The figure-eight complement is a punctured-torus bundle

In $S^3=\{(z_1,z_2)\in {\mathbb{C}}^2:\mathrm{\mid }{z}_1{\mathrm{\mid }}^2+\mathrm{\mid }{z}_2{\mathrm{\mid }}^2=1\}$, the standard Hopf link is $H=\{z_1{z}_2=0\}$. On its complement, $f(z_1,z_2)=z_1{z}_2\mathrm{/}\mathrm{\mid }{z}_1{z}_2\mathrm{\mid }$ is a submersion to $S^1$. Writing $r=\mathrm{\mid }{z}_1{\mathrm{\mid }}^2$, $\theta =\arg z_1$, and $\varphi =\arg z_2$, a fiber is $0<r<1$, $\varphi =-\theta +\text{constant}$, hence an annulus. For base angle $s\in [0,2\pi ]$, transport by ${\theta }_s=\theta +\chi (r)s$, ${\varphi }_s=\varphi +(1-\chi (r))s$, where $\chi =1$ near $r=0$ and $\chi =0$ near $r=1$, fixes both boundary circles at $s=2\pi$ and makes one full annular twist between them. The sign is determined by the signed linking of the positive normal push-off of the core with the core: changing the handedness of the Hopf band reverses the twist. With $a{\cdot }_{F}b=+1$, the positive twist about $a$ acts by $b\mapsto a+b$, while the negative twist about $b$ acts by $a\mapsto a+b$. Hence the two band monodromies have the stated actions $R_a$ and $L_b$.

The square plumbing is the Murasugi sum in Construction 5.37 of [F3]. Theorem 5.38 of [F3] gives a fibration with fiber $F$ and return map the composition of the two Hopf-band monodromies extended by the identity; in the stipulated order, step 1.1 identifies it as $R_a\circ L_b$. Example 5.39 of [F3] identifies the boundary of the opposite-handed sum as the standard figure-eight knot. Deleting the binding neighborhood as in [F1] therefore gives the stated exterior as the mapping torus.

Each annulus has Euler characteristic $0$, and the plumbing square has Euler characteristic $1$, so $\chi (F)=0+0-1=-1$. The plumbing is connected, oriented, and has one boundary component by the stated gluing; hence $\chi (F)=1-2g(F)$ gives $g(F)=1$. Thus $F$ is a once-punctured torus.

In the ordered basis $([a],[b])$, the columns of $R_a$ are $(1,0)$ and $(1,1)$, while the columns of $L_b$ are $(1,1)$ and $(0,1)$. Their product is $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$. Inclusion of the once-punctured torus into the torus obtained by capping its sole boundary induces an isomorphism on $H_1$, since attaching a disk along the boundary adds no 1-cycle or 2-cycle relation in degree one. Extending the boundary-fixing monodromy by the identity on that disk therefore has the same displayed action.