The quotient space $S^3/S^1$ is homeomorphic to $S^2$

Write $S^3=\{(z_1,z_2)\in\mathbb{C}^2:|z_1|^2+|z_2|^2=1\}$. Let $S^1$ act on $S^3$ by $z\cdot(z_1,z_2)=(zz_1,zz_2)$.

(a) Show that $f:S^3\rightarrow \mathbb{C}\times\mathbb{R}$ given by $f(z_1,z_2)=(z_1\bar{z_2},|z_1|^2)$ induces a homeomorphism $\bar f:S^3/S^1\rightarrow K$, where $K$ a compact subset of $\mathbb{C}\times\mathbb{R}$.

(b) Show that $\bar f(S^3/S^1)\cap(\mathbb{C}\times\{t\})$ is homeomorphic to a circle for all $t\in(0,1)$.

(c) Show that $S^3/S^1$ is homeomorphic to $S^2$.

For (a), the equivalence classes of $S^3/S^1$ are $[x]=\{y:\exists t\in[0,1)\;\; \mathrm{with}\;\; e^{2\pi it}y=x\}$. But I'm not sure how to show it is a homeomorphism. $K$ should be closed and bounded so that it is compact by Heine-Borel.

For the rest, I'm pretty stumped.

Solution 1:

I'll just address part (a). The key fact to use is:

If $g\colon K\to X$ is a continuous injective map with $K$ compact, then $g$ is a homeomorphism onto its image.

Now $f$ has been defined as a map from $S^3\subset \mathbb{C}^2$ into $\mathbb{C}\times\mathbb{R}$. You also have $S^1$ acting on $S^3$. It just so happens that $f$ is constant on the $S^1$-orbit of every point in $S^3$. To check this, you just need to verify that for every $(z_1,z_2)\in S^3$, and every $z\in S^1$, $f(z\cdot(z_1,z_2))=f(z_1,z_2)$ (the dot means use the action of $S^1$). Because of this, $f$ determines a map $\overline{f}\colon S^3/S^1\to \mathbb{C}^2$, which is defined by $\overline{f}([z_1,z_2])=f(z_1,z_2)$. The computation you just did ensured that this map was well defined (the value of $\overline{f}$ is independent of the representative you use in the equivalence class $[z_1,z_2]$).

To us the fact above, we first have to check that $S^3/S^1$ is compact. This follows immediately from the fact that quotient maps are continuous, since the continuous image of a compact space is compact. Now, we have to check that $\overline{f}$ is injective. Suppose $\overline{f}([w_1,w_2])=\overline{f}([z_1,z_2])$. By definition, this means $w_1\overline{w_2}=z_1\overline{z_2}$ and $|w_2|^2=|z_2|^2$. The second equation says that $w_2$ and $z_2$ have the same modulus. So there exists $z\in S^1$ such that $zw_2=z_2$. Now take the first equation and plug in $\overline{z}\overline{w_2} $ for $\overline{z_2}$. Then solve for $z_1$ and you get that $z_1=zw_1$. We just showed that $$z_1=zw_1\qquad\text{and}\qquad z_1=zw_2$$ for some $z\in S^1$. This exactly means that $(z_1,z_2)=z\cdot(w_1,w_2)$, so $[z_1,z_2]=[w_1,w_2]$ and $\overline{f}$ is injective.