Where does the Berry phase of $pi$ come from in a topological insulator?

The Berry connection and the Berry phase should be related. Now for a topological insulator (TI) (or to be more precise, for a quantum spin hall state, but I think the Chern parities are calculated in the same fashion for a 3D TI). I can follow the argument up to defining the Chern parity $nu$, where the Berry connection $textbf{A}$ is explicitly within this equation where it gets integrated over half the Brillouin zone as it was described in Liang Fu’s thesis from 2009 on page 31 to calculate the Chern parity or the $Z_2$ invariant(http://repository.upenn.edu/dissertations/AAI3363356/):

$nu = frac{1}{2pi} left( ointlimits_{partial(A + B)} dtextbf{k} cdot textbf{A} – iintlimits_{A + B} dk_x dk_y nabla times textbf{A} right) text{mod } 2 $

It is known that the Berry phase for the surface state is $pi$ and it is also known how the Berry connection $textbf{A}$ leads to a Berry phase (http://en.wikipedia.org/wiki/Berry_connection_and_curvature). But I can’t see how this is equal to $pi$:

$left( ointlimits_{partial(A + B)} dtextbf{k} cdot textbf{A} – iintlimits_{A + B} dk_x dk_y nabla times textbf{A} right) text{mod } 2 = pi$ ?

Is there a simple way to show that the surface states of a topological insulator must have a $pi$ Berry phase? Maybe
by invoking time-reversal symmetry on this equation?

For the Integer Quantum Hall Effect the conductance was used to show how the Berry phase is connected to the Berry connection. But
I would like to avoid using charge polarization, I just want to see the direct relation between the Berry connection on the BZ and the resulting Berry phase of the wavefunctions that is exactly equal to $pi$, e.g. a fermion must undergo two complete rotations to acquire a phase of $2pi$. Is there a simple argument to relate this equation with the time reversal contraint to the Berry phase?