Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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...

share|improve this answer
1  
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 $X$.

share|improve this answer
    
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 $(D^2 \times S^2)/E$ ??

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.