In R.C.Penner "Decorated Teichmuller theory of boarded surface", on Page 7 and 8, it says that (without proof) the Teichmuller space of surface with $s$ labelled punctures and $r$ labelled boundary components and one marked point on each boundary is homeomorphic to an open ball of dimension $6g6+2s+4r$, where is the proof of that?I never see the version of T space that has marked points on the boundary, is also that they all homeomorphic to a ball like usual? Is there any reference about this?Thank you!
2 Answers
Just look up Teichmuller theory on the Internet; there are plenty of references. For example see these notes by Curtis McMullen.

$\begingroup$ @Kevin, I know Teichmuller space, but I don't see the one could have marked points on the boundary, what's the proof that it is a ball? $\endgroup$– HaoSep 23, 2010 at 10:01

1$\begingroup$ Ordinary Teichmüller space is (6g6)dimensional. Each puncture contributes 2 dimensions, so that accounts for the 2s. A boundary circle comes from removing a disc: a disc has a center point (2 dimensions) and a radius (1 dimension). A marked point on a boundary circle contributes 1 more dimension. 4r = (2+1+1)r. $\endgroup$ Sep 23, 2010 at 13:42

1$\begingroup$ @Kevin, that is the dimension, but why topologically it is a ball?Any proof? $\endgroup$– HaoSep 23, 2010 at 21:32

$\begingroup$ Take the proof you know for when there is no boundary, and then add boundary and marked points. It's the same proof. $\endgroup$ Sep 24, 2010 at 2:48
You could check the following nice and rather elementary paper, which I believe contains answers to your your questions as well as nice coordinate systems on this Teichmüller space (if there is at least one marked point).
Dual Teichmuller and lamination spaces, V.V. Fock, A.B. Goncharov, arXiv:math/0510312
(However as pointed out by Kevin Lin those are wellknown questions and I'm sure there are many other possible references.)