= Solution
Use $c=1$ and signature $+---$. The positively curved <FLRW metric> in the usual radial chart is
$$
ds^2=dt^2-a^2(t)\left[\frac{dr^2}{1-r^2}+r^2(d\Theta^2+\sin^2\Theta\,d\phi^2)\right].
$$
Set $r=\sin\chi$ and introduce <conformal time> by $dt=a\,d\eta$. Then $dr^2/(1-r^2)=d\chi^2$ in the chart, and the spherical polar coordinate $\chi\in[0,\pi]$ extends the spatial description to the full <three-sphere>. Thus
$$
\boxed{ds^2=a^2(\eta)\left[d\eta^2-d\chi^2-\sin^2\chi(d\Theta^2+\sin^2\Theta\,d\phi^2)\right].}
$$
The $r$ coordinate itself is not one-to-one across the equator, which is why the new polar coordinate is useful.
For $\Lambda=0$ and $k=1$, the <Friedmann equation> is $\dot a^2=(8\pi G/3)\rho a^2-1$. Since $a'=a\dot a$, it becomes
$$
\boxed{(a')^2=\frac{8\pi G\rho}{3}a^4-a^2.}
$$
For <pressureless matter>, the background <continuity equation> gives $\rho a^3=\rho_0a_0^3$. Define the conserved normalization $M=(4\pi/3)\rho_0a_0^3$. It makes the equation $(a')^2=2GMa-a^2$. The <closed matter-dominated Friedmann solution> is the cycloid
$$
\boxed{a(\eta)=GM(1-\cos\eta),\qquad t(\eta)=GM(\eta-\sin\eta).}
$$
Indeed $a'=GM\sin\eta$ and $2GMa-a^2=(GM)^2\sin^2\eta$. Integrating $dt/d\eta=a$ gives the second expression, with the <Big Bang> at $t=\eta=0$. It expands to $a=2GM$ at $\eta=\pi$ and recollapses at $\eta=2\pi$. Here $M$ is the constant appearing in the Friedmann equation; it is not the total mass of the entire closed spatial slice, whose volume is $2\pi^2a^3$.
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