

*$a_1\sim b_1$ and $a_2\nsim b_2$
Now, I want to extend the definition for three or more graphs. I am confused about how to proceed in this case? Could anyone help me in this case?
My attempt for three graphs. I got two options here:
Vertices  $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are adjacent if
Case 1.

*

*$a_1=a_2$ OR

*$a_1\sim a_2$, $b_1\nsim b_2$ and $c_1\nsim c_2$
Case 2.

*

*$a_1=a_2$ OR

*$a_1\sim a_2$, $b_1\sim b_2$ and $c_1\nsim c_2$
https://en.wikipedia.org/wiki/Graph_product#:~:text=In%20mathematics%2C%20a%20graph%20product,1%20and%20G2%2C%20respectively.
 A: The general answer to questions like these is usually:

*

*If you have a reason to generalize the definition, you probably want the generalization to have some nice property. What is that property? There is probably only one natural generalization that continues to have it.

*If you don't have a reason to generalize the definition, don't do it.

In this case, the situation is much the same, but we can look a bit more closely.
The binary definition of the homomorphic product is motivated by the following fact:

For graphs $G$ and $H$, there is a homomorphism $G \to H$ if and only if $G \ltimes H$ contains an independent set of size $|V(G)|$.

Proof. Because $(a_1, a_2) \sim (b_1, b_2)$ whenever $a_1 = b_1$, an independent set of size $|V(G)|$ must include one pair $(v,w)$ for every $v \in V(G)$. For any such set $S$, define $f_S \colon V(G) \to V(H)$ by taking $f_S(v)$ to be the unique $w$ such that $(v,w) \in S$. (We can also go the other way: given $f$, define $S = \{(v, f(v)) : v \in V(G)\}$.)
The condition for $S$ to be an independent set and for $f_S$ to be a homomorphism is the same: whenever $u \sim v$ in $G$, we want $f_S(u) \sim f_S(v)$ in $H$. For $f_S$, that's just the definition of a homomorphism. For $S$ to be an independent set, we want the negation of "$a_1 \sim a_2$ and $b_1 \not\sim b_2$", which is precisely "$a_1 \sim a_2 \implies b_1 \sim b_2$". $\qquad\square$

"How do we extend this to three or more graphs?" does not have a good answer, because there's no notion of homomorphism between three or more graphs.
Occasionally, we ask the question: is there a homomorphism from some graph product of $G$ and $H$ to some third graph $K$? For example, when $K$ is a complete graph, this is asking about the chromatic number of this graph product, which is a well-studied problem.
If the product is the tensor product $G \times H$, then looking at homomorphisms $G \times H \to K$ motivates the triple product $(G \times H) \ltimes K$. Here, $(a_1, a_2, a_3) \sim (b_1, b_2, b_3)$ if either $a_1=b_1 \land a_2=b_2$, or $a_1 \sim b_1 \land a_2 \sim b_2 \land a_3 \not\sim b_3$. We get similar definitions for other products of $G$ and $H$, but none of them correspond to the two definitions proposed in the question.
One of those corresponds to a different triple product. The graph where $(a_1, a_2, a_3) \sim (b_1, b_2, b_3)$ if either $a_1=b_1$ or $a_1 \sim b_1 \land a_2 \not\sim b_2 \land a_3 \not\sim b_3$ is the triple product $G \ltimes (H * K)$, where $H * K$ is the co-normal product. We derive this by negating "$a_2 \not\sim b_2 \land a_3 \not\sim b_3$" to get "$a_2 \sim b_2 \lor a_3 \sim b_3$", and seeing what kind of product that is. Whether this is an interesting product or not depends on whether homomorphisms $G \to H * K$ are interesting to you.
