Let $G$ be a discontinuous group (this means that it acts discontinuously with finite stabilizers) of homeomorphisms of a simply connected, locally compact metric space $X$. Let $p:X\rightarrow X/G$ be the orbit map. Choose a base point $x_0\in X$ and let $\pi_1(X/G,p(x_0))$ be the fundamental group of $X/G$.

Define $\phi:G\rightarrow \pi_1(X/G,p(x_0))$ as follows - for any $g\in G$ let $\alpha$ be a path from $x_0$ to $g(x_0)$. Send $g$ to the homotopy class of loops $p\circ\alpha$. This map is independent of the choice of path as $X$ is simply connected. It is also easy to check that it is a group homomorphism.

We can show that for a discontinuous group of homeomorphisms of a locally compact metric space, $p:X\rightarrow X/G$ has the path lifting property. This helps show that $\phi$ is surjective.

What I want to know is what is the kernel of $\phi$?

I know that $\pi_1(X/G,p(x_0)) \cong G/N$ where $N$ is the normal subroup generated by all those elements of $G$ that have fixed points. So my guess is that $\ker\phi\cong N$ but I am not sure how to show this. If $g\in G$ has a fixed point say $g(x)=x$ then we can join $x_0$ to $x$ by a path $\gamma$ and use $\gamma*(g\circ\gamma)^{-1}$ to join $x_0$ to $g(x_0)$. Then I have to show $p\circ(\gamma*(g\circ\gamma)^{-1})$ is null homotopic in $X/G$. But what about the other way that is given a null homotopic loop in $X/G$ why should it come from a $g$ with has a fixed point?


  • $\begingroup$ It seems that any proof showing that $\pi_1(X/G,p(x_0))\cong G/N$ would also prove that $\ker\phi\cong N$. Then as you said, $N$ is generated as a normal subgroup by elements with fixed points. This does not imply that any element in $N$ has a fixed point itself. For example, let $X=\mathbb R$ and let $G$ be generated by the reflections at $0$ and $1$. Then $X/G\cong[0,1]$ and $G=N$, but $G$ contains translations by even integers that have no fixed points. $\endgroup$ Nov 16, 2015 at 8:15
  • $\begingroup$ @SebastianGoette The proof that $\pi_1(X/G,p(x_0))\cong G/N$ is from a paper by M A Armstrong. He first proves that if $G$ is discontinuous and generated by its elements having fixed points and $X$ is simply connected, locally compact then $X/G$ is simply connected. Then he proves that if $G$ is discontinuous and acts freely then $\pi_1(X/G)\cong G$. And finally factors the action of $G$ into a discontinuous action of $N$ on $X$ followed by a free discontinuous action of $G/N$ on $X/N$. He doesn't directly prove $\ker\phi\cong N$ $\endgroup$
    – R_D
    Nov 16, 2015 at 8:46
  • $\begingroup$ Do you want G to act by homeomorphisms on X or by isometries? $\endgroup$
    – NWMT
    Nov 16, 2015 at 18:19
  • $\begingroup$ Also if X is a CW complex and G acts combinatorially (maps cells to cells) there is a nice answer to your question. $\endgroup$
    – NWMT
    Nov 16, 2015 at 18:21
  • $\begingroup$ @NWMT, by homeomorphisms. Not isometries $\endgroup$
    – R_D
    Nov 18, 2015 at 1:36

1 Answer 1


Note that if a group $G$ acts on a space $X$ then $G$ also acts on the fundamental groupoid $\pi_1(X)$.

The study of the fundamental groupoid of an orbit space is developed in Chapter 11 of Topology and Groupoids, (T&G). An older version of this Chapter is available at arXiv:math/0212271.

The main result is to link the fundamental groupoid of an orbit space to the notion of orbit groupoid. The analysis of orbit groupoids uses fibrations of groupoids, which are also applied in Chapter 7 of T&G to yield operations of fundamental groupoids and exact sequences.

Here is a main result on the orbit groupoid $\Gamma //G$ of an action of a group $G$ on a groupoid $\Gamma$. (It comes from J. Taylor, "Quotients of a groupoid by the action of a group", Math. Proc. Camb. Phil. Soc 103 (1988) 239-249.)

11.5.2 The orbit morphism $p : \Gamma \to \Gamma // G$ is a fibration whose kernel is generated as a subgroupoid of $\Gamma$ by all elements of the form $\gamma - g\cdot \gamma$ where $g$ stabilises the initial point of $\gamma$. Furthermore,

(a) if $G$ acts freely on $\Gamma$, by which we mean no non-identity element of $G$ fixes an object of $\Gamma$, then $p$ is a covering morphism;

(b) if $\Gamma$ is connected and $G$ is generated by those of its elements which fix some object of $\Gamma$, then $p$ is a quotient morphism; in particular, $p$ is a quotient morphism if the action of $G$ on $Ob(\Gamma)$ has a fixed point;

(c) if $\Gamma$ is a tree groupoid, then each object group of $\Gamma // G$ is isomorphic to the factor group of $G$ by the (normal) subgroup of $G$ generated by elements which have fixed points.

You can also find out more on applications of fibrations of groupoids from arxiv:1207.6404.


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