Help with moduli spaces of four-marked spheres I'm thinking about the moduli space of the four-punctured sphere where some of the removed points are distinguishable and some are indistinguishable. I believe there should be some covering maps between some of these spaces and have evidence to back that up, but that doesn't square with my intuition of what these objects actually are.
Consider first $\mathcal{M}(A,B,C,D)$, which I use to denote the moduli space of spheres with four distinguishable (distinctly labeled) points. Thinking of these points as lying in $\mathbb{CP}^1$, there is a unique Mobius transformation taking $(A,B,C)$ to $(0,1,\infty)$. The fourth point $D$ maps to the cross-ratio of this quadruple, and is now free to wander around wherever it wishes in the complement of the other three points. Thus $\mathcal{M}(a,b,c,d) \cong \mathbb{CP}^1-\{0,1,\infty\}$.
Now consider $\mathcal{M}(A,A,C,D)$ in which one pair of the points is indistinguishable. Now the Mobius transformation which takes $(A,B,C)$ to $(0,1,\infty)$ can be postcomposed with the map $z \mapsto 1-z$ to give an identical element of this moduli space. So we quotient $\mathbb{CP}^1-\{0,1,\infty\}$ by this involution to see that $\mathcal{M}(A,A,C,D)$ is a twice-punctured sphere with an order 2 cone point. You might think of this as the $(2,\infty,\infty)$ turnover orbifold.
Last, consider $\mathcal{M}(A,A,A,D)$. Now our unique Mobius transformation can be postcomposed with any element of an order 6 dihedral group, generated by $z \mapsto 1-z$ and $z \mapsto (z-1)/z$. We wind up with a $(2,3,\infty)$ turnover orbifold for this moduli space.
There should be a map $\mathcal{M}(A,A,C,D) \to \mathcal{M}(A,A,A,D)$ whereby you just "forget" the uniqueness of the third point. Generically, this map should be $3$-to-$1$. Question: is this an (orbifold) covering map?

Evidence to support a covering map: look at (orbifold) fundamental groups.
We have $\pi_1(\mathcal{M}(A,A,A,D)) = \langle a,b \mid a^2=b^3=1 \rangle$, which can also be seen as the quotient of the 3-strand braid group $B_3 = \langle \sigma_1,\sigma_2 \mid \sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2 \rangle$ by its center $ZB_3 = \langle \sigma_1\sigma_2\sigma_1\sigma_2\sigma_1\sigma_2 \rangle$ if we take $a=\sigma_1\sigma_2\sigma_1$ and $b=\sigma_1\sigma_2$.

Now $\pi_1(\mathcal{M}(A,A,C,D))$ should likewise come from the 3-strand braid group, only you can't permute all three points. One can see that $\pi_1(\mathcal{M}(A,A,C,D)) = \langle a,(b^{-1}a)^2 \mid a^2=b^3=1 \rangle$ where $b^{-1}a=\sigma_1$. The map on moduli spaces I mentioned should respect the clear inclusion of fundamental groups, and a quick check with Magma verifies that this is an index 3 subgroup.
More support for covering map: the (orbifold) euler characteristic of $(2,\infty,\infty)$ ($=-\frac12$) is three times that of $(2,3,\infty)$ ($=-\frac16$).

Evidence against a covering map (and my whole issue): I personally don't see how $(2,\infty,\infty)$ could possess an order 3 symmetry, so I don't see how it could be an order 3 covering space of anything.
 A: It doesn't have to be a regular covering space. The deck transformation group might even be trivial. Equivalently, that index 3 subgroup need not be a normal subgroup.

As for constructing an actual nonregular covering map, here's a description. I don't have good drawing software (does anyone have any suggestions?), but I've sketched this picture out and am describing my sketch.
First let's visualize the orbifold $\mathcal M(A,A,A,D)$ as a sphere with one point $\infty$ removed, with two interior points $P$ labelled $\mathbb Z/2$ and $Q$ labelled $\mathbb Z/3$. Subdivide the sphere into two triangles by cutting along three arcs $R,B,G$: a red arc $R = \overline{\infty Q}$, a blue arc $B = \overline{\infty P}$, and a green arc $G = \overline{QP}$. But now, without actually changing the topology of the underlying orbifold, I'd like to stretch out the point $\infty$ to be an actual entire circle "at infinity", and so $\mathcal M(A,A,A,D)$ becomes a closed 2-disc with its boundary circle "at infinity", and thought of as being removed. That boundary circle is subdivided at two points that I'll label $\xi_R$, $\xi_B$, which subdivide that circle into two unnamed arcs "at infinity". The disc is subdivided into two "triangles" by cutting along three arcs $R = \overline{\xi_R Q}$, $B = \overline{\xi_B P}$, and $G = \overline{QP}$.
Think of $\mathcal M(A,A,C,D)$ as an annulus $S^1 \times [-1,1]$ with its two boundary circles "at infinity" and therefore removed. I will define a subdivision of $\mathcal M(A,A,C,D)$ into six "triangles", subdividing alon three red arcs, three blue arcs, and three green arcs. There will also be one finite vertex mapping to $Q$, and two finite vertices mapping to $P$.
The boundary circle $S^1 \times (-1)$ will be a 1-to-1 cover of the boundary circle of $\mathcal M(A,A,A,D)$ and thus will be incident to one blue arc and one red arc (to be described -- TBD). The other boundary circle $S^1 \times (+1)$ will be a 2-to-1 cover of the boundary circle of $\mathcal M(A,A,A,D)$ and thus will be incident to two blue arcs and two red arcs (TBD).
In $S^1$ let $X,Y$ be a pair of antipodal points.
The unique preimage of $Q$ will be the point $X \times 0 \in S^1 \times [-1,+1]$, and so $X \times 0$ will have a neighborhood which maps 3-to-1 around a neighborhood of $Q$ (except 1-to-1 at the center). Thus $X \times 0$ will be incident to three red arcs and three green arcs (TBD).
The two pre-images of $P$ will be the points $Y \times 0$ and $X \times \frac{1}{2}$. The first pre-image $Y \times 0$ will have a neighborhood which maps 2-to-1 onto a neighborhood of $P$ (except 1-to-1 at the center), thus $Y \times 0$ will be incident to two green arcs and two blue arcs (TBD). The second preimage $Y \times \frac{1}{2}$ will have a neighborhood that maps 1-to-1 onto a neighborhood of $P$, and therefore $Y \times \frac{1}{2}$ will be the $\mathbb Z / 2$ orbifold point of $\mathcal M(A,A,C,D)$, and it will be incident to one green arc and one blue arc (TBD).
So now all I have to do is describe the three red, three green and three blue arcs of $\mathcal M(A,A,C,D)$. The points $X \times 0$ and $Y \times 0$ subdivide the circle $S^1 \times 0$ into two green arcs, and the third green arc is $X \times [0,\frac{1}{2}]$. The point $Y \times 0$ subdivides the arc $Y \times [-1,+1]$ into two blue arcs, and the third blue arc is $X \times [\frac{1}{2},1]$. Finally, one red arc is $X \times [-1,0]$, and the other two red arcs connect $X \times 0$ to two points on the circle $S^1 \times 1$, which I won't name and you can probably sketch in.
If you get the sketch right, it will now be obvious that these arcs subdivide $\mathcal M(A,A,C,D)$ into six "triangles" each with one red side, one blue side, and one green side. This now defines a nonregular three-to-one covering map $\mathcal M(A,A,C,D) \to \mathcal M(A,A,A,D)$ with trivial deck transformation group.

One last point I'll make is that the "graph of groups" perspective gives you a direct picture of the index 3 subgroup. In the $\mathcal M(A,A,A,D)$ orbifold, the unique green edge is a graph of groups with one $\mathbb Z/3$ and one $\mathbb Z/2$ vertex, exhibiting the presentation $\langle a,b \mid a^2 = b^3 = 1 \rangle$, and showing explicitly the isomorphism with the free product $\mathbb Z/2 * \mathbb Z/3$. In the $\mathcal M(A,A,C,D)$ orbifold, the three green edges constitute a graph of groups homotopy equivalent to a circle with two trivially labelled vertices and one vertex labelled $\mathbb Z/2$, thus exhibiting the isomorphism with the free product $\mathbb Z * \mathbb Z/2$. Also, the map from the three-edge graph of groups to the one-edge graph of groups exhibits the index 3 group monomorphism, taking the $\mathbb Z$ free factor to the infinite cyclic group generated by $ab$.
