# 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.

• Recall a continuous bijection with compact domain and Hausdorff codomain is a homeomorphism. So show your map onto $S^2$ satisfies this and you're gold. May 1, 2017 at 16:24
• Or: The continuous image of a compact space is compact. May 1, 2017 at 16:28
• To be honest, I'm not too sure what $\bar f$ represents, or how it is induced. What are some references for topological groups? Munkres doesn't say much May 1, 2017 at 17:01

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.