What is the surface by identifying antipodal points of a 2-torus embedded in $\mathbb{R}^3$? We define antipodal points in $\mathbb{R}^3$ as $(x,y,z)\to (-x,-y,-z)$. As we all know, the 2-torus can be expressed as 
$$
\left(\sqrt{x^2+y^2}-R\right)^2+z^2=r^2,
$$
where $0<r<R$. 
Some were saying it was a Klein bottle. So I am considering that since torus can be regarded as a square with opposite edges identified in the same order. By identifying the antipodal points, we should also identify the interior of the identified square, so it is not quite the same as we have for Klein bottle. Can anyone hint on this? Thanks!
 A: This identification yields a Klein bottle. In particular, you can think of slicing the torus by a half-space passing through the origin, then identifying the circles cut by the boundary of this plane in an appropriate way, and seeing that the two circles one is supposed to identify are oriented in opposite directions.
More formally, let us define a torus by coordinates $(\theta,\psi)$ sending a pair of these to the coordinates $$(R\sin(\theta)+r\cos(\psi)\sin(\theta),R\cos(\theta)+r\cos(\psi)\cos(\theta),r\sin(\psi)).$$
One can observe that your identification identifies pairs $(\theta_1,\psi_1)$ and $(\theta_2,\psi_2)$ where $\theta_2-\theta_1=\pi$ and $\psi_1=-\psi_2$, taking these equations mod $2\pi$. In particular, one can quickly see that, every point in this quotient space has a unique representative with $\theta\in [0,\pi)$ and $\psi\in [0,2\pi)$ and we can see that points of the form $(\pi,\psi)$ correspond to ones of the form $(\pi,-\psi)$ and ones of the form $(\theta,2\pi)$ correspond to $(\theta,0)$. In particular, one sees that this space is the square $[0,\pi]\times [0,2\pi]$ achieved by identifying the top and bottom edges in the same direction, and the left and right edges in opposite directions, which is a Klein bottle.
