\geq f(Q)$. Therefore, $f(N(u_i)) > f(N(v))$.

Now, we check for edges into the clique. First consider an edge $(u_j, u_k)$ such that $u_j, u_k \in T$. Then, $r_j \neq r_k$ and $f(N(u_j)) = f(Q) - (q-t+1) + (q-t+1).r_j \neq f(Q) - (q-t+1) + (q-t+1).r_k = f(N(u_k))$. Finally consider an edge $(u_j, u_k)$ such that $j \in [q-t]$ and $j < k$. Let $\alpha = f(u_k) - f(u_j)$. Note that $1 \leq \alpha \leq q-t$. Then, $f(N(u_j)) - f(N(u_k)) = \alpha + (q-t+1).(r_j - r_k)$. Suppose that $(q-t+1).(r_j - r_k) = \alpha$. Hence, $1 \leq (q-t+1).(r_j - r_k) \leq q-t$. This contradicts $r_j - r_k \in \bbbz$. Therefore, $f(N(u_j)) \neq f(N(u_k))$.

Observe that $\eta(G) \leq |Q| \leq \chi(G)$, so the conjecture holds for split graphs.\
The bound given in Theorem \[COTASUPSPLIT\] is tight on several families of graphs. We give three of them.

-   *Splits graphs with additive 1-coloring*: Let $G$ be a split graph with maximal clique $Q$ and maximal set $T \subset Q$ having vertices with different degree. Then, $T=Q$ characterizes those graphs with additive 1-coloring: $T=Q$ implies $\eta(G) = 1$ by Theorem \[COTASUPSPLIT\] while the converse is obtained by Observation \[ADDITIVE1\].

-   *Splits graphs with maximal clique of 