Context
Real Projective Plane(contains the definition I'm about to mention) I'm interested in visualizing transformation between closed cross-capped disk, sliced cross-capped disk, and the self-intersecting disk, where the latter two objects are homeomorphic. So we have the following diagram to describe the relationship between these objects. Let $C$ denote the closed cross capped disk, $C_S$ denote the sliced cross-capped disk, and $D_I$ denote the self intersecting disk, homeomorphic to the sliced one, then,
$$\require{AMScd} \begin{CD} [0,2\pi]^2/R_1 @>p\text{ (quotient)}>> [0,2\pi]^2/R_2\\ @Vq_1VV @Vq_2VV\\ C_S @>f\text{ (quotient)}>> C\\ @Vg(\cong)VV \\ D_I \end{CD} $$ where $f:C_S\to C$ is a continuous quotient maps, and $g$ is a homeomorphism, and $R_1$ is generated by a $(u,0)\sim(2\pi-u,2\pi)$, while $R_2$ is generated by the relation in $R_1$ as well as $(0,v)\sim(2\pi,2\pi-v).$ where $u,v\in[0,2\pi]^2$
Questions:
1. What is the parametric equation for a sliced cross-capped disk? This question induced the folowwing one.
2. How to construct an explicit associative map for $k=f。g^{-1}$, that is $k_t=(f。g^{-1})_t:D_I\times[0,1]\to C$, where $k_{1/2}(u,v)=C_S$? (I'm curious about this because I'm trying to use computer program to simulate the process in order to have a better visual understanding about real projective plane and also classification theorem)
My Attempt:
Recall the parametric equation for $C$ $$ \begin{cases} x=r(1+\cos(v))\cos(u)\\ y=r(1+\cos(v))\sin(u)\\ z=-r\tanh(u-\pi)sin(v)\\ \end{cases} $$ and for $D_I$. $$ \begin{cases} x_1=\dfrac{rv\cos(2u)}{2\pi}\\ y_1=\dfrac{rv\sin(2u)}{2\pi}\\ z_1=\dfrac{rv\cos(u)}{2\pi}\\ \end{cases} $$
The first one: $$k_t(u,v)=(x-(x-x_1)(1-t),y-(y-y_1)(1-t),z-(z-z_1)(1-t))$$ But my program gives me a sequence of chaotic frames that show the discontinuous behavior of this transformation. You can see that the bottom of the surface is disconnected (the white part at the bottom). (see the figure below)
Seems like this is not valid.
The second one: $$ \begin{array} \\x(u,v,t)=r\dfrac{\cos(v)-(1-t)(\cos(v)-v)}{1-2(1-t)(1/2-\pi)}\big\{\cos{u}\big[1+2(1-t)(\cos(u)-1/2)\big]-(1-t)\big\};\\ y(u,v,t)=r\dfrac{\cos(v)-(1-t)(\cos(v)-v)}{1-2(1-t)(1/2-\pi)}(2-t)\sin(u)[1+(\cos(u)-t\cos(u)-(1-t)];\\ z(u,v,t)=? \end{array} $$ I don't know how to deal with that hyperbolic tangent function......
Any idea will be greatly appreciated. Thanks for your time and efforts. If there is anything unclear because of this long post, please let me know. :)