Branched covering of $S^3$ over $S^1$ I am currently thinking about branched covering spaces and want to visualise how exactly we create a covering of $S^3$ branched over $S^1$.
I can see how if we start with a unit complex disc and apply the map $z \mapsto z^k $ to its boundary $S^1$, we can suspend it, therefore obtaining a covering of $\Sigma S^1 = S^2$ branched over the poles $S^0$ with branching index $k$. However, I don't see why, if we suspend $S^2$ in the same manner to obtain $S^3$, we get a covering that is branched over $S^1$. I thought that since we collapse the $S^2$-cylinder at the bases, we get branching over $S^0$ again, and I am struggling to imagine the picture (or understand this in some other way).
Thanks!
 A: Suspending $z \mapsto z^k$
Think about the case of $\Sigma S^1 = S^2$ with the map $\Sigma f: S^2 \to S^2$, where $f : S^1 \to S^1$ is given by $f(z) = z^k$.
The fact that such $\Sigma f$ is branched along two points follows from the fact that $f$ is not branched at all, i.e. it is an honest covering.
This is why $\Sigma f$ is a covering outside the two points collapsed in the suspension. Denote these points by $x, y$.
Now, if you consider $\Sigma \Sigma f : S^3 \to S^3$, it is no longer true that $\Sigma \Sigma f$ is a covering outside the two points that come from collapsing $0 \times S^2$ and $1 \times S^2$ in $I \times S^2$. In fact, outside these two points, it is a branched covering, branched along two intervals. 
Let me elaborate on that. We have $S^3 = I \times S^2 / \sim$, where $\sim$ is a relation collapsing $0 \times S^2$ and $1 \times S^2$ to two different points $z,w$. Now, $S^3 \setminus \{z,w\}$ is homeomorphic to $(0,1) \times S^2$, and on this set, we have $\Sigma \Sigma f (t,p) = (t, \Sigma f (p))$. It follows immediately that such map is a covering outside $A=(0,1) \times \{x,y\}$, and branched along $A$, since $\Sigma f$ is branched along $\{x,y\}$. 
Summing up (sets), you get that $\Sigma \Sigma f$ is a covering outside $B = A \cup \{z,w\}$. Now, notice that the two segments of $A$ are just disjoint segments connecting $z$ and $w$, so $B$ is in fact a circle! Thus $\Sigma \Sigma f$ is branched along a circle.
Branched covering of $S^3$ via identification $S^3 \setminus S^1 \simeq \mathring{D}^2 \times S^1$
The way I usually imagine the coverings of $S^3$ along $S^1$ is this. Think of $S^3$ as a one-point compactification of $\mathbb{R}^3$. Consider a circle $S \simeq S^1$ in $\mathbb{R}^3$, embedded in a trivial way, say via the map $\theta \mapsto (\cos \theta, \sin \theta, 0)$. Let $D_0$ be a disk whose boundary is $S$. Imagine or draw this on a piece of paper. Now, we may start pushing the interior of $D_0$ up or down, without moving the boundary, and obtaining a family of disks $D_t$ intersecting only on their common boundary $S$. This gives us a local foliation of the complement of $S^1$. We may continue to push the disks further and further away on both sides, with the ''limiting'' disk being the complement of $D_0$ in the plane $\{ (a,b,0):a,b \in \mathbb{R}\}$. This is in fact an honest disk in $S^3$, containing the point in infinity. Denote it as $D_\infty$.
This way one obtains a family of disks $\{D_t\}$ in $S^3$, parametrized by $t \in \mathbb{R}$. Since the ''limiting disk'' you obtain by pushing upwards or downwards is the same $D_\infty$, you can think of this family parametrized by $z \in S^1$, and denote it $\{D_z\}_{z\in S^1}$. So, we have just ''shown'' that $S^3 \setminus S^1 \simeq \mathring{D}^2 \times S^1$, and now you may define a map $g$ analogous to $\Sigma \Sigma f$ above (in fact, it will be the same map modulo a suitable homeomorphism) by setting $g$ equal to the identity on $S^1$ and $g(x,z) = (x,z^k)$ on $\mathring{D}^2 \times S^1$.
Of course, you could use your favorite way to prove that $S^3$ can be obtained by gluing two copies of $D^2 \times S^1$ along the boundary, then collapse disks of one of the copies to get $S^1$, obtaining $S^3$ again, with the decomposition $S^3 = S^1 \cup \mathring{D} \times S^1$ as above. However, I like the $\mathbb{R}^3$ image because you can see how this branched covering works: indeed, it just pushes the disk $D_z$ to $D_{z^k}$ (in an appropriate manner).
