First, a bit of background on orbifolds:



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> Let $X$ be a connected (effective) orbifold. To every point $x \in X$, we associated a group $G\_x$ called the *isotropy group*. The singular locus $\Sigma X$ is the set of points $x$ for which $G\_x \neq 1$, and these points are called *singular points*. Non-singular points are called *regular*. The singular locus is a closed, nowhere dense subset of $X$, and the set of regular points $X^r$ is open and dense in $X$.
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> The orbifold $X$ has a *stratificiation* into *strata* of points of equal type, i.e. $X$ is the disjoint union of connected components along which the isotropy group is constant (up to isomorphism). In particular, the strata of codimension $0$ form the regular points, and the strata of codimension $\geq 1$ form the singular locus.
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Now, my question: **is the set of regular points $X^r$ connected?**


I've done a bit of background research, but I seem to find contradictory statements. Moreover, the original definition of orbifolds (i.e. Satake's V-manifolds) assumed that there were no codimension $1$ strata, in which case the statement is clearly true. To add to the confusion, some authors still use this definition.


So, what have I found so far:


**Suggesting "Yes":**


* "This stratum is a connected manifold ..." [*Riemannian orbifolds with non-negative curvature*, Dmytro Yeroshkin]
* "The regular points of an orbifold form the top-dimensional stratum. It is open, dense and path-connected." [*The Topology of locally volume collapsed 3-Orbifolds*, Daniel Faessler]
* "Observe that the set of regular points is a dense *connected* open subset of the topological space underlying a connected orbifold" [*Seifert Fibred 3-Orbifolds*, Bonahon & Siebenmann]
* "Observe that the set of regular points is a dense connected open subset of $X$" [*Lusternik-Schnirelmann category of Orbifolds*, Hellen Colman]
* "The regular part of an orbifold is connected." (**[proof given, which uses that the regular part is locally connected](https://arxiv.org/pdf/1801.03472.pdf)**) [*Orbifolds from a metric viewpoint*, Christian Lange]


**Suggesting "No":**


* "Having done this, the set of special points has codimension at least 2. The set of regular points is therefore a connected manifold." [*Differential Topology, Foliations, and Group Actions*, Paul A. Schweitzer]
* "...we require that
the fixed-point set is of codimension at least two. [...] This requirement has the consequence that the non-fixed-point set is locally connected." [*Stringy geometry and topology of orbifolds*, Yongbin Ruan]



Do regular points of an orbifold form a connected set?
Part of the reason that people consider orbifolds whose singular points have codimension $\ge 2$ is because they are restricting their attention to oriented orbifolds. For example, if you are working with $n$-orbifolds over the complex numbers $\mathbb{C}$ then each element of each isotropy group is a holomorphic transformation, and every holomorphic transformation preserves orientation; in that situation, the singular points have codimension $\ge 2$.


If you do not restrict to oriented orbifolds, there is only one way that a codimension 1 stratum can arise: the isotropy group is the order 2 cyclic group $\mathbb{Z}/2\mathbb Z$, acting by reflection across some codimension 1 plane in Euclidean space. From this, it follows that for each point $p$ in a codimension 1 stratum, there is an orbifold neighborhood of $p$ homeomorphic to
$$\{x=(x\_1,...,x\_n) \in \mathbb{R}^n \mid \|x\| < 1, x\_n \ge 0\}
$$
such that each point of that neighborhood corresponding to $x\_n=0$ is on a codimension 1 stratum with isotropy group $\mathbb{Z}/2\mathbb{\mathbb Z}$, and each point of that neighborhood corresponding to $x\_n>0$ is a regular point. Thus, the union of the regular points and the codimension 1 strata has the structure of an $n$-manifold with boundary, the codimension 1 strata are precisely the boundary points of the manifold, the regular points are precisely the interior of the manifold, and all the rest of the singular locus has codimension $\ge 2$.


