= Solution
Put $\Gamma(\eta)=a\bar n_e\sigma_T\ge0$, the <Thomson scattering> rate per unit <conformal time>. The <cosmological optical depth> is the integrated scattering rate from emission time to observation, so a Poisson scattering process has survival probability
$$
\boxed{E(\eta)=e^{-\tau(\eta)}=\Pr(\text{no further scattering between }\eta\text{ and }\eta_0)}.
$$
Since $\dot\tau=-\Gamma$, the <cosmological visibility function> is
$$
\boxed{g(\eta)=\Gamma(\eta)e^{-\tau(\eta)}=\frac{dE}{d\eta}}.
$$
A photon scatters in an interval $d\eta$ with rate factor $\Gamma d\eta$ and then reaches us without another scattering with probability $E$, so $g\,d\eta$ is its probability of last scattering in that interval. For an optically thick initial epoch,
$$
\int_{\eta_i}^{\eta_0}g(\eta)d\eta=1-e^{-\tau(\eta_i)}\simeq1.
$$
If the initial optical depth is finite, the missing weight represents photons that have already stopped scattering before $\eta_i$.
Before <cosmological recombination>, the optical depth to the present is enormous: $E\simeq0$, and $g$ is also small because almost every scattering is followed by another. Through <photon decoupling>, $E$ rises rapidly and $g$ has its dominant positive peak. After decoupling, without <reionization>, $E$ is close to one and reaches exactly one at $\eta_0$, while $g$ is small because the remaining electron density and scattering rate are small. Residual ionization can give a weak tail; there is no second reionization peak. The peak marks the <Cosmic microwave background last-scattering surface>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-312-visibility-sketch.png]
{title=The photon visibility function during recombination}
{description=Normalized schematic without reionization. The right curve is the derivative of the left curve and has unit area; the horizontal scale and width are illustrative, not a numerical recombination calculation.}
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