information of the surface diagram $F^*$), i.e., the set of triple points of $F^*$ of type $(1,2,3)$. Since the orientation of $D_{12}^{\ell}$ is parallel to the orientation of $D_{12}^{\ell *}$, the sign of an intersection of $D_{12}^{\ell}$ and $\cup_{\lambda \in [0, R]}
t_\lambda(K_3)$ coincides with the sign of the corresponding intersection of $D_{12}^{\ell*}$ and $K_3^*$, which is the sign of the triple point (see Figure 2). Thus we have the result.  

Proof of Theorem 1.1
====================

For a compact oriented $n$-manifold $M$ with $\{A,B\}= \{\partial M, \emptyset\}$, we denote by $$\cdot_M \quad : H_p(M,A) \times H_q(M,A) \rightarrow
H_{p+q-n}(M,A)$$ the intersection map, which is defined by $$x \cdot_M y = P_M( P_M^{-1}(x) \cup P_M^{-1}(y) )$$ where $P_M: H^{\ast}(M,B) \rightarrow H_{n-\ast}(M,A)$ is the Poincaré duality isomorphism (see [@Massey], page 391). We will use $\cdot$ and $P$ instead of $\cdot_M$ and $P_M$ when their meanings are obvious in context.

Let $F=K_1\cup K_2\cup K_3$ be a $3$-component surface-link. For simplicity of argument, we assume that $F$ is embedded in the $4$-sphere $S^4={\bf R}^4\cup\{\infty\}$. For a regular neighborhood $N_i$ of $K_i$ in $S^4$, we put $$E_i={\rm Cl}(S^4 \backslash N_i), \
E_{ij}={\rm Cl}(S^4 \backslash (N_i\cup N_j))
\mbox{ for $i \ne j$, and }
E= {\rm Cl}(S^4 