The Einstein static universe metric is $$ds^2=-dt^2 + d\chi^2 + \sin(\chi)^2d\Omega^2$$ where $-\infty<t<\infty$ , $0<\chi<\pi$ and $d\Omega^2$ is the metric on a $S^2$. It describes the topology of $R\times S^3$. Suppressing the $S^2$ coordinates, why is it always represented as an infinite cylinder if $\chi$ only goes from $0$ to $\pi$? to be a cylinder it would have to go from $-\pi$ to $\pi$ where the end points are identified. So shouldn't it just be half an infinite cylinder (i.e. just a an infinite strip)? If anything it seems to me that the cylinder would be a double covering of $R\times S^3$.
Here a picture to visualise (they use $r′$ instead of $\chi$): https://www.researchgate.net/profile/Alberto_Rozas-Fernandez/publication/267338977/figure/fig1/AS:295687481774090@1447508797819/Fig-2-The-flat-FRW-spacetime-filled-with-a-k-essential-scalar-field-is-conformal-to-the.png