We consider surface-links $F=K_1\cup K_2\cup K_3$ in which each $K_i$ is not necessarily connected. Such a surface-link is called a [*$3$-partitioned*]{} surface-link. The definition of the triple linking of $F=K_1\cup K_2\cup K_3$ is generalized directly for 3-partitioned surface-links, and all results and proofs in Sections 2 and 3 are valid for 3-partitioned surface-links. Theorem 1.5 is a special case of the following:

If two $3$-partitioned surface-links $F=K_1\cup K_2\cup K_3$ and $F'=K'_1\cup K'_2\cup K'_3$ are link homologous, then ${\mbox{\rm Tlk}}(K_i,K_j,K_k)={\mbox{\rm Tlk}}(K'_i,K'_j,K'_k)$.

[*Proof.*]{} It is sufficient to prove ${\mbox{\rm Tlk}}(K_1, K_2, K_3) =
{\mbox{\rm Tlk}}(K'_1, K'_2, K'_3)$ in a special case that $K_i=K'_i$, $K_j= K'_j$ and $K_k$ is homologous to $K'_k$ in $S^4 \backslash (K_i \cup K_j)$, where $\{i,j,k\}=\{1,2,3\}$. If $k=2$, then ${\mbox{\rm Tlk}}_4(K_1, K_2, K_3)= {\mbox{\rm Tlk}}_4(K_1, K'_2, K_3)$ by definition. If $k=1$, then ${\mbox{\rm Tlk}}_3(K_1, K_2, K_3)= {\mbox{\rm Tlk}}_3(K'_1, K_2, K_3)$. (This is seen as follows: Let $M_1$ be a 3-chain with $\partial M_1=K_1$. Since $K'_1$ is homologous to $K_1$ in $S^4 \backslash (K_2\cup K_3)$, there is a 3-chain $B$ in $S^4 \backslash (K_2\cup K_3)$ with $\partial B = K'_1 - K_1$. Let $M'_1=
M_1 +B$, which is a 3-chain with $\partial M'_1= K'_1$. Then $[M_1 \cap K_2]_{E_3}= [M'_1 \cap 