Incidence function $\phi: E(G) \rightarrow V(G) \times V(G)$ of union of graphs $G = F \cup H$.
Suppose $F,H$ are graphs. Then according to two books, I've been studying, we can define the $G = F \cup H$ as the union with $V(G) = V(F) \cup V(H)$ and $E(G) = E(F) \cup E(H)$.
However nothing is said about the incidence funcion $\phi$. Is this because it is arbitrary (left unspecified - to be specified by the person performing the operation) or because it is implicit that endpoints of edges in $G$ remains intact ? That is for any edge $e \in E(G)$, the endpoints is respect to $\phi_F$ if $e$ is an edge of $E(F)$ only - the same applies to an edge of $H$ only. But what if $e \in E(F)$ and $e \in E(H)$ ? How should the endpoints of $e$ in $G$ be specified ?
If $e$ is a common edge in $F$ and $H$ it should not matter. Since according to the definitions the incidence function of $G$ can also be stated as
$$ \phi ; E(F) \cup E(H) \rightarrow V(F)\cup V(H) \times V(F)\cup V(H)$$
The incidence function remains well-defined for union graphs. If $e$ is a common edge to both $F$ and $H$ it would not matter since $e \in E(F) \cup E(H) $ occurs only once. Now you talk about varying endpoints of an edge. This is impossible. Since an edge is defined by the endpoints.I don't know which book you use but say for an example Deistel says an edge is a $2$-subset of $V(G)$ and hence is defined by the endpoints. If $e \in E(F) \cap E(H)$ then the endpoints of $e$, say $v_1$ and $v_2$ are also common to both $V(F)$ and $V(H)$. So the function remains unaffected. Hope I helped.