formula $$\left(
\frac
{(1,x_T-x_M,y_T,-z_M)}
{\sqrt{1+ (x_T-x_M)^2+ y_T^2 + z_M^2}},
\frac
{(1,x_M,-y_B,z_M-z_B)}
{\sqrt{1+ x_M^2 +y_B^2 +(z_M-z_B)^2}}
\right).$$ The map $L'$ is injective and hence it is a homeomorphism onto its image. Its (local) degree is $+1$ or $-1$ which depends only on $\varepsilon$. Since the degree of $L$ is the sum of the (local) degrees of $L'$ for all triples $(x_1, x_2, x_3)$ in the preimage $L^{-1}((1,0,0,0),(1,0,0,0))$, this number agrees up to sign with the triple linking number ${\rm Tlk}_1(f(F_1), f(F_2), f(F_3)).$  

By Lemmas 3.1–3.5, we have Theorem 1.1.

Proof of Theorems 1.2–1.5
=========================

To prove Theorem 1.2, it is useful to change $\partial N_2$ in the definition of ${\mbox{\rm Tlk}}_5$ for $\partial E_2$.

${\mbox{\rm Tlk}}_5(K_1,K_2,K_3)
=- \varepsilon_{\partial E_2}(
M_1|_{\partial E_2} \cdot
M_2|_{\partial E_2} \cdot
M_3|_{\partial E_2})$, where the intersections are taken in $\partial E_2$.

[*Proof.*]{} Since $\partial N_2$ and $\partial E_2$ are the same $3$-submanifold of $S^4$ with opposite orientations, $[\partial N_2] = - [\partial E_2]$ in $H_3(\partial N_2)= H_3(\partial E_2)$. Thus, in $H_0(\partial N_2)=H_0(\partial E_2)$, $$\begin{array}{ll}
M_1|_{\partial N_2} \cdot
M_2|_{\partial N_2} \cdot
M_3|_{\partial N_2}
&=
(u_1|_{\partial N_2} \cup
u_2|_{\partial N_2} \cup
u_3|_{\partial N_2}) \cup [\partial N_2]  \\
&=
(u_1|_{\partial E_2} \cup
u_2|_{\partial E_2} \cup
u_3|_{\partial E_2}) \cup (-[\partial E_2])  