I have the following basic questions for the tensor product of directed graphs:
Consider two directed graphs $G_1=(V_1,E_1)$ and $G_2=(V_2,E_2)$. Is it true that in their tensor product graph, $G_1 \otimes G_2$, a cycle is exactly a "product" of cycles in $G_1$ and $G_2$?
This seems like a basic question but I have not dealt with products of graphs. I should note that the graph $G_1\otimes G_2$ is defined with vertex set $V=V_1\times V_2$ and $((v_1,v_2),(w_1,w_2))\in E_{G_1\otimes G_2}$ if and only if $(v_1,w_1)\in E_1$ and $(v_2,w_2)\in E_2$.
This seems like it follows from basic definitions, but I have been wrong before on basic subjects so I thought to verify. I mean projecting a cycle to either graph should give us a cycle, and it seems like any basic cycles can be "crossed" to a cycle in $G_1\otimes G_2$. Is this statment true? and is it as trivial as it seems?
The relationship you want holds for closed walks: $(v_1, w_1), (v_2, w_2), \dots, (v_k, w_k), (v_1, w_1)$ is a closed walk in the tensor product $G_1 \times G_2$ if and only if $v_1, v_2, \dots, v_k,v_1$ is a closed walk in $G_1$, and $w_1, w_2, \dots, w_k, w_1$ is a closed walk in $G_2$. Note the requirement on lengths.
For cycles, things are more complicated. It's possible for a cycle in $G_1 \times G_2$ to project to closed walks in $G_1$ and in $G_2$ that are not cycles. For example, in the tensor product of two cycle graphs $C_4 \times C_4$, where the vertex set of $C_4$ is $\{1,2,3,4\}$, the cycle $(1,1), (2,2), (3,1), (4,2), (1,1)$ projects down to a cycle $1,2,3,4,1$ in the first factor, but it projects to $1,2,1,2,1$ in the second factor.
The product of two cycles in $G_1, G_2$ of the same length will always be a cycle in $G_1 \times G_2$, but as the example above shows, this does not necessarily give us all possible cycles in $G_1 \times G_2$.