Consider the quotient of $T^3 = \mathbb{S}^1\times \mathbb{S}^1\times \mathbb{S}^1$ by the $\mathbb{Z}/3\mathbb{Z}$ action $(a,b,c) \mapsto (c,a,b)$, i.e., cycling the coordinates.
Is this diffeomorphic to $\mathbb{S}^3$? I was told it was, but am having difficulty showing this is true.
I have shown that the third symmetric power of $\mathbb{S}^1$ is the orientable disk bundle $\mathbb{S}^1\times\mathbb{D}^2$, I was hoping to describe my space as some sort of branched cover.
The proof consists of two steps, I will you to work out the details.
Consider a flat 2-dimensional torus $T^2$, the quotient of the Euclidean plane $E^2$ by an isometric discrete action of the group ${\mathbb Z}^2$ which is normalized by the group $\Gamma={\mathbb Z}_3$ be acting by rotations around the origin. Then this action descends to an isometric action of $\Gamma$ on $T^2$ itself. Show that $T^2/\Gamma$ is homeomorphic to $S^2$. (The easiest way to solve this is to find a hexagonal fundamental polygon $X$ for the action of ${\mathbb Z}^2$ on $E^2$ which is also invariant under the action of the rotational group $\Gamma$ and work out identifications of points in $X$ by the generators of ${\mathbb Z}^2$ and of $\Gamma$. Alternatively, prove that whenever $\Gamma$ is a finite group of orientation-preserving diffeomorphisms acting non-freely on the torus, then $T^2/\Gamma$ has Euler characteristic $2$ by using, say, the Riemann-Hurwitz formula.)
Now, let's turn to your question. The action of $\Gamma={\mathbb Z}_3$ on $T^3$ lifts to an isometric action on the universal cover $E^3$ of the torus $T^3$. This action normalizes the group $G\cong {\mathbb Z}^3$ of covering transformations of the covering map $E^3\to T^3$. Next, note that this action of $\Gamma$ fixes pointwise a straight line $L$ in $E^3$ (the one covering the diagonal in $S^1\times S^1\times S^1$ also fixed by $\Gamma$). Let $E^2\subset E^3$ denote a plane perpendicular to the line $L$. You will next observe that the stabilizer $H$ of $L$ in $G$ is isomorphic to ${\mathbb Z}^2$. The same is true for every plane parallel to $E^2$, the stabilizer will be the same subgroup $H< G$. Note that the product decomposition $E^3=E^2\times L$ is preserved by $G$, hence, descends to a $\Gamma$-invariant product decomposition of $T^3$ as $T^2\times S^1$. The torus factors $T^2\times \{t\}$ in this decomposition will be quotients by $H$ of the planes parallel to $E^2$.
Now, Step 1 shows that for each $t\in S^1$, the quotient of $T^2\times \{t\}$ by the action of $\Gamma$ is $S^2$. From this, you conclude that $T^3/{\mathbb Z}_3$ is homeomorphic to $S^2\times S^1$.
