(Trying to clarify the question; the answer given below is wrong.)

If ${\cal C}$ is a collection of subsets of a set $X$, we associate to ${\cal C}$ a graph $G_{\cal C} = (V,E)$ where $V = {\cal C}$ and $$E = \big\{\{A,B\}: A\neq B\in {\cal C} \land A\cap B \neq \emptyset\big\}.$$

If $G$ is a simple, undirected graph, we define its *intersection number* $\iota(G)$ to be the smallest $n\in\mathbb{N}$ such that there is a collection ${\cal C}$ of subsets $[n]:=\{1,\ldots,n\}$ such that $G_{\cal C} \cong G$.

Let $G, H$ be finite, simple, undirected graphs. Their *tensor product* $G\times H$ is given by $V(G\times H) = V(G) \times V(H)$ and $$E(G\times H) = \{\{(u, u'), (v,v')\}: \{u,v\} \in E(G) \text{ and } \{u',v'\} \in E(H)\}.$$

It is easy to see that $\iota(G\times H) \leq \iota(G)\dot \iota(H)$. (We give a proof of this just after the question.)

**Question**. Do we have $\iota(G\times H) = \iota(G) \iota(H)$ for all finite simple graphs $G, H$?

Proof that $\iota(G\times H) \leq \iota(G) \iota(H)$. Suppose $\iota(G) = n$ and $\iota(H) = m$, and ${\cal C}$ is a set of subsets of $[n]$ such that $G_{\cal C} \cong G$, and and ${\cal D}$ is a set of subsets of $[m]$ such that $G_{\cal D} \cong H$. Set $${\cal C} \times {\cal D} := \{C\times D: C\in{\cal C} \text{ and } D\in {\cal D}\},$$ notice that ${\cal C}\times {\cal D}$ is a set of subsets of $[m]\times [n]$ (which has $mn$ elements), and check that $G_{{\cal C}\times {\cal D}} \cong G\times H$.

irreflexiverelation $E$, and the obvious definition $x E y$ implies $f(x) E' f(y)$ would preclude the existence of a map from a pair of vertices with an edge between them to the one-point graph, and in fact the category has no terminal object and has fairly poor categorical properties. $\endgroup$ – Todd Trimble♦ Mar 10 '16 at 13:25