= Solution
Take the causally unwrapped <universal cover> of <Anti-de Sitter spacetime>, with curvature radius $L$. In global coordinates,
$$
ds^2=\frac{L^2}{\cos^2\rho}\bigl[-dT^2+d\rho^2+\sin^2\rho\,d\Omega_{d-2}^2\bigr],
\qquad 0\le\rho<\pi/2,\quad T\in\mathbb R.
$$
The ordinary radial <Penrose diagram> is a half-strip: the left edge $\rho=0$ is a regular center and the right edge $\rho=\pi/2$ is a timelike <conformal boundary>. For displaying signed spatial motion, the figure doubles the radial diameter to $-\pi/2<\rho<\pi/2$. Its two ends represent antipodal spatial directions of the higher-dimensional boundary; they are not a claim that the full AdS boundary has two disconnected components.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-75-ads-boundary.png]
{title=Conformal strip of the universal cover of anti-de Sitter spacetime, with timelike conformal boundaries and past/future timelike ends}
The vertical edges are spatial infinity at every finite global time, forming the timelike boundary $\mathbb R\times S^{d-2}$. The arrows mark the future and past timelike ends $T\to\pm\infty$, conventionally $i^+$ and $i^-$. There are \b[no separate null-infinity edges like Minkowski $\mathscr I^\pm$]: radial null rays satisfy $dT=\pm d\rho$ and reach the timelike boundary in finite global time. Their physical affine parameter is nevertheless infinite there. The drawing is a finite excerpt of an infinitely tall strip, not an identification of its top and bottom. On the original hyperboloid global time is periodic and there are <closed timelike curves>; unwrapping it is essential to the usual physical maximally extended AdS spacetime.
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