Good evening.
I recently stumbled across the claim $$\mathbb{R}P^3/\mathbb{S}^1 \cong\mathbb{S}^2,$$ which is not really clear to me. Here, $\mathbb{R}P^3$ denotes real projective space of dimension $3$.
I know that $\mathbb{R}P^3 \cong\mathbb{S}^3/\mathbb{Z}_2$ (more or less per definition) and $\mathbb{S}^3/\mathbb{S}^1 \cong\mathbb{S}^2$ (due to Hopf), but I'm unsure how exactly to proceed.
Edit The context in which $\mathbb{R}P^3/\mathbb{S}^1$ arises is the following: The space of parametrized great circles on $\mathbb{S}^2$ is diffeomorphic to $\mathbb{R}P^3$ (which can be shown by identifying it first with the unit tangent bundle $S\mathbb{S}^2$, and then using $S\mathbb{S}^2 \cong SO(3) \cong \mathbb{R}P^3$) and admits a $\mathbb{S}^1$-action by time-shift, i.e. by unparametrizing the great circles.
Let $p$ be a point of the unit sphere $S^{2}$ embedded in Euclidean three-space, let $S_{p}$ denote the oriented, unit-speed parametrized unit circle in the tangent plane $T_{p}S^{2}$, so that $C_{p} = S_{p} - p$ is an oriented great circle on $S^{2}$. As noted in the question, the union of these circles may be viewed as $SO(3) \simeq \mathbf{RP}^{3}$, and the circle action described in the question rotates $S_{p} \subset T_{p}S^{2}$ about $p$ compatibly with its orientation.
The total space $\mathbf{RP}^{3}$ with this group action is a principal circle bundle whose projection mapping $\mathbf{RP}^{3} \to S^{2}$ sends each $S_{p}$ to $p$. That is, the quotient space is $\mathbf{RP}^{3}/S^{1} \simeq S^{2}$.