To first order, the radial and transverse comoving separations are
$$
\Delta\chi_\parallel=\frac{\Delta z}{H(z)}
=\frac{f(z,\Omega_{m,0})}{H_0}\Delta z,
\qquad
\Delta\chi_\perp=\chi(z)\Delta\theta
=\frac{I(z;\Omega_{m,0})}{H_0}\Delta\theta,
$$
where
$$
f(z,\Omega_{m,0})=
\frac1{\sqrt{\Omega_{m,0}(1+z)^3+1-\Omega_{m,0}}}.
$$
The local spatial geometry is that of <Euclidean space>, so the <Pythagorean theorem> gives
$$
\Delta\chi
=\frac{f}{H_0}
\sqrt{(\Delta z)^2+\left(\frac I f\right)^2(\Delta\theta)^2}.
$$
Consequently the requested function is
$$
\boxed{y(z,\Omega_{m,0})=
\frac{I(z;\Omega_{m,0})}{f(z,\Omega_{m,0})}
=H(z)\chi(z)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 1 / d / Solution

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For a right triangle with perpendicular side lengths $a,b$ and hypotenuse $c$, the Pythagorean theorem states $a^2+b^2=c^2$.


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