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I am trying to show that $S^2$ and $\mathbb{C}P^1$ are homeomorphic making use of the following result - see e.g. Jack Lee Introduction to topological manifolds.

Let $Y \xrightarrow{\pi_1} X_1 $, $Y \xrightarrow{\pi_2} X_2$ be quotient maps which are constant on each other fibres. Then there exist a unique homeomorphism $f:X_1\rightarrow X_2$ such that $f\circ\pi_1=\pi_2$. In fact $f=\bar \pi_2$, the map induced by $\pi_2$ on the quotient, $\bar \pi_2(x)=\pi_2(y)$ where $y$ is any point of $Y$ such that $\pi_1(y)=x$.

Now let \begin{equation}Y=S^3=\{(z^1,z^2)\subset\mathbb{C}^2 : |(z^1)|^2+|(z^2)|^2=1\},\end{equation} \begin{equation} X^1=S^2=\{(z,t)\in \mathbb{C}\times\mathbb{R}:|z|^2+t^2=1\},\end{equation} \begin{equation} X^2=\mathbb{C}P^1=S^3/ \sim, \end{equation} with the equivalence relation being $[z^1,z^2]=[u^1,u^2]$ if $(u^1,u^2)=\exp(i\ a)(z^1,z^2)$ for some real number $a$. Let $\pi_2$ be the quotient projection onto $\mathbb{C}P^1$ which we equip with the quotient topology. Let $\pi_1$ be the Hopf projection \begin{equation} \pi_1(z^1,z^2)=(2z^1 \bar{z}^2, |z^1|^2-|z^2|^2). \end{equation} The Hopf projection is onto, continuous and, since it is a fibre bundle projection, open. Hence it is a quotient map. So both $\pi_1$ and $\pi_2$ are quotient maps and one can check that they have the same fibres.

It should follow that \begin{equation} \begin{split} \bar\pi_2&:S^2\rightarrow \mathbb{C}P^1, \quad \bar\pi_2(z,t)=[z,t],\\ \bar\pi_1&:\mathbb{C}P^1\rightarrow S^2, \quad \bar\pi_1([z^1,z^2])=(2 z^1\bar z^2,|z^1|^2-|z^2|^2) \end{split} \end{equation} are one the inverse of the other. But they are not.

One can check that the inverse of $\bar\pi_1$ is given by $\bar\pi_1^{-1}(z,t)=1/\sqrt{2(1-t)}(z,1-t)$. However what I am really interested in is understanding where the argument goes wrong and $\bar\pi_1^{-1}$ is not given by $\bar\pi_2$.

UPDATE

As pointed out by John Hughes, my computation of $\bar \pi_2$, the inverse of $\bar \pi_1$ was wrong. In case anyone is interested if $(z,t)\in S^2$, $z=|z|\exp(i \phi)=\sqrt{1-t^2}\exp(i\phi)$, then \begin{equation} \bar\pi_2(z,t)= \left[ \sqrt{ \frac{1 + t}{2}} \mathrm{e} ^{ i \, \frac{\phi }{2}}, \sqrt{ \frac{1 - t}{2}} \mathrm{e} ^{ -i \, \frac{\phi }{2}} \right]. \end{equation}

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  • $\begingroup$ they could both be $f:X\to X$ and be homeomorphic but not isomorphic? $\endgroup$ – JonMark Perry Jul 4 '15 at 12:16
  • $\begingroup$ @JonMarkPerry sorry, I do not understand your comment $\endgroup$ – GFR Jul 4 '15 at 12:25
  • $\begingroup$ $\bar\pi_1$ and $\bar\pi_2$ do not need to be the inverse of each other for $S^2$ and $\mathbb{C}P^1$ to be homeomorphic. $\endgroup$ – JonMark Perry Jul 4 '15 at 12:27
  • $\begingroup$ @JonMarkPerry True, but one can show that if $\bar{\pi}_2$ is the map induced on the $\pi_1$ quotient by $\pi_2$, and $\bar \pi_1$ is the map induced on the $\pi_2$ quotient by $\pi_1$ then they are one the inverse of the other. See e.g. introduction to topological manifolds by Jack Lee theorem 3.75. $\endgroup$ – GFR Jul 4 '15 at 12:33
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This doesn't make sense to me. Suppose that $Y = X_1 = X_2 = S^1$, with the map $\pi_1$ being the identity, and the map $\pi_2$ being $\theta \mapsto \theta + 1$. The each $\pi_i$ is a quotient map where the fibers are single points, so they're constant on each other's fibers. The claim that "there's a unique homeomorphism from $X_1$ to $X_2$" is clearly nonsense as written, since $\theta \mapsto \theta + a$ is a homeomorphism for any $a$ (where everything is taken mod $2 \pi$, of course).

I think perhaps you've misread the theorem or missed some hypothesis. Since I don't have the book, I can't really check this. But until you can resolve the apparent contradiction in this simple example, I won't try to resolve the more complicated $CP^1$ vs. $S^2$ question.

Post-comment addition

Your computation of $\bar{\pi}_2$ seems wrong to me. To compute
$$ \bar{\pi}_2(z, t) $$ you should (1) find a pair $(z_1, z_2)$ in $S^3$ that projects to $(z, t)$ under $\pi_1$, and (2) apply $\pi_2$ to it.

You've assumed that the solution to the first step is the pair $(z_1, z_2) = (z, t)$, which certainly is a point of $S^3$...but it's not a point that's sent to $(z, t) \in S^2$ by the map $\pi_1$, because $$ \pi_1(z, t) = (2zt, |z|^2 - t^2). $$

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  • $\begingroup$ You are absolutely right, the statement in the book is that there is a unique homeomorphism such that the diagram commute. I.e. $\bar\pi_2\circ\pi_1=\pi_2$. I will edit the question. Everything else seems to be right though, so the problem is still there! $\endgroup$ – GFR Jul 4 '15 at 12:47
  • $\begingroup$ Additional answer provided above; it seems you're computing $\bar{\pi}_2$ incorrectly. $\endgroup$ – John Hughes Jul 4 '15 at 13:22
  • $\begingroup$ You are right, cannot believe I could not see that! Thank you so much, as soon as I have time I will update my question with the correct formula for reference, $\endgroup$ – GFR Jul 4 '15 at 13:53

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