The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(k)U(k) & =E(k)U_{p}U(k)\\ H(-k)U_{p}U(k) & =E(k)U_{p}U(k)\\ H(k)U_{p}U(-k) & =E(-k)U_{p}U(-k) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-k)=U(k)$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}

and Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\nonumber \\ & =-\intop_{-\pi}^{\pi}dk\,\mathcal{A}(-k)\nonumber \\ & =-\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\nonumber \\ & =-\gamma\label{eq:-17} \end{align}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Is it really true that if $\gamma$=$-\gamma$ we can deduce it is $\pi$ or $0$.