What is a representation of a (branched) cover? I'm reading a paper about finite-order homeomorphisms of a closed oriented surface, say $f: M \rightarrow M$, $f^n = id$. I know that from such an $f$ we get a covering $P: M \rightarrow M_f$, where $M_f$ is the orbit space of $M$. (For simplicity, I am just assuming that this has no branch points.)
But then the author continually refers to "the representation of $P$", supposedly a map $\rho: \pi_1(M_f) \rightarrow \mathbb Z/n\mathbb Z$. I just do not see what this map is supposed to be, or how it is induced by $f$.
Here is an example I am trying to explicitly work through. Let $M$ be the genus-$4$ surface in the shape of a triangle, and let $f$ is a rotation of $2\pi/3$ about the center (imagine a fidget spinner..). Then $M_f$ is the genus-$2$ surface, with fundamental group
$$\pi_1(M_f) = \langle a_1,b_1,a_2,b_2 \mid [a_1,b_1][a_2,b_2] = 1\rangle.$$
What is $\rho$, a map from this into $\mathbb Z/3\mathbb Z$, supposed to be?
A search through Google and Hatcher (my reference text) did not yield any definition. Please let me know if any more details would be helpful.
 A: The key words you need are "deck transformation group" or "covering transformation group", and "regular covering space" or "normal covering space"; there's other variations on this terminology as well. If you look for the discussion on these topics in Hatcher, then you will find what you need.
Here's an outline.
Let $n$ be the least positive integer such that $f^n = \text{Id}$. Let  $\langle f \rangle = \{f^i \mid i \in \mathbb Z\}$. This is a group under composition, and in fact it is a finite cyclic group generated by $f$ and isomorphic to $\mathbb Z / n \mathbb Z$, the isomorphism given by $f^i \mapsto i / n\mathbb Z \in \mathbb Z / n \mathbb Z$.
Now, you assumed that $f$ has no branch points, but let me make the required assumption a bit more explicit:  What we must assume is that $\langle f \rangle$ acts freely, which means that if $i \in \{0,...,n-1\}$ and if $f^i$ has a fixed point then $i=0$. Without this assumption, the representation $\rho$ you want need not exist.
It follows from the assumption of freeness that the quotient map $P : M \to M_f$ is a regular covering map, that the deck transformation group (or covering transformation group) is the finite cyclic group $\langle f \rangle$, that the induced homomorphism $P_* : \pi_1(M) \to \pi_1(M_f)$ is injective, that its image $\text{image}(P_*) < \pi_1(M_f)$ is normal, and that the quotient group $\pi_1(M_f) \, / \, \text{image}(P_*)$ is isomorphic to the deck transformation group $\langle f \rangle$ which, as said, is isomorphic to $\mathbb Z / n \mathbb Z$. Putting this altogether, we get the desired representation
$$\pi_1(M_f) \mapsto \pi_1(M_f) \, / \, \text{image}(P_*) \approx \langle f \rangle \approx \mathbb Z / n \mathbb Z
$$
I could add to this a description of the example you requested, if you like, but for now perhaps you might like to work it out, knowing about the general approach.
