Positive and negative winding number related by time-reversal symmetry

I am now reading some articles about Dirac fermions in condensed matter physics and the most famous example is graphene. I am now trying to understand page 5 in this article : https://arxiv.org/abs/1410.4098

In the literature it is stated that since the wavefunction around the $bf K$ and $ bf K’$ points are related by time-reversal symmetry, the Berry phases (or winding numbers) have opposite sign. That is, if the Berry phase is $+pi$ around $bf K$ then the Berry phase will be $-pi$ around $bf K’$.

I am now trying to prove this property. I first use the time-reversal property to relate the states around $K$ with those around $bf K’$.

If I write the states around $bf K$ as $left| {{bf K} + {bf q}} rightrangle$ where $|bf q|$ $<<$ $|bf K|$. Then the time-reversal partner with the same energy will be $Theta left| {{bf K} + {bf q}} rightrangle approx left| {-{bf K} – {bf q}} rightrangle$. Now, use the property that ${bf K’} + {bf K}$ is a reciprocal lattice vector in graphene, the state $left| {-{bf K} – {bf q}} rightrangle$ is the same as $left| {{bf K’} – {bf q}} rightrangle$.

The Berry phase at $bf K$ is
begin{equation}
phi_{bf K} = ointlimits_{around bf K} {leftlangle {{bf{K}} + {bf{q}}} right|{nabla _{bf{q}}}left| {{bf{K}} + {bf{q}}} rightrangle } cdot dbf q
end{equation}
while the Berry phase at $bf K’$ is
begin{equation}
phi_{bf K’} = ointlimits_{around bf K’} {leftlangle {{bf{K’}} + {bf{q}}} right|{nabla _{bf{q}}}left| {{bf{K’}} + {bf{q}}} rightrangle } cdot dbf q
end{equation}

I know that it will be trivial if I explicitly write down the spinor expression of the graphene $approx left[ {begin{array}{*{20}{c}}
{1}
{e^{iphi_{bf q}}}
end{array}} right]$.

But I would like to seek some more general proof for the statement that the Berry phase (or winding number) has opposite sign for time-reversal related Dirac point without the explicit expression of the wavefunction. Is it possible?

I am also confused with another point. I actually don’t know how to well-define the path for the line-integral. For example, if I use counterclockwise $bf q$-path and get the Berry phase $=pi$, then I can reverse the $bf q$-path to be clockwise then I will get Berry phase $= -pi$. Therefore I can use a clockwise path for $bf K$ while a counterclockwise path for $bf K’$ and it will lead to same Berry phase. Does this make sense?

I am still new in this field and will be extremely grateful for any suggestions! 🙂