How can we show the spaces $M_{g}(n)$ and $M_{g, n}$ are homotopy equivalent?

Question

How can we prove that the moduli space,$M_{g}(n)$, of genus $g$ Riemann surface with $n$ boundary components is homotopy equivalent to $M_{g,n}$, that is ,the moduli space of genus $g$ Riemann surface with $n$ punctures? Thanks! (It is very intuitive, but it seems that I can't make it)

Answer 1

A compact Riemann surface of genus $g$ with $n$ boundary components has a unique realization as a hyperbolic surface with geodesic boundary. One may see this by reflecting through the boundary and uniformizing. The uniqueness of the uniformization implies it is invariant under reflection, and therefore the fixed point set is geodesic.

Thus, the moduli space of genus $g$ Riemann surfaces with $n$ boundary components is equivalent to the space of hyperbolic surfaces with totally geodesic boundary. One may now insert a punctured disk into each boundary component, to obtain a Riemann surface with punctures. I don't know of a canonical way to do this, but for example for a boundary component of length $l$, one may attach isometrically the boundary of a punctured Euclidean disk of circumference $l$. The important thing is that this gluing only depends on $l$, and that it induces a conformal structure on the punctured surface. This gives a map between the spaces. Since the mapping class groups are the same, it induces a homotopy equivalence (in the category of orbifolds). Of course, there are some technical details one must carry out to make this argument rigorous. There are several other ways to fill in a punctured disk.

Another possible approach is to use the Weil-Petersson metric on moduli space. One can take the WP nearest point in the Deligne-Mumford compactification of moduli space, which is finite distance away since the WP metric is incomplete. Because the metric is CAT(0), a unique nearest point exists.