# 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!

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In the situation you stated, I don't think we can say much more than that $\pi_1(E)$ is isomorphic to a subgroup of $\pi_1(X)$. (To compute $\pi_1(X)$, I believe you usually have to refer to the universal cover of $X$ and prove something manually, or use Seifert-van Kampen theorem and known fundamental groups.) – Tunococ Oct 16 '12 at 8:47
Just to echo @Tunococ, there's a covering space for every conjugacy class of subgroups of $\pi_1(X)$, and one approach to computing $\pi_1(X)$ once you've found $E$ to be the universal cover is as the set $p^{-1}(x_0)$ with the group structure of the deck transformations on $E$. – Kevin Carlson Oct 16 '12 at 9:34

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.

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With regard to the case $G$ is a topological group, your statement is OK if $G$ is path connected. Otherwise there is an obstruction in $H^3(\pi_0 G, \pi_1(G,e))$ whose vanishing is necessary and sufficient for the universal cover. See [82] R. Brown and O. Mucuk, Covering groups of non-connected topological groups revisited'', Math. Proc. Camb. Phil. Soc, 115 (1994) 97-110 from my publication list, which gives earlier references. – Ronnie Brown Jun 11 '13 at 9:39
Thank you, professor Ronnie Brown. The result you mentioned seems to be very interesting! You are right about the correction! I meant $G$ being a path connected topological group... I guess I was talking about basic covering theory, where, usually, we assume base and total spaces to be path connected. – Fernando Jul 29 '13 at 19:27

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.

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