= Solution
Use a radial <null geodesic> of the <FRW metric> in the form $ds^2=-dt^2+a(t)^2d\chi^2$ along a fixed spatial direction, with $c=1$. For an emitter and observer at fixed <comoving coordinates>, the comoving separation is
$$
\chi_o-\chi_e=\int_{t_e}^{t_o}\frac{dt}{a(t)}.
$$
Apply this to two adjacent wave crests. Since their endpoint separation is the same, subtraction to first order gives $\delta t_o/a(t_o)=\delta t_e/a(t_e)$. These intervals are the local periods measured by the <comoving observers>, so the received wavelength is stretched by $a(t_o)/a(t_e)$. Consequently the <cosmological redshift> obeys
$$
\boxed{1+z=\frac{\lambda_o}{\lambda_e}=\frac{a(t_o)}{a(t_e)}=\frac1{a(t_e)}}
$$
when $a(t_o)=1$. This relation assumes comoving emitter and observer; additional <peculiar velocities> supply kinematic redshift factors.
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