Think of the three-sphere as given by $\lbrace|z|^2+|w|^2=1, \;z,w\in \mathbb{C}^2\rbrace$. We can regard it in terms of Hopf coordinates
\begin{align\*}
z&= \cos(\theta/2)e^{i(\phi+\psi)}\\
w&= \sin(\theta/2)e^{i\psi}
\end{align\*}
where $0\leq\theta\leq \pi$ and $0\leq \phi,\psi<2\pi$. Now I want to consider the orbifold obtained when we divide $S^3$ by the following action:
$$(z,w)\mapsto (z,e^{2\pi i/k}w)$$
for $k$ some integer. In terms of the Hopf coordinates, this identifies
$$(\phi,\psi)\sim (\phi-2\pi /k,\psi+2\pi /k)$$
Is the resulting space a Seifert fibration with 
$$(b,g;(p\_1,q\_1),(p\_2,q\_2))=(1,0;(k,-1),(-k,1))$$
where $b$ is the Euler number of the circle fibration, $g$ the genus of the base and $(p\_i,q\_i)$ label the fixed points of the action described above?
Also, applying the same orbifolding to the Lens spaces $L(n,1)$, is it correct that the resulting orbifolds are the Seifert fibrations
$$(b,g;(p\_1,q\_1),(p\_2,q\_2))=(n,0;(k,-n),(-k,n))$$
And if this is true, are these orbifolds also known in another way than just by their Seifert "invariants"?



Orbifold of the three-sphere (and lens spaces)
The quotient of $S^3$ by your action is again $S^3$. You can see that by dividing the sphere into regions where (say) $|w| \leq 1/2$ and $|w| \geq 1/2$. The quotient of each of these is a solid torus, and you can check that they are glued together to give a 3-sphere. A sanity check is that the fixed point set is the circle $w=0$, and that the projection gives the standard branched cover of $S^3$ over itself. The quotient orbifold is $3$-sphere with an unknot branch set of order $k$.


I'm not sure what you mean by `applying the same orbifolding to the Lens spaces...'. 


