# Mapping from X to $S^4$

I found this question in a book (Topology II: homotopy and homology: classical manifolds)

Show that the quotient space $X = S^2 \times S^2 / [(x_1,x_2) \sim (Rx_1,Rx_2)]$ where R is the reflection in the equatorial plane, is homeomorphic to $S^4$.

I am still in the process of learning topology and I really don't think I can prove this result (or even understand a proof if someone were generous enough to provide me with one). I apologize in advance if this is trivial, but it will be of great help if someone could give a homeomorphism. I would really like to use the mapping of the space $X$ to $S^4$ in my work. If you are aware of the proof and would provide it in your answer, I will definitely make an effort to understand it.

Thank you for your time.

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I think it might be helpful to restrict your $x_1$ coordinate to be on or above the equator. As long as $x_1$ isn't on the equator, this gives a unique representative for every point, so you get $\{x_1\}\times S^2$. When $x_1$ is on the equator, you can apply $R$ freely to the second coordinate, so you get $\{x_1\}\times D^2$. The puzzle is how to fit these together...

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I would add that thinking about why $S^1\times S^1$ modulo simultaneous reflection in each coordinate is equal to $S^2$ is helpful for developing intuition about the question. – Grumpy Parsnip Jan 24 '11 at 13:16
Thanks for the suggestions and comments. I did think about the $S^1 \times S^1$ case and you are right, it does seem very intuitive that $S^1 \times S^1 / (x_1,x_2) \sim (Rx_1,Rx_2)$ be homeomorphic to $S^2$. It seemed obvious enough that I guessed the $S^2 \times S^2$ case would be straightforward. But I could never figure it out. – Srikanth Jan 24 '11 at 17:03

I don't have enough rep points to respond directly to Eric, but that should not be the case, since $S^4$ doesn't have a boundary either. I think you're thinking of $S^2 / (x \tilde{} Rx) \times S^2 / (x \tilde{} Rx),$ which is not exactly the same space as $X$.

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Thanks. I was thinking exactly as you said and just realized it and deleted my comment. – Eric O. Korman Jan 24 '11 at 1:17

Trying to follow up on Aaron's hints.

So one way to look at $X$ is as the product space $D^2 \times S^2$ with an equivalence relation $E$ on the boundary $\partial D^2 \times S^2 = S^1 \times S^2$.

The Equivalence relation $E$ on the boundary is : $({ b_1 \in S^1 }, { b_2 \in S^2 }) \sim ({ b_1 \in S^1 }, { Rb_2 \in S^2 })$

SO $X = (D^2 \times S^1) /E$.

Now consider the following subspace $Y$ of $X$:
$Y = { [-1,1] \times S^2 }$ with $E_1 = (-1;y \in S^2) \sim (-1;Ry \in S^2)$ and $E_2 = (1;y \in S^2) \sim (1;Ry \in S^2)$ as equivalence relations on $Y$

$I=[-1,1]$ is a line passing through the origin of $D^2$. I think I could prove $Y$ to be homeomorphic to $S^3$.

Please check: This is probably a very crude way to show the homeomorphism, but this is the only way I know how to do this. May be there are some algebraic methods to prove this result ?

Visualize $Y$ as a combination two solid cylinders of $I \times D^2$. So I have two solid cylinders, $C_1$ and $C_2$, when joined along the cylindrical surfaces form $Y$. So each point on the cylindrical surface of $C_1$ has an equivalent point on the cylindrical surface of $C_2$, since they belong to the same sphere $S^2$.

The disks that belong to the top and bottom surfaces of cylinder $C_1$ are identified with the disks of cylinder $C_2$ due to equivalence relations $E_1,E_2$ on the boundaries of subspace $I \times S^2$ that it inherits from $E$. The relations are:

$E_1 = (-1;y \in S^2) \sim (-1;Ry \in S^2)$ and $E_2 = (1;y \in S^2) \sim (1;Ry \in S^2)$. \

Hence I have two cylinders $C_1$ and $C_2$ where any point on the surface of $C_1$ has one and only one equivalent point on the surface of $C_2$. A solid cylinders is homeomorphic to a solid ball, hence the subspace $Y$ is equivalent to two solid spheres where the surface points of one solid sphere are identified with surface points of the other solid sphere. This space is equivalent to $S^3$.

May be there is some way to get $S^4$ by extending the subspace $(I \times S^2)/[E_1,E_2]$ to the entire $(D^2 \times S^2)/E$ ??

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