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Let $S_{g,n}$ denote a genus $g$ surface with $n$ punctures. There is a map $F$ from the Teichmüller space of the punctured surface $T(S_{g,n})$ to the Teichmüller space of the compact surface $T(S_{g})$ obtained by filling in the punctures. Is this a fiber bundle? Is the pre-image of this forgetful map the configuration space of $n$ points?

Intuitively, this makes sense since there are no conformal automorphisms of $X$ (if there are sufficiently many punctures) and so different configurations of points result in distinct points in the Teichmüller space of the punctured surface. We also have the composition $$T(S_{g})\times\operatorname{Conf}_{n}(S_{g})\xrightarrow{\Phi}T(S_{g,n})\xrightarrow{F}T(S_{g})$$ is just the projection onto the first factor; here $\Phi$ punctures the Riemann surface. What is known about $\Phi$?

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    $\begingroup$ Not addressing the question but I find the notation $M_{g,n}$ for a surface particularly confusing, given that $\mathcal M_{g,n}$ usually denotes the moduli space of Riemann surfaces. $\endgroup$
    – Nicolast
    Commented Mar 6 at 21:52
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    $\begingroup$ The answer is yes, the forgetful map is a fiber bundle. But its fiber is the universal cover of the configuration space of $n$ points. $\endgroup$
    – Nicolast
    Commented Mar 6 at 21:54
  • $\begingroup$ Yes, the notation $S_{g,n}$ is much more common for the surface. $\endgroup$
    – Sam Nead
    Commented Mar 6 at 21:54
  • $\begingroup$ @Nicolast, that is interesting. Do you have a reference that the fiber is the universal cover? $\endgroup$
    – Yousuf Soliman
    Commented Mar 6 at 22:00
  • $\begingroup$ @Yousuf - you are confusing the moduli space and the Teichmuller space. The capping off map exists in both settings, and is a fibre bundle in both cases. But the fibre is a configuration space of points in the surface only in moduli space setting. $\endgroup$
    – Sam Nead
    Commented Mar 6 at 22:05

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Edited to make it right:

The “capping off” map indeed is a fibre bundle map. However, the fibre is not the configuration space you suggest. For, consider the case of $n = 1$. Here the configuration space would be a copy of the surface, and the larger Teichmuller space would have nontrivial fundamental group! This is obviously wrong.

When $n = 1$, what is actually going on is that the fibre over $X$ is a copy of the hyperbolic plane. This plane is the space of points of $X$, equipped with a choice of homotopy class of path back to a (fixed earlier) base point. Thus the fibre over $X$ is the universal cover of $X$.

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  • $\begingroup$ Thank you. Do you have a reference or a proof sketch that $\Phi$ is a homeomorphism? $\endgroup$
    – Yousuf Soliman
    Commented Mar 6 at 20:31
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    $\begingroup$ Not only it is not a homeomorphism, it is not even well-defined. $\endgroup$
    – Moishe Kohan
    Commented Mar 6 at 21:30
  • $\begingroup$ Oops! I’ll delete my answer and rewrite. $\endgroup$
    – Sam Nead
    Commented Mar 6 at 21:41
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    $\begingroup$ @YousufSoliman: Do you know the definition of the Teichmuller space? You are attempting to define inverse to a universal covering map. $\endgroup$
    – Moishe Kohan
    Commented Mar 6 at 22:33
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    $\begingroup$ @SamNead Thanks! I see right away that this doesn’t even work for $n=1$ $\endgroup$
    – Yousuf Soliman
    Commented Mar 6 at 23:06

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