**What it means for a covering of a sphere to be branched:** Let $f:X \to Y$ be a map of Riemann surfaces. We are particularly interested in the case that $Y$ is $\mathbb{CP}^1$; in this case, $Y$ has the topology of a sphere. At most points $y$ in $Y$, there will be a neighborhood $V$ of $y$ so that $f^{-1}(V)$ is just a union of $n$ disjoint copies of $U$, each mapping isomorphically to $V$. These are the points where there is no branching.

At a few points of $y$, something different will happen. Let $x$ be a preimage of $y$, let $V$ be a small neighborhood of $y$ and let $U$ be the connected component of $f^{-1}(V)$ containing $x$. At these points, the map $f$ looks like $t \mapsto t^e$, as a map from the unit disc in $\mathbb{C}$ to itself. This is called branching.

The generic situation is that, at finitely many points of $Y$, one of the preimages is branched with $e=2$ and the other $n-2$ preimages are unbranched.

Algebraically, if $t$ is a local coordinate on $X$, then we have branching where $\partial f/\partial t$ vanishes. Even if we can't explicitly write $f$ as a function of $t$, if we can find a polynomial relation $P(f,t)=0$, then we have $(\partial P/\partial f)(\partial f/\partial t) = \partial P/\partial t$, so branching will occur when $\partial P/\partial t=0$.

**The relation between branched covers and the symmetric group**: Let $f: X \to \mathbb{CP}^1$ be a branched cover. Let $R \subset \mathbb{CP}^1$ be the points over which branching occurs. Then $f^{-1}(\mathbb{CP}^1 \setminus R) \to \mathbb{CP}^1 \setminus R$ is a cover, in the sense of algebraic topology.

As you probably know, connected covers of a space $U$ are classified by subgroups of $\pi_1(U)$. In particular, degree $n$ covers are classified by index $n$ subgroups. I like to recast this and say that degree $n$ connected covers of $U$ are classified by transitive actions of $\pi_1(U)$ on an $n$-element set. This has the advantage that, more generally, we can say that degree $n$ covers of $U$ are classified by actions of $\pi_1(U)$ on an $n$-element set.

Now, in this case, $\pi_1(\mathbb{CP}^1 \setminus R)$ is isomorphic to the group generated by $R$, modulo the relation $\prod{r \in R} [r]=1$. You should be warned that this isomorphism depends on some choices. First of all, we need to choose a base point $y$ in $\mathbb{CP}^1 \setminus R$! Even once we've done that, we need to choose loops based at $y$, circling each of the elements of $R$, and disjoint from each other away from $y$. These will then be the classes $[r]$. The order that the product above is taken is related to the circular order at which these loops come in to $y$.

So, to covers of $\mathbb{CP}^1 \setminus R$ correspond to maps from this group to $S_n$. To give such a map, we choose an element $b(r)$ of $S_n$ for each $r$ in $R$; these must obey $\prod_{r \in R} b(r)=1$. (I am being very sloppy about when two such maps give isomorphic covers, and, indeed, what it means to say two covers are isomorphic.)

The relation between the geometry of the cover, and the map $b(r)$ is the following: If the permutation $b(r)$ has cycles of lengths $e_1$, $e_2$, ..., $e_k$, then $f^{-1}(r)$ contains $k$ points, which are branched with degrees $e_1$, $e_2$, ..., $e_k$.

**Useful facts to know:** The cover is connected if and only if the action of the $b(r)$ on $[n]$ is transitive.

The genus of $X$ is given by, the Riemmann-Hurwitz formula:
$$2g-2 = -2n+\sum_{r \in R} (n-\#\mbox{cycles of $r$}).$$

It is very difficult to obtain an explicit equation for $X$ from the data of $R$ and $b:R \to S_n$.