Klein Bottle Embedding on $\mathbb{R}^4$. First of all, I am aware of the question in How to embed Klein Bottle into $R^4$ , which was inconclusive. Anyway, I've made some progress, but I still have a question.
I am using Do Carmo's Riemannian Geometry, and struggling to solve a problem.
The problem is:
Show that the mapping $G:\mathbb{R}^2\to\mathbb{R}^4$ given by
$$G(x,y)=((r\cos (4\pi y)+a)\cos (4\pi x),(r\cos (4\pi y)+a)\sin (4\pi x),r\sin (4\pi y)\cos (2\pi x),r\sin (4\pi y)\sin (2\pi x)))$$
induces an embedding of the Klein bottle into $\mathbb{R}^4$ (It is a slightly different function from the one in the book, but works in the same way).
First of all, it's not hard to see that 
$$G(x+n,y+m)=G(x,y)\text{ whenever }m,n\in\mathbb{Z}.$$
Therefore, this mapping is well-defined over the torus $\mathbb{T}^2$. What I need now is to show that $G(-x,-y)=G(A(x,y))=G(x,y)$, where $A$ is the antipode mapping. If this were true, then the mapping G would be well-defined over the Klein Bottle, but it's obvious that this is false.
Am I working wrong here somewhere?
 A: 
Consider the embedding $\psi:\Bbb R\times\Bbb R\big/{\Bbb Z\times\Bbb Z}\hookrightarrow\Bbb R^3$
$$\psi([\theta,\tau])=
\begin{pmatrix}
\cos(2\pi\theta) &-\sin(2\pi\theta) & 0\\
\sin(2\pi\theta) &-\sin(2\pi\theta) & 0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
2+\cos(2\pi\tau)\\
\sin(2\pi\tau)\\
0
\end{pmatrix}$$
Then one verifies 
(on the picture and then by doing the math) that $-\psi([\theta,\tau])=\psi([\theta+\frac12,-\tau])$
Thus the Klein bottle is the quotient of $\Bbb R\times \Bbb R$ by the (non-commutative) group $G=\langle v,t\rangle$ of homeomorphisms generated by
vertical displacement by one
$$v(\theta,\tau)=(\theta,\tau+1)$$
and a twist
$$t(\theta,\tau)=(\theta+\frac12,-\tau)$$
(Notice that $t^2=h$ the horizontal displacement by one $h(\theta,\tau)=(\theta+1,\tau)$)
You'll verify that $G(v(\theta,\tau))=G(\theta,\tau)=G(t(\theta,\tau))$ so $G$ descends to a map $\tilde{G}:K\to \Bbb R^4$. $\tilde{G}$ is an injective (easy verification) immersion ($G$ already was). By compactness of $K$, it is an embedding.
