Input Output formalism for a single-sided cavity in quantum optics, all modes same direction?

I have been trying to understand the input output formalism for a cavity by reading chapter 7 of Walls and Milburn (D. F. Walls and G. J. Milburn. Quantum optics. Springer, 2006.). There is something that I am struggling to understand. This input output formalism describes quantum mechanically the interaction between the internal field of a single-mode cavity, and the extra-cavity fields, namely the cavity inputs and outputs. However, when describing the model for a single-sided cavity, he said explicitly that all the fields considered are propagating in the same direction, the positive x direction. This is the quote right above equation 7.1 of the book: "We will assume that the only modes that are excited have the same plane polarisation and are all propagating in the same direction, which we take to be the positive x-direction." This puzzles me greatly. He surely wants to describe both input and output to the single-sided cavity. If all the fields considered are propagating in the same direction, how can there be both input and output? Shouldn't the input and output fields for a single-sided cavity necessarily be propagating in opposite directions?

In his model, the field operator describing the extra-cavity field is as follows: $$b(x,t)=e^{-i \Omega (t-x/c)} \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} d \omega b(\omega) e^{-i \omega (t-x/c)}.$$ where each $$b(\omega)$$ is an annihilation operator for an extra-cavity mode with frequency $$\omega$$ (actually $$\omega+\Omega$$, where $$\Omega$$ is the carrier frequency), $$x$$ is the position, $$t$$ means time and $$c$$ is the propagation velocity. His way of describing input and output is as follows, which, because of the above doubt that I have, I find quite hard to understand. He defined a specific combination of the extra-cavity field at an early time $$t_0$$ as the input and another specific combination of the extra-cavity fields at a late time $$t_1$$ as the output: $$a_{IN}(t)=- \frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega e^{-i \omega (t-t_0) b_0(\omega)},$$ $$a_{OUT}(t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega e^{-i \omega (t-t_1) b_1(\omega)}.$$ where $$b_0(\omega)$$ and $$b_1(\omega)$$ are the time-dependent heisenberg picture operator $$b(\omega, t_0)$$ and $$b(\omega, t_1)$$, respectively. This sounds quite natural to me, since I think he probably has the whole thing in mind as a scattering process. The field at a very early time is the input and the field at a very late time is the output. He shows that under some very reasonable approximations one can get two very simple equations $$\dot{a}(t)=-\frac{i}{\hbar} [a(t),H_{SYS}] + \frac{\gamma}{2}a(t) -\sqrt{\gamma}a_{IN}(t),$$ and $$a_{IN}(t)+a_{OUT}(t)= \sqrt{\gamma}(t),$$ which then lead to quite reasonable physical predictions. However, both these input and output field operators are constructed using the mode annihilation operators $$b(\omega)$$ that he introduced at the beginning, which - he stated explicitly as mentioned above - are all modes propagating in the positive $$x$$-direction! So whatever interaction these modes may have with the cavity, any kind of propagation have to be in the positive $$x$$-direction, right? How then can one possibly describe reflection?