I asked the same question on math.stackexchange recently (<https://math.stackexchange.com/questions/2134978/is-it-possible-to-orbifold-torus-td-into-a-sphere-sd-using-mathbbz-2>), but it didn't receive much attention there, so I decided to move the question to this forum.


Consider a torus $T^d$ constructed as a hypercube $[-1,1]^d$ with identified opposite faces. Then additionally identify pairs points $\boldsymbol{r}$ and $-\boldsymbol{r}$ within this cube. For $d>1$, what is the result of this orbifold quotient? Specifically, is the underlying space homeomorphic to sphere $S^d$ or not?


For $d=2$ it is easy to visualize this procedure simply by taking half of a square $[-1,1]^2$ and by properly "gluing" the edges. It turns out that indeed $T^2/\mathbb{Z}\_2\cong S^2(2,2,2,2)$, which has underlying space $S^2$. I am wondering whether such a statement (or a similar one) also generalizes to $d\geq 3$.



Does the torus $T^d$ 2-fold cover an orbifold $Q^d$ with underlying space $S^d$?
Given the specific quotient described in the question, the answer is no. Under this quotient there is a fixed point at the origin and the neighborhood of this point quotients to a cone over $RP^{d-1}$. For $d=2$, $RP^1=S^1$ so the cone has underlying space a disk, however for $d>2$, the underlying cone will not be homeomorphic a $d$-ball and so it is not even a manifold (a necessary condition to be $S^d$). 


For $d\geq 3$, even if we allow more interesting fixed points sets (say knots and links), we still cannot obtain $S^d$ as the 2-fold quotient (or in fact cyclic) of $T^d$. This is more commonly said in reverse that $T^d$ is not the cyclic-branched cover over $S^d$, which follows from:



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> Hirsch, U., & Neumann, W. D. (1975). On cyclic branched coverings of spheres. Mathematische Annalen, 215(3), 289-291.
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