continuous action on the sphere

Question

I am wondering if the following action defines a continuous action on (a quotient) of $S^2$.

Consider the rotation action of $S^1$ on $S^2$ along the vertical axis containing the north and south pole, which will be a continuous action with fixed points precisely the poles.

Now let $X$ be the space obtained by identifying the north and south poles into a single point. Is there a well defined continuous action of $S^1$ on $X$, with a single fixed point now?

Answer 1

A continuous action is a continuous function $f:S_1\times S_2\to S_2$.

You want a continuous function $\tilde f:S_1\times (S_2/\sim)\to S_2/\sim$, where $\sim$ identifies the poles.

Let $X=S_1\times S_2$, and define $\approx$ on $X$ by $(x,y)\approx (x',y')$ when $y\sim y'$. I claim that this is equivalent to finding a continuous function $\tilde f:X/\approx \;\to S_2/\sim$. To prove this, it suffices to prove $X/\approx$ is homeomorphic to $S_1\times (S_2/\sim)$, which seems tedious at worst.

We can show $\tilde f$ exists, and in fact that it makes the below diagram commute: $$ \require{AMScd} \begin{CD} X @>{f}>> S^2\\ @V\pi_1 VV @VV\pi_2 V \\ X/\approx @>>{\tilde f}> S^2/\sim; \end{CD} $$ The map $\tilde f$ is unqiely defined as $\pi_2\circ f\circ \pi_1^{-1}$. Note that $\pi_1^{-1}$ is not a well defined map, as the preimage under $\pi_1$ may have multiple values. In order for $\tilde f$ to be well defined, all you need is that $\pi_2\circ f(x)$ is the same for all $x\in \pi_1^{-1}(x')$, for all $x'\in X/\approx$. In your case, this is easy to show. The only points with multiple preimages are of the form $(x,\{\text{poles}\})$, and $\pi_2\circ f$ of both poles is $\{\text{poles}\}$, for any $x$.

All that remains is to show $\tilde f$ is continuous. For this, we use the universal property of topological quotients, which implies that $g:X/\approx\;\to Z$ is continuous iff $g\circ \pi:X\to Z$ is continuous. In our case, $\tilde f\circ \pi_1=\pi_2\circ f$, and the latter is a composition of continuous functions, so $\tilde f\circ \pi_1$ is continuous, so $\tilde f$ is.

Answer 2

Yes, the quotient action is continuous.

One way to embed the picture into $\mathbb{R}^3$: the resulting space is the surface obtained by revolving a vertical circle tangent to the $z$ axis around said axis.

Let that circle have equation $(x-\frac12)^2+z^2=\frac14$, or equivalently $x^2+z^2=x$. The rotated surface then has equation $r^2+z^2=r$ in cylindrical coordinates. For nonzero $r$, we map the point $(r,\theta,\sqrt{1-r^2})$ on the unit sphere to $(r,\theta,\sqrt{r-r^2})$ on the surface and its reflection $(r,\theta,-\sqrt{1-r^2})$ to $(r,\theta,-\sqrt{r-r^2})$. Since we kept $r$ and $\theta$ the same, the rotation action commutes with this map.