the quotien space of $ S^1\times S^1$

Question

Let $X=S^1\times S^1,Y=(S^1\times S^1)/{\sim}$,the equivalent relation is :$(x,y)\sim (ix,iy),x,y\in S^1$.Is $Y$ still a torus?

Answer 1

You can verify that $S^1\times S^1\rightarrow X$ is a finite covering map, since the euler number of $S^1\times S^1$ is zero, the euler number of $X$ is zero. The manifold $X$ is oriented since $(x,y)\rightarrow (ix,iy)$ respects the orientation of $S^1\times S^1$ we deduce that $X$ is the torus. To see that $X$ is oriented, you can also write $(ie^{2\pi x},ie^{2i\pi y})=(e^{2i\pi(x+1/4)},e^{2\pi(y+1/4)})$ and remark that $X$ is the quotient of $\mathbb{R}^2$ by $(x,y)\rightarrow (x+1,y), (x,y)\rightarrow (x,y+1)$ and $h:(x,y)\rightarrow (x+1/4,y+1/4)$ and notice that $h$ preserves the canonical orientation of $\mathbb{R}^2$ which induces the orientation of the torus.