Covering space and Fundamental group Let $p:E\to X$ be a covering space and $\pi_1(E)$ be a fundamental group of $E$.
Can you give me a recept for calculating a fundamental group $\pi_1(X)$ (may be for some special cases)?
Thanks a lot!
 A: A result in Hatcher (Proposition 1.39) connects the group of deck transformations of the universal covering of a space $X$ to $\pi_1(X)$. When $p:E \rightarrow X$ is a covering space for $X$ that is path-connected, and $X$ is path-connected and locally path-connected, then $G(E) \approx \pi_1(X)$, where $G(E)$ is the group of deck transformations of $E$.
Thus, one way to compute $\pi_1(X)$ from a universal covering space $p:E \rightarrow X$ is to compute $G(E)$. Do do this, you must find all homeomorphisms $f:E \rightarrow E$ such that $p \circ f = p$.
Example: Compute $\pi_1(S^1)$.
Put the base point of $S^1$ at $(1,0)$ in the complex plane. The universal cover of $S^1$ is $\mathbb{R}$, with covering map $p(t) = e^{2\pi i t}$ and basepoint $\{0\}$. Then if $f_n(t) = t + n$, we have $$(p \circ f_n)(t) = p(t+n) = e^{2\pi i (t + n)} = e^{2\pi i n}e^{2\pi i t} = e^{2 \pi i t} = p(t)$$ and so $f_n$ is a deck transformation for each $n$. Since $\mathbb{R}$ is a universal cover, then a deck transformation is completely determined by where is sends the basepoint. But for each $n \in p^{-1}(1,0)$, we have a corresponding $f_n$, and so the collection $\{f_n\}_{n \in \mathbb{Z}}$ gives the group $G(\mathbb{R})$. Hence, since $G(\mathbb{R}) \approx \mathbb{Z}$, then $\pi_1(S^1) \approx \mathbb{Z}$ also.
A: Obs: I'm assuming all spaces to be path connected.
You cannot talk in such a generality.... 
In general, calculating the fundamental group is a "difficult" problem. There are no algoritms such as you were searching for.
Related with coverings, there are a lot of theorems that help you to calculate in special cases.
$ \pi _ 1 (B, b)/ p _\ast (\pi _ 1(E,e)) $ is a $G$-set isomorphic to the fiber and so on.
But there are a lot of nice special cases. I will give a nice one - and I hope you enjoy as I enjoyed when I figured it out for the first time.
If $ G $ is a topological group, so is any covering space of $G$. This is an exercise of liftings: in this exercise, you should prove also that the covering is itself a homomorphism of topological groups...
But, if $p: E\to G $ is a covering, you get a nice understanding of the fundamental group of $G$. 
First of all, you know this is a regular cover - since the fundamental group of any topological group is abelian (this is an exercise of Eckman-Hilton clock).
Second, it is easy to prove that the group of the automorphisms of $ p $ is isomorphic to the Kernel of $ p $. So, you got that $\pi _ 1(G,g)/p _ \ast (\pi _1 (E,e)) $ is isomorphic to the kernel of the covering.
Look at the particular case of universal coverings - you get that the kernel is the fundamental group of the base. It is the case of the universal covering of $ S^1 $.
If you don't think it's nice enough, you should investigate more the particular case of topological groups and get more nice results... And you can specialize your question and get more interesting results. 
