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.

While reading a note on hopf fibration we came proving $\mathbb{CP}^1$ is homeomorphic to $\hat{\mathbb{C}}$ author says like $\mathbb{CP}^1$ has the quotient topology $\mathbb{CP}^1=S^3/\sim$. We know $\hat{\mathbb{C}}$ is homeomorphic to $S^2$, "now consider the map $f:S^3\rightarrow S^2$ which maps $(z_1,z_2)\in S^3$ corresponding to $z_1/z_2\in\hat{\mathbb{C}}$. then $f$ is continous since it is just division of two complex number,Given that $f$ is continous map from $S^3$(compact) to $S^2$(hausdorf), $f$ must be a quotient map"

I do not understand what is going on here, I am not getting what is the range of the function is it $\hat{\mathbb{C}}$? or $S^2$? and why quotient map? please tell me what is going on here? Need reference for hopf fibration in detail, please!

share|improve this question
3  
It is very unclear what you are asking. "I am not getting what is the range of the function $\widehat{\mathbb{C}}$?" This question doesn't make any sense... If this question means "is the range $\widehat{\mathbb{C}}$ or $S^2$," the paragraph above says "We know $\hat{\mathbb{C}}$ is homeomorphic to $S^2$" so I don't understand the confusion. –  Matt Oct 11 '12 at 4:40
    
Hi you are right, I am quite bad in my english, my confusion was from where to where the map is? and how it is continous from $S^3\rightarrow S^2$? –  El Angel Exterminador Oct 11 '12 at 5:15

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.