I am studying orbifold fundamental group (or more generally orbifold homotopy groups). In a nutshell, my questions is: what are they intuitively? In what follows I give definitions and more precise questions. My definition of orbifold fundamental group is via classifying space of groupoid, which is explained in the next paragraph (so you may want to skip it if you know the definition). 


Let $\mathcal{G}$ be a topological groupoid consisting a topological spaces $G\_{0}$ of $objects $ and $G\_{1}$ of $arrows$ together with usual continuous structure maps. Let $|\mathcal{G}|$ denote the associated topological space $G\_{0}/G\_{1}$. Let $G\_{n}$ be the iterated fibered product $G\_{n}=G\_{1}\times\_{s,t} G\_{n-1}$. These $G\_{n}$ have the structure of a simplicial topological space, called the $nerve$ of $\mathcal{G}$. Face operads $d\_{i}:G\_{n}\rightarrow G\_{n-1}$ for $i=0,\dots,n$ are given by
$$
d\_{i}(g\_{1},\dots,g\_{n})=(g\_{1},\dots,g\_{i}g\_{i+1},\dots,g\_{n})
$$
for $i=1,\dots,n-1$ and 
$$
d\_{0}(g\_{2},\dots,g\_{n})=(g\_{2},\dots,g\_{n}), \ \ 
d\_{n}(g\_{1},\dots,g\_{n})=(g\_{2},\dots,g\_{n-1}). 
$$
The classifying space $B\mathcal{G}$ of $\mathcal{G}$ is then defined as 
$$
B\mathcal{G}=\bigsqcup\_{n}(G\_{n}\times \Delta^{n})((d\_{i}(g),x)\sim(g,\delta\_{i}(x))),
$$
where $\Delta^{n}$ is the topological $n$-simplex and $\delta\_{i}:\Delta^{n-1}\rightarrow \Delta^{n}$ is the standard facemap.\


The $n$-th orbifold homotopy group of $\mathcal{G}$ based at $x\in |\mathcal{G}|$ is defined to be 
$$
\pi\_{n}^{orb}(\mathcal{G},x)=\pi\_{n}(B\mathcal{G},y),
$$
where $y\in G\_{0}$ maps to $x$ under the quotient map $G\_{0}\rightarrow |\mathcal{G}|$. 


The following are my questions:


1. Why is this a reasonable definition? Any manifold $M$ can be thought of topological groupoid via its chart i.e. $G\_{0}=\bigsqcup\_{i}U\_{i}$ and $G\_{1}=\bigsqcup\_{i,j}U\_{i}\times\_{M} U\_{j}$. It is not clear to me that the definition above reproduce $\pi\_{n}(M)$.
2. I am aware of an explicit description the orbifold fundamental groups of the orbifold Riemann surface $\Sigma\_{g,n,k}$ of genus $g$ and $n$ orbifold points $p\_{i}$ of order $k\_{i}$: 
$$
\pi\_{n}^{orb}(\Sigma\_{g,n,k})
=\langle \alpha\_{i},\beta\_{i},\sigma\_{j} \ (1\le i \le g,1\le j \le n)\ | \ 
\sigma\_{1}\dots\sigma\_{n}\prod\_{i=1}^{g}[\alpha\_{i},\beta\_{i}]=1,,\sigma\_{i}^{k\_{i}}=1\rangle 
$$
Is it easy to see this explicit presentation by the definition above?
3. It seems there are several ways to define the fundamental group of an orbifold, such as covering space etc. How should one understand orbifold fundamental groups?


Thank you for your assistance. 



How should one understand orbifold fundamental groups? 
The definition of an orbifold in terms of a groupoid is flexible and technically useful and gives very clean definitions, but it's not so close to geometric intuition. Perhaps it is once you've mastered the art of thinking simplicially, which I probably haven't. I tend to think of orbifolds like this: the simplest orbifolds are the global quotients $[X/G]$. In this case the formalism has been cooked up so that geometry of $[X/G]$ is exactly the same as $G$-equivariant geometry on $X$ (whatever this means in a given context). Another catchphrase for this is that all group actions behave like free group actions. All other orbifolds can be obtained by gluing together ones of the above form. This leads to the definition in terms of coordinate patches of the form $[U/G]$.


The easiest example of an orbifold fundamental group is when your orbifold has the form $[X/G]$ with $X$ simply connected. Since every group action behaves like a free one, the map $X \to [X/G]$ is a covering map, which exhibits $X$ as the universal cover of the orbifold and $G$ as the group of deck transformations. So $\pi\_1([X/G]) = G$.


The second simplest example is, I think, an (effective) orbifold Riemann surface. Fortunately you asked precisely about this one. Here you can really think concretely about loops on your surface. Fix an orbifold point $x$ of order $n$. The idea is that an orbifold point of order $n$ is $(1/n)$th of a point, so that an orbifold point is something inbetween an ordinary point and a puncture. The higher order stabilizer the point has, the closer it is to being an actual puncture. More concretely, what this means is that *a loop on your surface that winds exactly $n$ times around $x$ can be shrunk across $x$*. It's like the $n$ turns together add up to one whole point, which your loop can then slide across.


To see this slightly more formally, think in a chart centered on $x$, where your orbifold looks like $[D/\mu\_n]$, with $D$ the unit disk and $\mu\_n$ the group of $n$th roots of unity. Any loop in $D$ around the origin can be shrunk to a point, which should imply that the image of this loop in our orbifold is also homotopically trivial. But the image is just a loop that goes $n$ times around $x$. To see that no fewer than $n$ turns suffice, you need to think a bit about the definition of a loop on an orbifold. In any case, once we accept this fact we can obtain the presentation of the fundamental group that you gave in your question, in the same way as for the ordinary fundamental group of a punctured Riemann surface.


An example is $[S^2 / \mu\_n]$ with $\mu\_n$ acting by rotations along the equator. You have orbifold points at the north and south pole. Either of the two descriptions above immediately imply that the fundamental group is cyclic of order $n$: in terms of the second description, a generator is a simple loop around one of the poles, which becomes trivial when it is wound around itself $n$ times.


You might also find my recent question, and the answer by Jeffrey Giansiracusa, useful: [Homotopy theory of topological stacks/orbifolds](https://mathoverflow.net/questions/104757/)


