# Attaching map os two squares onto a sphere in isomorph to the union of sphere and a ellipsoid.

Consider the topological spaces $$X = \partial ([0, 1]^3) \subset \mathbb{R}^3$$ (the boundary of the unity cube in $$\mathbb{R}^3$$) and Y the unit sphere $$\mathbb{S}^2 \subset \mathbb{R}^3$$. Let $$A \subset X$$ be the union on the saquares $$[0,1] \mbox{x}[0,1]\mbox{x}{0}$$ and $$[0,1] \mbox{x}[0,1]\mbox{x}{1}$$, and consider the attaching map $$f:A \rightarrow Y$$ sending the first square to the north pole and the second one to the south pole. I wanna show that: $$X \space \space U_{f} \space Y \cong Z$$ Where $$Z \subset \mathbb{R}^3$$ is the union of the unit sphere and the ellipsoid E(2,2,1).

I thought about centralizing this first cube in $$(0, 0, 0)$$ and duplicate its side by $$2$$, after that I think that I'll need to take this squares and maping it into the circular section of the ellipsoid by the plane $$XoY$$ by some function like $$(x,y,z) \longmapsto \left(2x \frac{\sqrt{1- z^2}}{\sqrt{x^2 + y^2}},2y \frac{\sqrt{1- z^2}}{\sqrt{x^2 + y^2}}, z\right)$$ but I have a difficult to formalize it.

Note: I know also that this map that I suggested isn't defined for the poles of the ellipse, but for this case we can sent the center of each square into the respective pole of the ellipse.

Can someone help me?

Thanks for the attention.

If I understand what you are asking, you just want to produce a homeomorphism between $$X$$ and $$E(2,2,1)$$. And letting $$W$$ be the boundary of the cube of side length $$2$$ centered at $$(0,0,0)$$, it's sufficient to find a homeomorphism $$f : W \to E(2,2,1)$$

For finding $$f$$, I would suggest a different approach. For each $$p \ne (0,0,0) \in \mathbb R^3$$ consider the ray $$R_p = \{rp \mid r \ge 0\}$$.

Define $$f(p) = R_p \cap E(2,2,1)$$ for $$p \in W$$.

Clearly $$f^{-1} : E(2,2,1) \to W$$ defined by $$f^{-1}(q) = R_q \cap W$$ is the inverse function of $$f$$, and therefore $$f$$ is a bijection. I'm using here that every ray intersects $$W$$ in exactly one point, and it intersects $$E(2,2,1)$$ in exactly one point.

Once you've shown that $$f$$ is continuous, it follows that $$f$$ is a homeomorphism, because every continuous bijection between compact Hausdorff spaces is a homeomorphism.

Furthermore, continuity of $$f$$ follows from the gluing theorem once you prove that its restriction to each of the six faces of $$W$$ is continuous.

Let's prove that $$f$$ is continuous on the face $$[-1,+1] \times [-1,+1] \times \{1\} \subset W$$ (for example). Given $$p=(x,y,1)$$ you have to solve for a scalar $$s_p$$, expressed in terms of $$x,y$$, such that $$s_p = s_p(x,y,1) = (s_px,s_py,s_p) \in E(2,2,1)$$ You can then define $$f(p) = s_p p$$.

You can easily find $$s_p$$ by plugging into the defining equation for $$E(2,2,1)$$ and solving: $$\frac{(s_px)^2}{4} + \frac{(s_py)^2}{4} + \frac{(s_p)^2}{1} = 1$$ $$s_p(x,y) = \frac{1}{\sqrt{\frac{x^2}{4} + \frac{y^2}{4} + 1}}$$ Clearly this is a continuous function in $$(x,y)$$.